Integral, materi kelas XII, materi kelas XII semester I, Uncategorized

Integral

A. Pendahuluan

Gagasan integral mula-mula berawal dari Metode yang digunakan Archimedes, seorang ilmuan bangsa Yunani dari Syracusa (287-212 M) dimana di dalam suatu daerah yang akan ditentukan luasnya maka dilukislah beberapa daerah poligon yang luasnya mendekati luas daerah itu, dikarenakan daerah poligon mudah dihitung. Selanjutnya supaya lebih presisi dipilihlah poligan yang lebih banyak yang merupakan pendekatan yang lebih baik untuk daerah itu. Jika hal ini dilakukan terus-menerus maka semua daerah poligon akan mencakupi daerah itu.

Sehingga arti secara fisis dari integral adalah limit jumlah, di mana sering dianggap seperti mencari luas daerah.

B. Integral Tak Tentu

\LARGE\boxed{\int f(x)\: dx=F(x)+C}

dengan

\left\{\begin{matrix} F(x) & adalah & fungsi & integral&umum&di&mana&F'(x)=f(x)\\ f(x) & adalah & integran \\ C & adalah & konstanta & integral \end{matrix}\right.

 Rumus-rumus integral

  • \LARGE{\int a\: x^{n} dx=\frac{a}{n+1}.x^{n+1}+C}, \: dengan\: \: n\neq -1
  • \int a\: dx=ax+C
  • \int \frac{1}{x}\: dx=\int x^{-1}\: dx=\ln x+C
  • \int \left | x \right |\: dx=\frac{1}{2}x\left | x \right |+C
  • \int \ln x\: dx=x\ln x-x+C
  • \int e^{x}\: dx=e^{x}+C
  • \int a^{x}\: dx=\frac{a^{x}}{\ln a}+C
  • \int \sin x\: dx=-\cos x+C
  • \int a\sin bx\: dx=-\frac{a}{b}\cos bx\: +C
  • \int \cos x\: dx=\sin x+C
  • \int a\cos bx\: dx=\frac{a}{b}\sin bx\: +C
  • \int e^{ax}\: dx=\frac{1}{a}.e^{ax}+C
  • \int (x^{m}+x^{n}+...+x^{p})\: dx=\int x^{m}\: dx+\int x^{n}\: dx+...+\int x^{p}\: dx

Sifat-sifat integral

  • \int dx=x+C
  • \int a\: f(x)\: dx=a\int f(x)\: dx
  • \int a\left ( f(x)+g(x) \right )\: dx=a\int f(x)\: dx\: +\: a\int g(x)\: dx
  • \int a\left ( f(x)-g(x) \right )\: dx=a\int f(x)\: dx\: -\: a\int g(x)\: dx

 

\LARGE\fbox{Contoh Soal}


 

1. \int x^{5}\: dx=\frac{1}{5+1}.x^{5+1}+C=\frac{1}{6}.x^{6}+C

2. \int 3x^{5}\: dx=\frac{3}{5+1}.x^{5+1}+C=\frac{1}{2}.x^{6}+C

3. \int \frac{1}{x^{4}}\: dx=\int x^{-4}\: dx=\frac{1}{-4+1}.x^{-4+1}+C=-\frac{1}{3}.x^{-3}+C=-\frac{1}{3x^{3}}+C

4. \int 5y\: dy=\frac{5}{1+1}y^{1+1}+C=\frac{5}{2}y^{2}+C.

5. \int e^{6x}\: dx=\frac{1}{6}.e^{6x}+C.

6. \int 2014^{x}\: dx=\frac{2014^{x}}{\ln 2014}.

7. \int \left ( 3x^{2}-x+2-\frac{1}{x}+ \frac{3}{x^{2}}\right )dx=\int 3x^{2}\: dx-\int x\: dx+2\int dx-\int \frac{1}{x}\: dx+3\int \frac{1}{x^{2}}\: dx=\frac{3}{2+1}x^{2+1}-\frac{1}{1+1}x^{1+1}+2x-\ln x\: +3\left ( \frac{1}{-2+1}x^{-2+1} \right )+C=\frac{2}{3}x^{3}-\frac{1}{2}x^{2}+2x-\ln x\: -\frac{3}{x}+C.

8. \int \left ( x^{2}-2xy+y^{2} \right )dx=\int x^{2}\: dx-2y\int x\: dx+y^{2}\int dx=\frac{1}{2+1}x^{2+1}-\frac{2y}{1+1}x^{1+1}+y^{2}.x+C=\frac{1}{3}x^{3}-x^{2}y+xy^{2}+C.

9. \int \left | x-1 \right |\: +\left | x-2 \right |\: dx=(x-1)\left | x-1 \right |\: +(x-2)\left | x-2 \right |+C.

10. \int 3\sin 4x\: dx=-\frac{3}{4}\cos 4x+C.

11. \int \frac{1}{3}\cos 5x\: dx=\frac{1}{15}\sin 5x+C.

\LARGE\fbox{Latihan Soal}.

Selesaikan soal berikut!

  1. \int 2014\: dx
  2. \int \frac{dx}{2014}
  3. \int 2014x\: dx
  4. \int -2014x^{2}\: dx
  5. \int \left ( x+2014 \right )dx
  6. \int \left (-2014x^{3}+2015x^{2}-2016 \right )dx
  7. \int x\sqrt{x}\: dx
  8. \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x}}}}}\: dx
  9. \int \frac{2014}{\sqrt[3]{x^{2}}}\: dx
  10. \int \frac{2014x}{\sqrt[3]{x^{5}}}\: dx
  11. \int \left (2014-2013t+t^{2} \right )dt
  12. \int \left (\frac{3}{t^{3}} +\frac{2}{t^{2}} +2014\right )dt
  13. \int \left (\sqrt{t} +\frac{1}{2\sqrt{t}} \right )dt
  14. \int \left (ay^{4} +by^{2} \right )dy
  15. \int \left (4ax^{3}+3bx^{2}+2cx+1 \right )dx
  16. \int \frac{x^{2}+2014}{x^{2}}\: dx
  17. \int \left (e ^{x}+e^{-x} \right )dx
  18. \int e^{2014x}dx
  19. \int \frac{dx}{e^{2014x}}
  20. \int \left ( \sqrt{10^{x}} \right )dx

 

C. Integral Tentu

\LARGE\boxed{\int_{a}^{b}f(x)\: dx=\left [ F(x) \right ]_{a}^{b}=F(x)|_{a}^{b}=F(b)-F(a)}

\LARGE\fbox{Contoh Soal}

  1. \int_{1}^{3}\left ( 2x-1 \right )\: dx=\left [ x^{2}-x \right ]_{1}^{3}=(9-3)-(1-1)=6.
  2. \int_{0}^{\frac{\pi }{2}}\sin x\: dx=\left [ -\cos x \right ]_{0}^{\frac{\pi }{2}}=\left ( -\cos \frac{\pi }{2} \right )-\left ( -\cos 0 \right )=1.
  3. \int_{0}^{5}\left | x-1 \right |+\left | x-2 \right |\: dx=\frac{1}{2}(x-1)\left | x-1 \right ||_{0}^{5}\: +\frac{1}{2}(x-2)\left | x-2 \right ||_{0}^{5}=\left ( \frac{1}{2}.4.4+\frac{1}{2}.3.3 \right )-\left ( \frac{1}{2}.-1.1+\frac{1}{2}.-2.2 \right )=8+4\frac{1}{2}+\frac{1}{2}+2=15.

 

Sifat-Sifat Integral Tentu

  • \int_{a}^{b}f(x)\: dx=-\int_{b}^{a}f(x)\: dx
  • \int_{a}^{b}k.f(x)\: dx=k\int_{a}^{b}f(x)\: dx
  • \int_{a}^{b}\left ( f(x)\pm g(x) \right )\: dx=\int_{a}^{b}f(x)\: dx\: \pm\int_{a}^{b}g(x) \: dx
  • \int_{a}^{b}f(x)\: dx=\int_{a}^{c}f(x)\: dx+\int_{c}^{b}f(x)\: dx\: ,\: \: dengan\: \: a<c<b
  • \int_{a}^{a}f(x)\: dx=0

 

D. Integral Substitusi (Bagian 1)

\LARGE\boxed{\int u^{n}.u'\: dx=\int u^{n}\: du=\frac{1}{n+1}u^{n+1}+C}

\LARGE\fbox{Contoh Soal}

1. \int \left ( x^{2}+3 \right )^{20}2x\: dx=

Jawab:

misalkan\left\{\begin{matrix} u & = & x^{2} &+&3&,&maka \\ du &= &2x & dx \end{matrix}\right.

sehingga

\int \left ( x^{2}+3 \right )^{20}2x\: dx=\int u^{20}\: du=\frac{1}{21}u^{21}+C=\frac{1}{21}\left ( x^{2}+3 \right )^{21}+C.

