Integral Fungsi Aljabar

Jika anda menggunakan HP maka gunakan mode LANDSCAPE agar equation tidak terpotong

A. PENGERTIAN

Misalkan $f\left( x \right)$ adalah turunan dari $F\left( x \right)$ maka dapat dinyatakan $F\left( x \right)$ sebagai anti turunan atau integral dari $f\left( x \right)$. Secara matematis ditulis:

$$\int {f\left( x \right)\;dx = F\left( x \right) + c}$$

dengan $c$ merupakan suatu konstanta.

B. RUMUS-RUMUS INTEGRAL TAK TENTU FUNGSI ALJABAR

$\begin{align} 1) &\int {{x^n}\;dx{\text{ }} = \frac{1}{{(n + 1)}}{x^{n + 1}} + c} ,{\text{ dengan}}\;\;n \ne - 1\\ 2) &\int {{{\left( {ax + b} \right)}^n}{\text{ }}dx{\text{ }} = \frac{1}{{a(n + 1)}}{{\left( {ax + b} \right)}^{n + 1}} + c} ,{\text{dengan }}n \ne -1\\ 3) &\int {\frac{1}{x}dx = \ln \left| x \right| + c} \\ 4) &\int {\frac{1}{{\left( {ax + b} \right)}}dx = \frac{1}{a}\ln \left| {ax + b} \right| + c} \end{align}$

Contoh Soal

Soal 1

$\displaystyle\int {3{x^2} - 8x + 6{\text{ }}dx} = ...$

Pembahasan:

$= \frac{3}{{\left( {2 + 1} \right)}}{x^{\left( {2 + 1} \right)}} - \frac{8}{{\left( {1 + 1} \right)}}{x^{\left( {1 + 1} \right)}} + 6x + c$

$= \boxed{{x^3} - 4{x^2} + 6x + c}$

Catatan: misalkan $k$ suatu konstanta (angka) maka $\int {k\;dx = kx + c}$. Itulah mengapa hasil integral $6$ adalah $6x$.  

Soal 2

$\displaystyle\int {x{{\left( {x + 1} \right)}^{\text{2}}}dx} = ...$

Pembahasan:

$\begin{align} &= \int {x\left( {{x^2} + 2x + 1} \right)} \;dx \hfill \\ &= \int {{x^3} + 2{x^2} + x\;dx} \hfill \\ &= \frac{1}{{\left( {3 + 1} \right)}}{x^{\left( {3 + 1} \right)}} + \frac{2}{{\left( {2 + 1} \right)}}{x^{\left( {2 + 1} \right)}} + \frac{1}{{\left( {1 + 1} \right)}}{x^{\left( {1 + 1} \right)}}\; + c \hfill \\ &= \boxed{\frac{1}{4}{x^4} + \frac{2}{3}{x^3} + \frac{1}{2}{x^2} + c}\ \end{align}$

Soal 3

$\displaystyle\int {2\sqrt x \;dx} = ...$

Pembahasan:

$\begin{align} &= \int {2{x^{\frac{1}{2}}}\;dx} \hfill \\ &= \frac{2}{{\left( {\tfrac{1}{2} + 1} \right)}}{x^{\left( {\frac{1}{2} + 1} \right)}} + c \hfill \\ &= \frac{4}{3}{x^{1\frac{1}{2}}} + c \hfill \\ &= \boxed{\frac{4}{3}x\sqrt x + c} \ \end{align}$

Soal 4

$\displaystyle\int {\frac{x}{{2\sqrt[3]{{{x^2}}}}}{\text{ }}dx} = ...$

Pembahasan:

$\begin{align} &= \frac{1}{2}\int {x \cdot {x^{ - \frac{2}{3}}}} \;dx \hfill \\ &= \frac{1}{2}\int {{x^{\frac{1}{3}}}} \;dx \hfill \\ &= \frac{1}{2} \cdot \frac{1}{{\left( {\tfrac{1}{3} + 1} \right)}}{x^{\left( {\frac{1}{3} + 1} \right)}} + c \hfill \\ &= \frac{3}{8}{x^{1\frac{1}{3}}} + c \hfill \\ &= \boxed{\frac{3}{8}x\sqrt[3]{x} + c} \hfill \\ \end{align}$

Soal 5

$\displaystyle\int {\frac{{{x^2} + 6\sqrt x - 7}}{{{x^2}}}{\text{ }}dx} =...$

Pembahasan:

