MATEMATIKA: April 2021

Hana Fahira (15) XI IPS 2

Untuk pembahasan kali ini, kita akan membahas luas dan volume daerah yang berkaitan dengan integral bersama dengan beberapa contoh soalnya. Mari kita simak!

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

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

Misalkan kita diberikan gambar berikut,

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

CONTOH SOAL

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

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

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

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

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

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

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

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

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

Sekian pembahasan yang bisa saya sampaikan. Semoga bermanfaat. Terima kasih

Daftar Pustaka:

https://ahmadthohir1089.wordpress.com/2015/08/30/insyaallah-25/

$x$ $l a t e x a \leq x \leq b$ $x$

$l a t e x f (x) \geq g (x)$ $l a t e x a \leq x \leq b$ $x$ $x$

$1$

$2$

$1$ $2$

$l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$ $l a t e x L = \int (x + 6 - x^{2}) d x = \frac{1}{2} x^{2} + 6 x - \frac{1}{3} x^{3} ∣_{- 2}^{3}$

$l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$ $l a t e x = (\frac{9}{2} + 18 - 9) - (2 - 12 + \frac{8}{3}) = 21 \frac{1}{2}$

$x$ $l a t e x a \leq x \leq b$ $x$

$l a t e x y = x^{2}$ $l a t e x y = x^{2}$ $l a t e x y = x^{2}$

$l a t e x y = x + 2$

$l a t e x x^{2} = x + 2$ $l a t e x x^{2} = x + 2$ $l a t e x x^{2} = x + 2$

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