2. \int \left ( x^{4}-x^{2} \right )^{5}\left ( 16x^{3}-8x \right )dx=.

Jawab:

misalkan\: \: \left\{\begin{matrix} u & = & x^{4} & - & x^{2}&,&maka\\ du & = & 4x^{3} & - &2x&dx \\ 4du & = & 16x^{3} & - & 8x&dx \end{matrix}\right..

\int u^{5}.4du=\frac{4}{6}u^{6}+C=\frac{2}{3}\left ( x^{4}-x^{2} \right )^{6}+C.

 

3. \int \frac{x+2}{x^{2}+4x+4}dx=

Jawab:

misalkan\: \: \left\{\begin{matrix} u & = & x^{2} & + & 4x&+&4&,&maka\\ du & = & 2x & + &4&dx \\ \frac{1}{2}du & = & x & + & 2&dx \end{matrix}\right..

\int \frac{1}{u}.\frac{1}{2}du=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln u+C=\frac{1}{2}\ln \left ( x^{2}+4x+4 \right )+C.

4. \int \frac{e^{3y}}{\left ( 1-2e^{3y} \right )^{2}}dy=.

Jawab:

misalkan\: \: \left\{\begin{matrix} u & = & 1 & - & 2e^{3y}& ,&maka\\ du & =&- & 6e^{3y} &dy &,&sehingga \\ -\frac{1}{6}du & = & e^{3y} & dy \end{matrix}\right..

\int -\frac{1}{6}\frac{1}{u^{2}}du=-\frac{1}{6}\int \frac{du}{u^{2}}=-\frac{1}{6}\left ( -1 \right )\left ( u^{-1} \right )+C=\frac{1}{6u}+C=\frac{1}{6}.\frac{1}{1-2e^{3y}}+C.

5. \int \sin 2x\: dx=

Jawab:

misalkan\: \: \left\{\begin{matrix} u & = & 2x&,&maka \\ du & = & 2 & dx&,& \\ \frac{1}{2}du & = & dx \end{matrix}\right..

\int \sin u.\: \frac{1}{2}du=\frac{1}{2}\left ( -\cos u \right )+C=-\frac{1}{2}\cos 2x+C.

6. \int \sin ^{2}x\: dx=.

Jawab:

\int \frac{1-\cos x}{2}\: dx= \frac{1}{2}\int \: dx-\frac{1}{2}\int \cos\: dx =\frac{1}{2}x-\frac{1}{2}\sin x+C.

7. \int \cos 3x\: dx=.

Jawab:

misalkan\: \: \left\{\begin{matrix} u & = & 3x & , & maka\\ du & = & 3 & dx &,&sehingga \\ \frac{1}{3}du & = & dx \end{matrix}\right..

\int \cos u.\: \frac{1}{3}du=\frac{1}{3}\sin u+C=\frac{1}{3}\sin 3x+C.

8. \int \cos ^{3}x\: dx=

Jawab:

perhatikan\: \: langkah\: \: berikut\: \: ini\\ \int \cos ^{3}x\: dx=\int \cos ^{2}x\: dx=\int \left ( 1-\sin ^{2}x \right ).\cos x.\: dx.

=\int \cos x\: dx-\int \sin ^{2}x\: .\cos x.\: dx=\sin x-\int \sin ^{2}x.\cos x.\: dx.

misalkan\: \: \left\{\begin{matrix} u & = & \sin x & , &maka \\ du & = & \cos x & dx \end{matrix}\right..

=\sin x\: -\: \int u^{2}\: du\\ =\sin x\: -\: \frac{1}{3}\sin ^{3}x+C.

 

\LARGE\fbox{Rumus Penting Trigonometri}

(1)

39

(2)

40

(3)

38

(4)

41

(5)

35

(6)

36

(7)

42

 

E. Integral Sunstitusi (Bagian 2)

44 

\LARGE\fbox{and}


43

\LARGE\fbox{Contoh Soal}

 

1. Tentukanlah \int \sqrt{4-x^{2}}\:\: \: dx

Jawab:

misalkan\: \: \left\{\begin{matrix} x = 2 & sin & t&,&maka&sin&t=&\frac{x}{2}\\ dx & = & 2 & cos & t&dt \end{matrix}\right.

sebagai ilustrasi

122

dari ilustrasi gambar diketahui bahwa

\left\{\begin{matrix} sin & t & = & \frac{x}{2} \\ cos & t & = & \frac{\sqrt{4-x^{2}}}{2} & \\ \sqrt{4-x^{2}} & = &2 &cos & t \end{matrix}\right.

Sehingga

\int \sqrt{4-x^{2}}.\: \: dx=\int 2\: \cos\: t.\: 2\: \cos \:t\: dt =\int 4\: \cos ^{2}\: t\: dt.

=\int \left (2+2\cos 2t \right )dt=2t+\sin 2t+C=2t+2\sin t\: \cos t+C.

\int \sqrt{4-x^{2}}\: \: dx=2\: \arcsin \frac{x}{2}+2.\frac{x}{2}.\frac{\sqrt{4-x^{2}}}{2}+C=2\arcsin \frac{x}{2}+\frac{x}{2}\sqrt{4-x^{2}}\: +\: C.

2. Tentukan \int \frac{x^{2}}{\sqrt{9-x^{2}}}\: \: dx

Jawab:

misalkan\: \: \left\{\begin{matrix} x = 3 & \sin t &,&maka&\sin t=\frac{x}{3} \\ dx =3 & \cos t & dt & \end{matrix}\right.

sebagai ilustrasi

123

dari ilustrasi gambar diketahui bahwa

\left\{\begin{matrix} \sin t & = & \frac{x}{3} \\ \cos t & = & \frac{\sqrt{9-x^{2}}}{3} \\ \sqrt{9-x^{2}} & = &3 & \cos t \end{matrix}\right.

\int \frac{x^{2}}{\sqrt{9-x^{2}}}\: \: dx=\frac{\left ( 3\sin t \right )^{2}}{3\cos t}.\: \: 3\cos t\: dt=\int 9\sin ^{2}t\: \: dt.

=\int \left ( \frac{9}{2}-\frac{9}{2}\cos 2t \right )dt=\frac{9}{2}t-\frac{9}{4}\sin 2t+C=\frac{9}{2}t-\frac{9}{2}\sin t.\cos t+C.

\int \frac{x^{2}}{\sqrt{9-x^{2}}}\: \: dx=\frac{9}{2}\arcsin \frac{x}{3}-\frac{9}{2}.\frac{x}{3}.\frac{\sqrt{9-x^{2}}}{3}+C=\frac{9}{2}\arcsin \frac{x}{3}-\frac{x}{2}\sqrt{9-x^{2}}\: +\: C.

3.  Tentukanlah \int \frac{dx}{x^{2}+4}=

Jawab:

misalkan\: \: \left\{\begin{matrix} x & = & 2 & \tan t &,&maka&\tan t &=&\frac{x}{2}\\ dx & = & 2 & \sec ^{2} t&dt \end{matrix}\right.

sebagai ilustrasi

124

dari ilustrasi gambar diketahui bahwa

\sqrt{x^{2}+4}=2\sec t

\int \frac{dx}{x^{2}+4}\: dx=\int \frac{2\sec ^{2}t\: \: dt}{4\sec ^{2}t}=\int \frac{1}{2}\: dt=\frac{1}{2}t+C.

=\frac{1}{2}\arctan \frac{x}{2}+C

\LARGE\fbox{Latihan Soal}

Tentukanlah hasil integral berikut ini

  1. \int \sqrt{9-x^{2}}\: dx
  2. \int \sqrt{16-x^{2}}\: dx
  3. \int \sqrt{9-4x^{2}}\: \: dx
  4. \int \frac{x^{3}}{\sqrt{x^{2}+1}}\: \: dx
  5. \int \frac{dx}{x^{2}\sqrt{9+x^{2}}}
  6. \int \frac{\sqrt{x^{2}-1}}{x}

 

F. Integral Parsial

\int u\: dv=uv-\int v\: du

  • Bentuk umum

\begin{array}{|l|c|}\hline &\int x^{n}\: dx=\displaystyle \frac{1}{n+1}\: x^{n+1}+C,\qquad n\neq -1\\\cline{2-2} \textrm{Integral}\: \textrm{tak}\: \textrm{tentu}&\int ax^{n}\: dx=\displaystyle \frac{a}{n+1}\: x^{n+1}+C,\qquad n\neq -1\\\cline{2-2} &\int a\: dx=ax+C\\\hline \textrm{Integral}\: \textrm{tentu}&\int_{a}^{b}f\left ( x \right )dx=\left [ F\left ( x \right ) \right ]_{a}^{b}=F\left ( b \right )-F\left ( a \right )\\\hline \end{array}  .

  • Teknik Pengintegralan

\begin{array}{|c|c|}\hline \textrm{dengan}\: \textrm{substitusi}\: \textrm{aljabar}&\int u^{n}\: u'dx=\int u^{n}\: du=\displaystyle \frac{1}{n+1}u^{n+1}+C,\quad n\neq -1\\\hline &\textrm{jika}\quad \sqrt{a^{2}-u^{2}},\qquad \textrm{maka}\quad u=a\sin \theta\\\cline{2-2} \textrm{dengan}\: \textrm{substitusi}\: \textrm{trigonometri}&\textrm{jika}\quad \sqrt{a^{2}+u^{2}},\qquad \textrm{maka}\quad u=a\tan \theta \\\cline{2-2} &\textrm{jika}\quad \sqrt{u^{2}-a^{2}},\qquad \textrm{maka}\quad u=a\sec \theta\\\hline \textrm{integral}\: \textrm{parsial}&\int u\: dv=uv-\int v\: du\\\hline \end{array}  .

  • Integral Tak Tentu Fungsi Trigonometri

.\qquad \begin{cases} \int \sin x\: dx & =-\cos x+C \\ \int \sin ax\: dx & =-\displaystyle \frac{1}{a}\cos ax+C \\ \int \sin \left ( ax+b \right )\: dx & =-\displaystyle \frac{1}{a}\cos \left ( ax+b \right )+C \\ \int \cos x\: dx & =\sin x+C \\ \int \cos ax\: dx & =\displaystyle \frac{1}{a}\sin ax+C \\ \int \cos \left ( ax+b \right )\: dx & =\displaystyle \frac{1}{a}\sin \left ( ax+b \right )+C \\ \int \tan x\: dx & =-\ln \left | \cos x \right |+C=\ln \left | \sec x \right |+C \\ \int \tan ax\: dx & =-\displaystyle \frac{1}{a}\ln \left | \cos ax \right |+C=\displaystyle \frac{1}{a}\ln \left | \sec ax \right |+C\\ \int \tan \left ( ax+b \right )\: dx & =-\displaystyle \frac{1}{a}\ln \left | \cos \left ( ax+b \right ) \right |+C=\displaystyle \frac{1}{a}\ln \left | \sec \left ( ax+b \right ) \right |+C \end{cases}   .

CONTOH SOAL INTEGRAL TAK TENTU

\begin{array}{ll}\\\fbox{1}&\textrm{Tentukanlah}\: \textrm{hasi}l\: \textrm{dari}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \left (x+1 \right )dx\\ &\textrm{b}.\quad \int \left (x^{3}+x^{2}+3x+2015 \right )\: dx\\ &\textrm{c}.\quad \int \left ( ax+b \right )^{2} dx\\ &\textrm{d}.\quad \int \left ( 4x-5 \right )^{2} dx\\ &\textrm{e}.\quad \int 4\left ( x-5 \right )^{2} dx\\ &\textup{f}.\quad \int x\left ( x-5 \right )^{2} dx\\ &\textrm{g}.\quad \int \left ( \displaystyle 8x^{2}-\frac{1}{x^{2}}+2015 \right )dx\\ &\textrm{h}.\quad \int \displaystyle \left ( x-\frac{1}{x} \right )^{2} dx\\ &\textrm{i}.\quad \int \left (\displaystyle x+\frac{1}{x} \right )dx\\ &\textrm{j}.\quad \int \displaystyle \sqrt{8x^{3}}\: dx\\ &\textrm{k}.\quad \int \displaystyle x\sqrt{x}\: dx\\ &\textrm{l}.\quad \int \displaystyle x^{2}\sqrt{x}\: dx \end{array}    .

Jawab:

\begin{aligned}1.\: \: a.\quad \int \left ( x+1 \right )dx&=\int x\: dx+\int 1\: dx\: \: ,&\textnormal{ingat bahwa }\qquad x=x^{1}\\ &=\displaystyle \frac{1}{\left (1+1 \right )}x^{\left (1+1 \right )}+x+C\: \: ,&\textnormal{perhatikan bahwa pangkat x dari 1 menjadi bertambah 1}\\ &=\displaystyle \frac{1}{2}x^{2}+x+C\: \: ,&\textnormal{sehingga pangkat x menjadi 2} \end{aligned}.

\begin{aligned}1.\: \: b.\quad \int \left ( x^{3}+x^{2}+3x+2015 \right )dx&=\displaystyle \frac{1}{\left (3+1 \right )}x^{\left (3+1 \right )}+\frac{1}{\left (2+1 \right )}x^{\left (2+1 \right )}+\frac{3x^{\left (1+1 \right )}}{\left (1+1 \right )}+2015x+C\\ &=\displaystyle \frac{1}{4}x^{4}+\frac{1}{3}x^{3}+\frac{3}{2}x^{2}+2015x \end{aligned}    .

\begin{aligned}1.\: \: c.\quad \int \left ( ax+b \right )^{2}dx&=\int \left ( ax+b \right ).\left ( ax+b \right )dx\\ &=\int \left (a^{2}x^{2}+abx+abx+b^{2} \right )dx=\int \left (a^{2}x^{2}+2abx+b^{2} \right )dx\\ &=\displaystyle \frac{a^{2}x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}+\frac{2abx^{\left (1+1 \right )}}{\left (1+1 \right )}+b^{2}x+C\\ &=\displaystyle \frac{a^{2}x^{3}}{3}+\frac{2abx^{2}}{2}+b^{2}x+C\: ,&\textnormal{atau dapat juga ditulis}\\ &=\displaystyle \frac{a^{2}}{3}x^{3}+abx^{2}+b^{2}x+C\end{aligned}.

\begin{aligned}1.\: \: d.\quad \int \left ( 4x-5 \right )^{2}dx&=\int \left ( 4x-5 \right ).\left ( 4x-5 \right )dx\\ &=\int \left (16x^{2}-40x+25 \right )dx\\ &=\displaystyle \frac{16x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{40x^{\left (1+1 \right )}}{\left (1+1 \right )}+25x+C\\ &=\displaystyle \frac{16x^{3}}{3}-\frac{40x^{2}}{2}+25x+C\: ,&\textnormal{atau dapat juga ditulis}\\ &=\displaystyle \frac{16}{3}x^{3}-20x^{2}+25x+C\end{aligned}.

\begin{aligned}1.\: \: e.\quad \int 4\left ( x-5 \right )^{2}dx&=\int 4.\left ( x-5 \right ).\left ( x-5 \right )dx\\ &=\int \left (4x^{2}-40x+25 \right )dx\\ &=\displaystyle \frac{4x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{40x^{\left (1+1 \right )}}{\left (1+1 \right )}+25x+C\\ &=\displaystyle \frac{4}{3}x^{3}-20x^{2}+25x+C\end{aligned}.

\begin{aligned}1.\: \: f.\quad \int x\left ( x-5 \right )^{2}dx&=\int x.\left ( x-5 \right ).\left ( x-5 \right )dx\\ &=\int \left (x^{3}-10x^{2}+25x \right )dx\\ &=\displaystyle \frac{x^{\left ( 3+1 \right )}}{\left ( 3+1 \right )}-\frac{10x^{\left (2+1 \right )}}{\left (2+1 \right )}+\frac{25x^{\left ( 1+1 \right )}}{\left (1+1 \right )}+C\\ &=\displaystyle \frac{1}{4}x^{4}-\frac{10}{3}x^{3}+\frac{25}{2}x^{2}+C\end{aligned}.

\begin{aligned}1.\: \: g.\quad \int \left ( 8x^{2}-\displaystyle \frac{1}{x^{2}}+2015 \right )dx&=\int 8x^{2}\: dx-\int \displaystyle \frac{1}{x^{2}}\: dx+\int 2015\: dx\\ &=\int 8x^{2}\: dx-\int x^{-2}\: dx+\int 2015\: dx\\ &=\displaystyle \frac{8x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{x^{\left (-2+1 \right )}}{\left (-2+1 \right )}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}-\frac{x^{-1}}{\left (-1 \right )}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}+x^{-1}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}+\frac{1}{x}+2015x+C\end{aligned}.

\begin{aligned}1.\: \: h.\quad \int \left ( x-\displaystyle \frac{1}{x}\right )^{2}dx&=\int \left ( \displaystyle x-\frac{1}{x} \right ).\left ( x-\frac{1}{x} \right ) dx\\ &=\int \displaystyle \left (x^{2}-2+\frac{1}{x^{2}} \right )dx\\ &=\int \displaystyle x^{2}\: dx-\int 2\: dx+\int \frac{1}{x^{2}}\: dx\\ &=\displaystyle \frac{1}{3}x^{3}-2x-\frac{1}{x}+C\: ,&\textnormal{lihat pembahasan sebelumnya pada No. 1. g}\end{aligned}.

\begin{aligned}1.\: \: i.\quad \int \left ( x-\displaystyle \frac{1}{x}\right )dx&=\int x\: dx-\int \displaystyle \frac{1}{x}\: dx\\ &=\displaystyle \frac{1}{2}x^{2}-\ln x+C\end{aligned}.

\begin{aligned}1.\: \: j.\quad \int \displaystyle \sqrt{8x^{3}}\: dx&=\int \displaystyle \sqrt{8}.x^{ \frac{3}{2}}\: dx\\ &=\displaystyle \frac{\displaystyle \sqrt{8}. x^{\left ( \frac{3}{2}+1 \right )}}{\left ( \displaystyle \frac{3}{2}+1 \right )}+C\\ &=\displaystyle \frac{\sqrt{8}x^{\frac{5}{2}}}{\displaystyle \frac{5}{2}}+C\quad ,&\textnormal{ingat bahwa}\: \sqrt{8}=2\sqrt{2}\\ &=\displaystyle\frac{4\sqrt{2}.\sqrt{x^{5}}}{5}+C\: ,&\textnormal{ingat juga bahwa }\: \sqrt{x^{5}}=\sqrt{x^{4}.x}=x^{2}\sqrt{x}\\ &=\displaystyle \frac{4}{5}x^{2}\sqrt{2x}+C\end{aligned}.

\begin{aligned}1.\: \: k.\quad \int \displaystyle x\sqrt{x}\: dx&=\int \displaystyle x^{1}.x^{\frac{1}{2}}\: dx=\int x^{1+\frac{1}{2}}\: dx=\int x^{\frac{3}{2}}\: dx\\ &=\displaystyle \frac{x^{\left (\frac{3}{2}+1 \right )}}{\left (\displaystyle \frac{3}{2}+1 \right )}+C=\frac{x^{\frac{5}{2}}}{\displaystyle \left ( \frac{5}{2} \right )}+C\\ &=\displaystyle \frac{2}{5}\sqrt{x^{5}}+C\\ &=\displaystyle \frac{2}{5}x^{2}\sqrt{x}+C\end{aligned}.

\begin{aligned}1.\: \: l.\quad \int x^{2}\sqrt{x}\: dx&=\int x^{2}.x^{\frac{1}{2}}\: dx=\int x^{\frac{5}{2}}\: dx\\ &=\displaystyle \frac{1}{\displaystyle \left (\frac{5}{2}+1 \right )}x^{\left ( \frac{5}{2}+1 \right )}+C\\ &=\displaystyle \frac{1}{\left ( \frac{7}{2} \right )}x^{\left (\frac{7}{2} \right )}+C\\ &=\displaystyle \frac{2}{7}x^{3}\sqrt{x}+C \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah}\: \textrm{hasil}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \sin 2x\: dx\\ &\textrm{b}.\quad \int \cos 3x\: dx\\ &\textrm{c}.\quad \int \tan 4x\: dx\\ &\textrm{d}.\quad \int \sin ^{2}x\: dx\\ &\textrm{e}.\quad \int \cos ^{2}2x\: dx\\ &\textrm{f}.\quad \int \tan ^{2}3x\: dx\\ &\textrm{g}.\quad \int \sin ^{3}4x\: dx\\ &\textrm{h}.\quad \int \cos ^{3}5x\: dx\\ &\textrm{i}.\quad \int \sin x\cos x\: dx\\ &\textrm{j}.\quad \int \left ( \sin x+\cos x\right )^{2}\: dx\\ &\textrm{k}.\quad \int \left ( x^{2}+\cos \left ( 2x+3 \right ) \right )\: dx \end{array}.

Jawab:

\begin{aligned}2.\: \: a.\quad \int \sin 2x\: dx&=-\displaystyle \frac{1}{2}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \: b.\quad \int \cos 3x\: dx&=\displaystyle \frac{1}{3}\sin 3x+C \end{aligned}.

\begin{aligned}2.\: \: c.\quad \int \tan 4x\: dx&=-\displaystyle \frac{1}{4}\ln \left | \cos 4x \right |+C \end{aligned}.

\begin{aligned}2.\: \: d.\quad \int \sin ^{2}x\: dx&=\int \left ( \displaystyle \frac{1-\cos 2x}{2} \right )\: dx=\int \displaystyle \frac{1}{2}\: dx-\int \displaystyle \frac{\cos 2x}{2}\: dx=\displaystyle \frac{1}{2}\int 1\: dx-\displaystyle \frac{1}{2}\int \cos 2x\: dx\\ &=\displaystyle \frac{1}{2}x-\displaystyle \frac{1}{2}\left ( \frac{1}{2}\sin 2x \right )+C\\ &=\displaystyle \frac{1}{2}x-\displaystyle \frac{1}{4}\sin 2x+C\end{aligned}.

\begin{aligned}2.\: \: e.\quad \int \cos ^{2}2x\: dx&=\int \left ( \displaystyle \frac{1+\cos 4x}{2} \right )\: dx=\displaystyle \frac{1}{2}\int 1\: dx+\displaystyle \frac{1}{2}\int \cos 4x\: dx\\ &=\displaystyle \frac{1}{2}x+\displaystyle \frac{1}{2}\left ( \frac{1}{4}\sin 4x \right )+C\\ &=\displaystyle \frac{1}{2}x+\displaystyle \frac{1}{8}\sin 4x+C\end{aligned}.

Sebagai pengingat    \LARGE\boxed{\cos 2x=2\cos ^{2}x-1=1-2\sin ^{2}x}.

\begin{aligned}2.\: \: f.\quad \int \tan ^{2}3x\: dx&=\int \left ( \sec ^{2}3x-1 \right )\: dx\\ &=\int \sec ^{2}3x\: dx-\int 1\: dx\\ &=\displaystyle \frac{1}{3}\tan 3x-x+C\end{aligned}.

\begin{aligned}2.\: \: g.\quad \int \sin ^{3}4x\: dx&=\int \left ( \displaystyle \frac{3}{4}\sin 4x-\displaystyle \frac{1}{4}\sin 12x \right )\: dx,&\textnormal{ingat bahwa:}\: \: \sin 3x=3\sin x-4\sin ^{3}x\\ &=\displaystyle \frac{3}{4}\int \sin 4x\: dx-\displaystyle \frac{1}{4}\int \sin 12x\: dx\\ &=\displaystyle \frac{3}{4}\left ( -\frac{\cos 4x}{4} \right )-\displaystyle \frac{1}{4}\left ( -\frac{\cos 12x}{12} \right )+C\\ &=-\displaystyle \frac{3}{14}\cos 4x+\displaystyle \frac{1}{48}\cos 12x+C\end{aligned}.

\begin{aligned}2.\: \: h.\quad \int \cos ^{3}5x\: dx&=\int \left ( \displaystyle \frac{3}{4}\cos 5x+\displaystyle \frac{1}{4}\cos 15x \right )\: dx,&\textnormal{ingat bahwa:}\: \: \cos 3x=4\cos ^{3}x-3\cos x\\ &=\displaystyle \frac{3}{4}\int \cos 5x\: dx+\displaystyle \frac{1}{4}\int \cos 15x\: dx\\ &=\displaystyle \frac{3}{4}\left ( \frac{\sin 5x}{5} \right )+\displaystyle \frac{1}{4}\left ( \frac{\sin 15x}{15} \right )+C\\ &=\displaystyle \frac{3}{20}\sin 5x+\displaystyle \frac{1}{60}\sin 15x+C\end{aligned}.

\begin{aligned}2.\: \:i.\quad \int \sin x\cos x\: dx&=\displaystyle \int \frac{1}{2}\sin 2x\: dx=-\displaystyle \frac{1}{4}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \:j\quad \int \left (\sin x+\cos x \right )^{2}\: dx&=\displaystyle \int \left ( \sin x+\cos x \right )\left ( \sin x+\cos x \right )\: dx\\ &=\int \left (\sin ^{2}x+2\sin x\cos x+\cos ^{2}x \right )\: dx\\ &=\int \left (\sin ^{2}x+\cos ^{2}x+2\sin x\cos x \right )\: dx=\int \left ( 1+\sin 2x \right )\: dx\\ &=x-\displaystyle \frac{1}{2}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \:k\quad \int \left ( x^{2}+\cos \left ( 2x+3 \right ) \right )\: dx&=\displaystyle \frac{1}{3}x^{3}+\displaystyle \frac{1}{2}\sin \left ( 2x+3 \right )+C \end{aligned}.

\begin{array}{ll}\fbox{3}.&\textrm{Tentukan}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \sin x\cos x\: dx\\ &\textrm{b}.\quad \int \sin x\sin 2x\: dx\\ &\textrm{c}.\quad \int \cos x\cos 3x\: dx\\ &\textrm{d}.\quad \int \sin x\cos 2x\: dx\\ &\textrm{e}.\quad \int \sin 2x\cos 2x\: dx\\ &\textrm{f}.\quad \int \cos x\sin 3x\: dx\\ &\textrm{g}.\quad \int \sin x\sin 2x\sin 3x\: dx\\ &\textrm{h}.\quad \int \cos x\cos 2x\cos 3x\: dx\end{array}.

Jawab:

\begin{aligned}3.\: \: a.\quad \int \sin x\cos x\: dx&=\int \displaystyle \frac{1}{2}\left ( \sin \left ( x+x \right )+\sin \left ( x-x \right ) \right )\: dx\\ &\textnormal{Kita dapat menggunakan formula}\: \begin{cases} 2\sin A\sin B & = \cos \left ( A-B \right )-\cos \left ( A+B \right )\\ 2\cos A\cos B & = \cos \left ( A-B \right )+\cos \left ( A+B \right )\\ 2\sin A\cos B & = \sin \left ( A+B \right )+\sin \left ( A-B \right )\\ 2\cos A\sin B & = \sin \left ( A+B \right )-\sin \left ( A-B \right ) \end{cases}\\ &=\displaystyle \frac{1}{2}\int \sin 2x\: dx\\ &=-\displaystyle \frac{1}{2}\left ( \frac{1}{2}\cos 2x \right )+C\\ &=-\displaystyle \frac{1}{4}\left ( 1-2\sin ^{2}x \right )+C\\ &=\displaystyle \frac{1}{2}\sin ^{2}x+C-\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{1}{2}\sin ^{2}x+C\\ &\textnormal{sebagai alternatif jawaban pada No. 2. i} \end{aligned}.

\begin{aligned}3.\: \: b.\quad \int \sin x\sin 2x\: dx&=\int \displaystyle \frac{1}{2}\left ( \cos \left ( x-2x \right )-\cos \left ( x+2x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left (\cos \left ( -x \right )-\cos 3x \right )\: dx\: &\textnormal{ingat}\: \: \cos \left ( -x \right )=\cos x\\ &=\displaystyle \frac{1}{2}\left ( \int \cos x\: dx-\int \cos 3x\: dx \right )\\ &=\displaystyle \frac{1}{2}\left ( \sin x-\displaystyle \frac{1}{3}\sin 3x \right )+C \end{aligned}.

\begin{aligned}3.\: \: c.\quad \int \cos x\cos 3x\: dx&=\int \displaystyle \frac{1}{2}\left ( \cos \left ( x-3x \right )+\cos \left ( x+3x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left (\cos \left ( -2x \right )+\cos 4x \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left ( \cos 2x+\cos 4x \right )\: dx\\ &=\displaystyle \frac{1}{2}\left ( \displaystyle \frac{1}{2}\sin 2x+\displaystyle \frac{1}{4}\sin 4x \right )+C\\ &=\displaystyle \frac{1}{4}\sin 2x+\displaystyle \frac{1}{8}\sin 4x+C\end{aligned}.

\begin{aligned}3.\: \: g.\quad \int \sin x\sin 2x\sin 3x\: dx&=\int \displaystyle \frac{1}{2}\left ( 2\sin x\sin 2x \right )\sin 3x\: dx\\ &=\displaystyle \frac{1}{2}\int \left ( \cos x-\cos 3x \right )\sin 3x\: dx\\ &=\displaystyle \frac{1}{2}\left (\int \cos x\sin 3x\: dx-\int \cos 3x\sin 3x\: dx \right )\\ &=\displaystyle \frac{1}{2}.\frac{1}{2}\int \left ( \sin \left ( x+3x \right )-\sin \left ( x-3x \right ) \right )\: dx-\displaystyle \frac{1}{2}.\frac{1}{2}\int \left ( \sin \left ( 3x+3x \right )-\sin \left ( 3x-3x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{4}\int \sin 4x\: dx-\displaystyle \frac{1}{4}\int -\sin 2x\: dx-\displaystyle \frac{1}{4}\int \sin 6x\: dx\\ &=-\displaystyle \frac{1}{16}\cos 4x-\displaystyle \frac{1}{8}\cos 2x+\displaystyle \frac{1}{24}\cos 6x+C\end{aligned}.

Untuk jawaban soal-soal lainnya silahkan dibuat sebagai latihan mandiri

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui}\: f'\left ( x \right )=6x^{2}-2x+6\: \textrm{dan}\: \textrm{nilai}\: \textrm{fungsi}\: f\left ( 2 \right )=-7.\: \textrm{Tentukan}\: \textrm{rumus}\: \textrm{fungsi}\: \textrm{tersebut}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}f\left ( x \right )&=\int f'\left ( x \right )\: dx\\ &=\int \left ( 6x^{2}-2x+6 \right )\: dx\\ &=2x^{3}-x^{2}+6x+C \end{aligned}\\\\ &\textrm{Karena}\: f\left ( 2 \right )=-7,\: \textrm{maka}\\\\ &\begin{aligned}f\left ( 2 \right )&=2.2^{3}-2^{2}+6.2+C\\ \Leftrightarrow -7&=16-4+12+C\\ \Leftrightarrow C&=-31 \end{aligned}\\\\ &\textrm{Jadi},\: f\left ( x \right )=2x^{3}-x^{2}+6x-31 \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \textrm{gradien}\: \textrm{suatu}\: \textrm{kurva}\: \textrm{adalah}\: \: \displaystyle \frac{dy}{dx}=2x-2.\: \textrm{Tentukan}\: \textrm{persamaan}\: \textrm{kurva}\: \textrm{tersebut}\: \textrm{jika}\: \textrm{melalui}\: \textrm{titik}\: \left ( 3,2 \right )\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\displaystyle \frac{dy}{dx}&=2x-2\\ dy&=\left ( 2x-2 \right )\: dx\\ \int dy&=\int \left ( 2x-2 \right )\: dx\\ y&=x^{2}-2x+C\end{aligned}\\\\ &\textrm{Karena}\: \textrm{kurva}\: \textrm{melalui}\: \left ( 3,2 \right ),\: \textrm{maka}\\\\ &\begin{aligned}2&=3^{2}-2.3+C\\ C&=-1 \end{aligned}\\\\ &\textrm{Jadi},\: \textrm{persamaan}\: \textrm{kurvanya}\: \textrm{adalah}\: y=x^{2}-2x-1 \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Carilah}\: y=f(x)\: \textrm{jika}\: \textrm{diketahui}\\ &\textrm{a}.\quad {f}''\left ( x \right )=2,\: {f}'\left ( 2 \right )=5,\: \textrm{dan}\: f\left ( 2 \right )=10.\\ &\textrm{b}.\quad {f}''\left ( x \right )=x^{2},\: {f}'\left ( 0 \right )=6,\: \textrm{dan}\: f\left ( 0 \right )=3.\\ &\textrm{c}.\quad {f}''\left ( x \right )=\displaystyle x^{-\frac{3}{2}},\: {f}'\left ( 4 \right )=2,\: \textrm{serta}\: f\left ( 0 \right )=0.\\ &\textrm{d}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=4-6x,\: \textrm{untuk}\: \: x=2,\: \displaystyle \frac{dy}{dx}=-4\: \: \textrm{dan}\: y=7.\\ &\textrm{e}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=25x-2,\: \textrm{untuk}\: \: x=2,\: \displaystyle \frac{dy}{dx}=25\: \: \textrm{dan}\: y=20.\\ &\textrm{f}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=6x-4,\: \textrm{kurva}\: \textrm{melalui}\: \textrm{titik}\: \left ( -1,2 \right )\: \textrm{dan}\: \left ( 1.4 \right ).\end{array}.

Jawab:

\begin{aligned}6.\: \: a.\qquad\qquad {f}''\left ( x \right )&=2\\ \int {f}''\left ( x \right )dx&=\int 2\: dx\\ {f}'\left ( x \right )&=2x+C_{1},&\textnormal{diketahui dari soal}\: {f}'\left ( 2 \right )=5,\: \textrm{maka}\\ 2x+C_{1}&={f}'\left ( x \right )\\ 2\left ( 2 \right )+C_{1}&={f}'\left ( 2 \right )=5\\ C_{1}&=5-4=1\\ \textrm{sehingga}\: \: {f}'\left ( x \right )&=2x+1 \end{aligned}\\\\ \textrm{Selanjutnya dengan cara yang kurang lebih sama kita akan mendapatkan f(x), yaitu}\\ \begin{aligned}{f}'\left ( x \right )&=2x+1\\ \int {f}'\left ( x \right )dx&=\int \left ( 2x+1 \right )\: dx\\ f\left ( x \right )&=x^{2}+x+C,&\textnormal{diketahui juga dari soal bahwa f(2)=10},\: \textrm{maka}\\ x^{2}+x+C&=f\left ( x \right )\\ \left (2 \right )^{2}+2+C&=f\left ( 2 \right )=10\\ C&=10-6=4\\ \textrm{sehingga}\: \:f\left ( x \right )&=x^{2}+x+4 \end{aligned}\\\\ \therefore f\left ( x \right )=x^{2}+x+4        .

\begin{aligned}6\: \: d.\quad \displaystyle \frac{d^{2}y}{dx^{2}}&=4-6x\\ \int \displaystyle \frac{d^{2}y}{dx^{2}}&=\int \left ( 4-6x \right )dx\\ \displaystyle \frac{dy}{dx}&=4x-3x^{2}+C_{1},&\textnormal{di soal diketahui}\: \: \displaystyle \frac{dy}{dx}=-4,\: x=2\: \textrm{maka} \\ -4&=4\left ( 2 \right )-3\left ( 2 \right )^{2}+C_{1}\\ C_{1}&=0\end{aligned}\\ \textrm{Sehingga}\: \: \displaystyle \frac{dy}{dx}=-3x^{2}+4x\\\\ \begin{aligned}\displaystyle \frac{dy}{dx}&=-3x^{2}+4x\\ \int \displaystyle \frac{dy}{dx}&=\int \left ( -3x^{2}+4x \right )\: dx\\ y&=-x^{3}+2x^{2}+C,&\textnormal{diketahui juga bahwa y=7, x=2},\: \textrm{maka}\\ 7&=-\left ( 2 \right )^{3}+2\left ( 2 \right )^{2}+C\\ C&=7\\ \textrm{Sehingga}\: y&=-x^{3}+2x^{2}+7 \end{aligned}\\\\ \therefore y=-x^{3}+2x^{2}+7          .

Untuk Soal yang lain silahkan dikerjakan sebagai laithan mandiri

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\textrm{Tentukan}\: \textrm{integralkan}\: \textrm{tak}\: \textrm{tentu}\: \textrm{berikut}\\ \begin{array}{lllll}\\ 1.&a.& \int dx& k.&\int -2x^{-5}\: dx\\ &b.&\int 3\: dz&l.&\int \left (m^{\frac{1}{3}}-\displaystyle \frac{1}{m^{\frac{1}{2}}} \right )\: dm\\ &c.&\int \left ( 3x^{2}-2 \right )\: \: dx&m.&\int \left ( t^{2}-\displaystyle \frac{4}{t^{2}} \right )\: dt\\ &d.&\int \left (x^{5}-3x+4 \right )\: dx&n.&\int \left ( x+5 \right )\left ( x-7 \right )\: dx\\ &e.&\int \left ( 4x^{3}-6x^{2}+3x-6 \right )\: dx&o.&\int \displaystyle \frac{x^{3}+5x^{2}-4}{x^{2}}\: dx\\ &f.&\int \left (6-2t+3t^{2}-8t^{3} \right )\: dt&p.&\int \sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\: dx\\ &g.&\int y\: dx&q.&\int x^{-6}\left ( x\sqrt{x}-x^{-2} \right )\: dx\\ &h.&\int x\: dy&r.&\int \left ( n^{2}-\displaystyle \frac{2}{n^{2}} \right )\: dn\\ &i.&\int \left ( x+y+z \right )\: dx&s.&\int \left ( y\sqrt{y}-\displaystyle \frac{1}{y\sqrt{y}} \right )\: dy\\ &j.&\int \displaystyle \frac{3}{x^{2}\sqrt{x}}\: dx&t.&\int \left ( x-1 \right )\left ( x-2 \right )\left ( x-3 \right )\: dx\end{array}.

\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \textrm{pada}\: \textrm{tiap}\: \textrm{titik}\: \left ( x,y \right )\: \textrm{gradien}\: \textrm{sebuah}\: \textrm{kurva}\: \textrm{ditentukan}\: \textrm{oleh}\: \: \displaystyle \frac{dy}{dx}=1-2x.\\ & \textrm{Jika}\: \textrm{nilai}\: \textrm{maksimum}\:\: y\:\: \textrm{adalah}\: \displaystyle \frac{25}{4},\: \textrm{maka}\: \textrm{persamaan}\: \textrm{kurva}\: \textrm{tersebut}\: \textrm{adalah}.... \end{array}.

\begin{tabular}{ll}\\ 3.&lihat pembahasan pada contoh soal No.3\\ &Silahkan kerjakan soal-soal yang belum dibahas\end{tabular}.

\begin{tabular}{ll}\\ 4.&lihat pembahasan pada contoh soal No.6\\ &Silahkan kerjakan juga soal-soal yang belum dibahas\end{tabular}.

 CONTOH SOAL INTEGRAL TENTU

\begin{array}{ll}\\ \fbox{1}.&\textrm{Hitunglah\: hasil\: dari\: integral-integral\: berikut}\\ &\textrm{a}.\quad \int_{1}^{3}\left ( 2y+1 \right )\: dy\\ &\textrm{b}.\quad \int_{-1}^{1}x^{3}-1\: dx\\ &\textrm{c}.\quad \int_{-1}^{1}\left ( 7x^{3}-3 \right )\: dx\\ &\textrm{d}.\quad \int_{-2}^{1}\left ( m+3 \right )^{2}\: dm\\ &\textrm{e}.\quad \int_{0}^{ \frac{\pi }{2}}\sin \theta \: d\theta \\ &\textrm{f}.\quad \int_{-\pi }^{\pi }\left ( 1+\cos x \right )\: dx \end{array}.

Jawab:

\begin{aligned}1.\: \: a.\quad \int_{1}^{3}\left ( 2y+1 \right )\: dy&=y^{2}+y\: |_{1}^{3}\\ &=\left ( 3^{2}+3 \right )-\left ( 1^{2}+1 \right )\\ &=12-2\\ &=10 \end{aligned}.

\begin{aligned}1.\: \: b.\quad \int_{-1}^{1}\left ( x^{3}-1 \right )\: dx&=\displaystyle \frac{1}{4}x^{4}-x|_{-1}^{1}\\ &=\left ( \displaystyle \frac{1}{4}.\left ( 1 \right )^{4}-1 \right )-\left ( \displaystyle \frac{1}{4}.\left ( -1 \right )^{4}-\left ( -1 \right ) \right )\\ &=\left ( \displaystyle \frac{1}{4}-1 \right )-\left ( \displaystyle \frac{1}{4}+1 \right )\\ &=-2 \end{aligned}.

\begin{aligned}1.\: \: e.\quad \int_{0}^{\frac{\pi }{2}}\sin \theta \: d\theta &=-\cos \theta \: |_{0}^{\frac{\pi }{2}}\\ &=\left ( -\cos \displaystyle \frac{\pi }{2} \right )-\left ( -\cos 0 \right )\\ &=\left ( 0 \right )-\left ( -1 \right )\\ &=1 \end{aligned}.

\begin{aligned}1.\: \: f.\quad \int_{-\pi }^{\pi }\left ( 1+\cos x \right )\: dx&=\left ( x+\sin x \right )|_{-\pi }^{\pi }\\ &=\left ( \pi +\sin \pi \right )-\left ( -\pi +\sin \left ( -\pi \right ) \right )\\ &=\left ( \pi +0 \right )-\left ( -\pi -0 \right )\\ &=2\pi \end{aligned}.

\begin{aligned}2.\: \: \textrm{Nilai}&\: \int_{0}^{\frac{\pi }{3}}\sin 2x\: dx=....\\ a.\quad \frac{3}{4}&\qquad b.\quad \frac{1}{2}\qquad c.\quad\frac{1}{3}\qquad d.\quad \frac{1}{4}\qquad e.\quad 0\qquad\qquad \textbf{(UN\: 2006)}\end{aligned}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\int_{0}^{\frac{\pi }{3}}\sin 2x\: dx&=-\displaystyle \frac{1}{2}\cos 2x|_{0}^{\frac{\pi }{3}}\\ &=\left ( -\displaystyle \frac{\cos \frac{2\pi }{3}}{2} \right )-\left ( -\displaystyle \frac{\cos 0}{2} \right )\\ &=\left ( -\displaystyle \frac{\frac{1}{2}}{2} \right )-\left ( -\displaystyle \frac{1}{2} \right )\\ &=\left ( -\displaystyle \frac{1}{4} \right )-\left ( -\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{1}{4} \end{aligned}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\begin{array}{ll}\\ 1.\quad \textrm{Hitunglah\: nilai\: tiap\: integral\: berikut\: ini} \end{array}\\ \begin{array}{lllll}\\ .\quad&a.&\int_{0}^{3}2x\: dx&i.&\int_{1}^{2}\sqrt{x^{5}}\: dx\\ &b.&\int_{2}^{3}2x^{2}\: dx&j.&\int_{1}^{2}\displaystyle \frac{1}{x^{2}}\: dx\\ &c.&\int_{3}^{7}\displaystyle \frac{1}{2}x^{3}\: dx&k.&\int_{4}^{9}3\sqrt{x}\: dx\\ &d.&\int_{0}^{3}\left ( x^{2}+2x-1 \right )\: dx&l.&\int_{4}^{6}\left ( \sqrt{t}+t^{2}-\sqrt[3]{t^{2}} \right )\: dt\\ &e.&\int_{0}^{3}\left ( x+3 \right )^{2}\: dx&m.&\int_{0}^{\pi }\cos x\: dx\\ &f.&\int_{-1}^{2}\left ( x+2 \right )\left ( x-1 \right )\: dx&n.&\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin x\cos x\: dx\\ &g.&\int_{1}^{3}\left ( t+t^{2}-3t^{3} \right )\: dt&o.&\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\sin x\: dx\\ &h.&\int_{1}^{4}\left ( \displaystyle \frac{x^{4}-x^{3}+\sqrt{x}-1}{x^{2}} \right )\: dx&p.&\int_{0}^{\frac{\pi }{6}}\left ( \sin 2x+\cos 3x \right )\: dx \end{array}.

\begin{aligned}2.\: \: \textrm{Nilai}\: \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}&\left ( 3\cos x-5\sin x \right )\: dx=....\textbf{(EBTANAS\: 1997)} \end{aligned}.

Penggunaan Integral

A. Luas Daerah yang Dibatasi Kurva

Untuk menghitung luas daerah yang dibatasi suatu kurva dengan sumbu x dapat kita gunakan konsep integral tentu

Perhatikan Ilustrasi berikut

268

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textbf{Luas Daerah}}\\\hline \textrm{Di Atas Sumbu X}&\textrm{Di Bawah Sumbu X}\\\hline &-\displaystyle \int_{a}^{b}f(x)\: \: dx\\ \displaystyle \int_{a}^{b}f(x)\: \: dx&atau\\ &\displaystyle \int_{b}^{a}f(x)\: \: dx\\\hline \end{array}.

Misalkan kita diberikan gambar berikut,

269

maka luas  A_{1}\: \textrm{dan}\: A_{2}  adalah:

L_{\displaystyle A_{1}\: \textrm{dan}\: \displaystyle A_{2}}=\displaystyle \int_{b}^{c}f(x)\: dx-\displaystyle \int_{a}^{b}f(x)\: dx.

B. Volume Benda Putar

\boxed{V=\pi \displaystyle \int_{a}^{b}\left ( f(x) \right )^{2}\: \: dx=\pi \displaystyle \int_{a}^{b}y^{2}\: \: dx}.

Perhatikanlah ilustrasi jika suatu bidang datar dirotasikan terhadap sumbu Y

270

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{lp{16.0cm}}\\ \fbox{1}.&\textrm{Tentukanlah luas daerah bidang berikut dan tentukan pula volumenya seandainya bidang yang diarsir tersebut diputar terhadap sumbu X} \end{array}\.

271

Jawab:

\begin{array}{lll}\\ \begin{aligned}L_{\textrm{Arsiran}}&=\displaystyle \int_{1}^{3}2x\: dx\\ &=\displaystyle \left [ x^{2} \right ]_{1}^{3}\\ &=\left ( 3 \right )^{2}-\left ( 1 \right )^{2}\\ &=9-1\\ &=8\quad \textbf{satuan luas}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\textbf{dan}&\begin{aligned}V_{\textrm{Benda putar}}&=\pi \displaystyle \int_{1}^{3}\left ( y \right )^{2}\: dx=\pi \displaystyle \int_{1}^{3}\left ( 2x \right )^{2}\: dx\\ &=\pi \displaystyle \int_{1}^{3}4x^{2}\: dx\\ &=\pi \left [ \displaystyle \frac{4x^{3}}{3} \right ]_{1}^{3}\\ &=\pi \left ( \displaystyle \frac{4\times 3^{3}}{3} \right )-\pi \left ( \displaystyle \frac{4\times 1^{3}}{3} \right )\\ &=36\pi -\displaystyle \frac{4}{3}\pi \\ &=34\displaystyle \frac{2}{3}\pi \quad \textbf{satuan volum} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: f(x)=\left ( x-2 \right )^{2}-4\: \: \textrm{dan}\: \: g(x)=-f(x),\: \textrm{maka luas daerah yang di batasi kurva \textit{f} dan \textit{g} adalah ....\textbf{(UAN 2003)}} \end{array}\\ \begin{array}{lll}\\\\ .\quad&a.&10\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\\\ &b.&21\displaystyle \frac{1}{3}\: \: \textrm{satuan luas}\\\\ &c.&22\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\\\ &d.&42\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\\\ &e.&45\displaystyle \frac{1}{3}\: \: \textrm{satuan luas} \end{array}.

Jawab:

Perhatikan Ilustrasi berikut

274

\begin{aligned}\displaystyle \int_{0}^{4}\left ( g(x)-f(x) \right )\: \: dx&=\displaystyle \int_{0}^{4}\left ( 4x-x^{2} \right )-\left ( x^{2}-4x \right )\: \: dx\\ &=\displaystyle \int_{0}^{4}\left ( 8x-2x^{2} \right )\: \: dx\\ &=\displaystyle \left [4x^{2}-\frac{2}{3}x^{3} \right ]_{0}^{4}\\ &=\displaystyle \left ( 4.4^{2}-\frac{2}{3}.4^{3} \right )-\left ( 4.0^{2}-\frac{2}{3}.0^{3} \right )\\ &=\displaystyle \left ( 64-\frac{2}{3}.64 \right )-0\\ &=\displaystyle \frac{64}{3}=21\frac{1}{3}\: \: \textrm{satuan luas} \end{aligned}.

Kita juga dapat menggunakan rumus   \displaystyle L=\frac{\displaystyle D\sqrt{D}}{\displaystyle 6a^{2}}.

\begin{array}{|l|}\hline \begin{aligned}f(x)&=g(x)\\ f(x)&=-f(x), &\textnormal{ingat g(x)\: =\: -f(x)}\\ 2f(x)&=0, &\textnormal{tidak boleh disederhanakan, }\\ 2\times \left (\left ( x-2 \right )^{2}-4 \right )&=0, &\textnormal{karena akan mempengaruhi hasil akhir}\\ 2\times \left ( x^{2}-4x \right )&=0\\ 2x^{2}-8x&=0,\quad \begin{cases} a=2,\: b=-8 & c=0 \\ D=b^{2}-4ac, & D=\left ( -8 \right )^{2}-4(2)(0)=64 \end{cases}\\ &\\ L_{\: \textbf{daerah}}&=\displaystyle \frac{\textbf{D}\sqrt{\textbf{D}}}{6\textbf{a}^{2}}\\ &=\displaystyle \frac{64\sqrt{64}}{6(2)^{2}}\\ &=\displaystyle \frac{64\times 8}{6\times 4}\\ &=\displaystyle \frac{64}{3}\\ &=21\displaystyle \frac{1}{3} \end{aligned}\\\hline\end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui parabola}\: \: f_{1}(x)=a_{1}x^{2}+b_{1}x+c_{1}\: \: \textrm{dan}\: \: f_{2}(x)=a_{2}x^{2}+b_{2}x+c_{2}.\\ &\textrm{Titik potong kedua para bola tersebut dapat cari dengan}\\ &\\ &f_{1}(x)=f_{2}(x)\: \: \Leftrightarrow \: \: a_{1}x^{2}+b_{1}x+c_{1}=a_{2}x^{2}+b_{2}x+c_{2}\\ &\: \, \, \qquad\qquad\qquad \Leftrightarrow \: ax^{2}+bx+c=0.\\ &\\ &\textrm{Jika kedua parabola berpotongan di dua titik, tunjukkan bahwa luas daerah antara} \\ &\textrm{kedua parabola tersebut dapat dinyatakan dengan}\: \: \: \displaystyle \textbf{L}=\frac{\textbf{D}\sqrt{\textbf{D}}}{\textbf{6a}^{\textbf{2}}} \\\end{array}.

Bukti:

ax^{2}+bx+c=0\: \begin{cases} &x_{1}=\displaystyle \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \\ & \\ &x_{2}=\displaystyle \frac{-b- \sqrt{b^{2}-4ac}}{2a} \end{cases}\\ \begin{aligned}L&=\displaystyle \int_{\frac{-b- \sqrt{b^{2}-4ac}}{2a}}^{\frac{-b+ \sqrt{b^{2}-4ac}}{2a}}\: \left ( ax^{2}+bx+c \right )\: \: dx=\left [ \displaystyle \frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx \right ]_{\frac{-b- \sqrt{b^{2}-4ac}}{2a}}^{\frac{-b+ \sqrt{b^{2}-4ac}}{2a}}\\ &=\left [ \displaystyle \frac{a}{3}\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right )^{3}+\displaystyle \frac{b}{2}\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right )^{2}+c\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right ) \right ]\\ &\quad -\left [ \displaystyle \frac{a}{3}\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right )^{3}+\displaystyle \frac{b}{2}\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right )^{2}+c\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right ) \right ]\\ &=\displaystyle \frac{a}{24a^{3}}\left [ \left ( \sqrt{D}^{3}-3\sqrt{D}^{2}b+3\sqrt{D}b^{2}-b^{3} \right )+\left ( \sqrt{D}^{3}+3\sqrt{D}^{2}b+3\sqrt{D}b^{2}+b^{3} \right ) \right ]\\ &\quad +\displaystyle \frac{b}{8a^{2}}\left [ \left ( b^{2}-2b\sqrt{D}+\sqrt{D}^{2} \right )-\left ( b^{2}+2b\sqrt{D}+\sqrt{D}^{2} \right ) \right ]+\displaystyle \frac{c}{2a}\left [ \left ( -b+\sqrt{D} \right )-\left ( -b-\sqrt{D} \right ) \right ]\\ &=\displaystyle \frac{1}{24a^{2}}\left [ 2\sqrt{D}^{3}+6\sqrt{D}b^{2} \right ]+\displaystyle \frac{b}{8a^{2}}\left [ -4b\sqrt{D} \right ]+\displaystyle \frac{c}{2a}\left [ 2\sqrt{D} \right ]\\ &=\displaystyle \frac{\sqrt{D}^{3}}{12a^{2}}+\frac{b^{2}\sqrt{D}}{4a^{2}}-\frac{b^{2}\sqrt{D}}{2a^{2}}+\frac{c\sqrt{D}}{a}=\displaystyle \frac{D\sqrt{D}}{12a^{2}}+\frac{b^{2}\sqrt{D}}{4a^{2}}-\frac{b^{2}\sqrt{D}}{2a^{2}}+\frac{c\sqrt{D}}{a}\\ &=\displaystyle \frac{\sqrt{D}}{12a^{2}}\left [ D+3b^{2}-6b^{2}+12ac \right ]\\ \end{aligned}

\begin{aligned}&=\displaystyle \frac{\sqrt{D}}{12a^{2}}\left [ \left ( b^{2}-4ac \right )-3b^{2}+12ac \right ]\\ &=\displaystyle \frac{\sqrt{D}}{12a^{2}}\left [ -2b^{2}+8ac \right ]=-\displaystyle \frac{\sqrt{D}}{6a^{2}}\left [ b^{2}-4ac \right ]=-\frac{\sqrt{D}}{6a^{2}}\left [ D \right ]\\ &=-\frac{D\sqrt{D}}{6a^{2}},\quad \textbf{luas tidak mungkin negatif}\\ L&=\displaystyle \frac{D\sqrt{D}}{6a^{2}}\quad \blacksquare \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Tentukan volume benda putar yang terbentuk, jika suatu daerah yang dibatasi oleh kurva }\\ &y^{2}=x\: \: \textrm{dan}\: \: y=x\: \textrm{diputar mengelilingi sumbu X} \\\end{array}.

Jawab:

Perhatikanlah ilustrasi gambar berikut ini

275.

\begin{array}{|r|l|l|}\hline \multicolumn{3}{|c|}{\textrm{Langkah-langkah}}\\\hline \textrm{Pertama (Mencari Batas)}&\textrm{Kedua (Menentukan Volumenya)}&\textrm{Keterangan}\\\hline \begin{aligned}y&=y\\ x^{2}&=x\\ x^{2}-x&=0\\ x\left ( x-1 \right )&=0\\ x=0\: \: \textrm{atau}\: \: x&=1\\ &\\ &\\ & \end{aligned}&\begin{aligned}V&=\pi \displaystyle \int_{a}^{b}\left ( y_{1}^{2}-y_{2}^{2} \right )\: \: dx\\ &=\pi \displaystyle \int_{0}^{1}\left ( x-x^{2} \right )\: \: dx\\ &=\pi \left [ \displaystyle \frac{1}{2}x^{2}-\frac{1}{3}x^{3} \right ]_{0}^{1}\\ &=\pi \left [ \displaystyle \frac{1}{2}-\frac{1}{3} \right ]\\ V&=\displaystyle \frac{1}{6}\pi \end{aligned}&\begin{aligned}&\textnormal{Perhatikan bahwa;}\\ &y^{2}=x\Rightarrow y=\sqrt{x},\: \textrm{dianggap sebagai}\: \: y_{1}\\ &\textnormal{Sehingga}\: y_{1}-\textrm{nya adalah}\: \: \sqrt{x}\\ &\textnormal{dan}\: \: y=x\: \: \textrm{dianggap sebagai}\: \: y_{2}\\ &\left ( y_{1}^{2}-y_{2}^{2} \right )=\left ( \left ( \sqrt{x} \right )^{2}-\left ( x \right )^{2} \right )=x-x^{2}\end{aligned} \\\hline \multicolumn{2}{|l|}{\textrm{Jadi, volume dari benda putar tersebut dalam satuan volum adalah}\: \: \displaystyle \frac{1}{6}\pi }&\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Tentukan volume benda putar yang terbentuk, jika suatu daerah yang dibatasi oleh kurva }\\ &y=2x\: ,\: y=x,\: x=1,\: \textrm{dan}\: \: x=3\: \textrm{diputar mengelilingi sumbu X} \\\end{array}.

Jawab:

Perhatikanlah ilustrasi gambar berikut

276

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Langkah-Langkah}}\\\hline \textrm{Batas}&\textrm{Menentukan Volumenya}\\\hline x=1\: \: \textrm{dan}\: \: x=3&\begin{aligned}V&=\displaystyle \pi \int_{a}^{b}\left ( f^{2}(x)-g^{2}(x) \right )\: \: dx\\ &=\displaystyle \pi \int_{1}^{3}\left ( \left ( 2x \right )^{2}-\left ( x \right )^{2} \right )\: \: dx\\ &=\displaystyle \pi \int_{1}^{3}3x^{2}\: \: dx\\ &=\displaystyle \pi \left [ x^{3} \right ]_{1}^{3}\\ &=\displaystyle \pi \left ( 3^{3} \right )-\pi \left ( 1^{3} \right )\\ &=27\pi -1\pi \\ V&=26\pi\: \textbf{Satuan Volum} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Tentukan volume daerah yang dibatasi oleh lingkaran }\\ &x^{2}+y^{2}=4\: ,\: \textrm{selang}\: -2\leq x\leq 2\: \textrm{dan}\: \: \textrm{diputar mengelilingi sumbu X} \\\end{array}.

Jawab:

277

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Langkah-Langkah}}\\\hline \textrm{Batas}&\textrm{Menentukan Volumenya}\\\hline x=-2\: \: \textrm{sampai}\: \: x=2&\begin{aligned}V&=\displaystyle \pi \int_{a}^{b}y^{2}\: \: dx\\ &=\displaystyle \pi \int_{-2}^{2}\left ( 4-x^{2} \right )\: \: dx\\ &=\displaystyle \pi \left [ 4x-\displaystyle \frac{x^{3}}{3} \right ]_{-2}^{2} \\ &=\displaystyle \pi \left ( 8-\displaystyle \frac{8}{3} \right )-\pi \left ( -8+\displaystyle \frac{8}{3} \right )\\ &=\displaystyle \pi \left ( 8+8-\frac{8}{3}-\frac{8}{3} \right )\\ V&=\displaystyle \frac{32}{3}\pi\: \textbf{Satuan Volum} \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah luas daerah yang diarsir berikut} \end{array}.

\qquad a.

278

\qquad b.

279

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tunjukkan bahwa luas ellips}\: \: \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\: \: \textrm{adalah}\: \: \pi ab \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Tentukanlah volume benda putar yang terjadi jika daerah dari hasil putar tersebut mengelilingi sumbu X serta dibatasi oleh} \end{array}\\ \begin{array}{ll}\\ .\: \: \qquad &a.\quad y=x+3,\: \textrm{sumbu x, garis x = 2, dan x = 4}\\ &b.\quad \displaystyle y=\frac{1}{2}x+2,\: \textrm{sumbu x, garis x = 0, dan x = 4}\\ &c.\quad \displaystyle y=\sqrt{x+2} ,\: \textrm{sumbu x, garis x = 2, dan x = 4}\\ &d.\quad y=4-2x,\: \textrm{sumbu x, garis x = 0, dan x = 4}\\ &e.\quad \displaystyle x^{2}+y^{2}=16,\: \textrm{dan sumbu X} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Tentukanlah volume benda putar yang terjadi jika daerah dari hasil putar tersebut mengelilingi sumbu X serta dibatasi oleh} \end{array}\\ \begin{array}{ll}\\ .\: \: \qquad &a.\quad y=2x-x^{2},\: \textrm{dan}\: \: y=0\\ &b.\quad \displaystyle y^{2}=x,\: \textrm{dan}\: \: y=2\\ &c.\quad \displaystyle y=x^{2} ,\: \textrm{dan}\: \: y=-x^{2}+4\\ &d.\quad y=7-2x^{2},\: \textrm{dan}\: \: y=x^{2}+4\\ &e.\quad \displaystyle y=x^{2},\: \textrm{dan}\: \: y^{2}=x \end{array}

 

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