$\begin{align} &= \int {\frac{{{x^2}}}{{{x^2}}}} + \frac{{6{x^{\frac{1}{2}}}}}{{{x^2}}} - \frac{7}{{{x^2}}}\;dx \hfill \\ &= \int {1 + 6{x^{ - \frac{3}{2}}} - 7{x^{ - 2}}} \;dx \hfill \\ &= x + \frac{6}{{\left( { - \tfrac{3}{2} + 1} \right)}}{x^{\left( { - \frac{3}{2} + 1} \right)}} - \frac{7}{{\left( { - 2 + 1} \right)}}{x^{\left( { - 2 + 1} \right)}} + c \hfill \\ &= x - 12{x^{ - \frac{1}{2}}} + 7{x^{ - 1}} + c \hfill \\ &= \boxed{x - \frac{{12}}{{\sqrt x }} + \frac{7}{x} + c} \hfill \\ \end{align}$

Soal 6

$\displaystyle\int {7{{(3x + 5)}^{20}}dx} =...$

Pembahasan:

$\begin{align} &= \frac{7}{{3 \cdot \left( {20 + 1} \right)}}{\left( {3x + 5} \right)^{\left( {20 + 1} \right)}} + c \hfill \\ &= \boxed{\frac{1}{9}{{\left( {3x + 5} \right)}^{21}} + c} \hfill \\ \end{align}$

Soal 7

Diketahui $f'\left( x \right) = 2x + 3$, jika $F(x)$ adalah fungsi integral dari $f'\left( x \right)$ dan $F(–2) = 5$ maka fungsi $F(x)$ tersebut adalah…

Pembahasan:

$\begin{align} F\left( x \right) &= \int {f'\left( x \right)} \\ &= \int {2x + 3} \;dx \\ &= {x^2} + 3x + c \\ \end{align}$

$\begin{align} &{\text{Diketahui }}F\left( { - 2} \right) = 5 \\ &{\left( { - 2} \right)^2} + 3\left( { - 2} \right) + c = 5 \\ &c = 7 \hfill \\ &\therefore {\text{ }}\boxed{F\left( x \right) = {x^2} + 3x + 7} \\ \end{align}$

Soal 8

$\displaystyle\int \frac{1}{4 x} d x=...$

Pembahasan:

$\displaystyle\int \frac{1}{4 x} d x=\frac{1}{4} \int \frac{1}{x} d x=\boxed{\frac{1}{4} \ln |x|+c}$

Soal 9

$\displaystyle\int \frac{4}{1-2 x} d x=...$

Pembahasan:

$\begin{aligned} \int \frac{4}{1-2 x} d x &=4 \int \frac{1}{1-2 x} d x \\ &=4\left(\frac{1}{-2}\right) \ln |1-2 x|+c \\ &=\boxed{-2 \ln |1-2 x|+c} \end{aligned}$

C. INTEGRAL TENTU FUNGSI ALJABAR

Integral tentu dengan fungsi awal $f(x)$ dan fungsi hasil integral adalah $F(x)$ dengan batas-batas $a$ sampai $b$ dapat dinotasikan sebagai berikut:
$$\int\limits_a^b {f(x){\text{ }}dx = } \left. {{\text{ }}F(x){\text{ }}} \right|_a^b = F(b) - F(a)$$ Sifat-Sifat Integral Tentu

$1){\text{ }}\displaystyle\int\limits_a^a {f(x) dx = 0}$
$2){\text{ }}\displaystyle\int\limits_b^a {f(x) dx = - \int\limits_a^b {f(x) dx} }$
$3){\text{ }}\displaystyle\int\limits_b^a {f( - x) dx = - \int\limits_{ - a}^{ - b} {f(x) dx} }$
$4){\text{ }}\displaystyle\int\limits_a^b {f(x) dx = \int\limits_a^p {f(x) dx + \int\limits_p^b {f(x)} } } dx$
$5){\text{ }}{\text{Untuk fungsi genap }}{\text{ }}(f(–x) = f(x)) {\text{ }}{\text{berlaku: }}$
${\text{ }}{\text{ }}{\text{ }}\displaystyle\int\limits_{ - a}^a {f(x)} \,dx = 2\int\limits_{ - a}^0 {f(x)\,dx = } \;2\int\limits_0^a {f(x)\,dx}$
$6){\text{ }}{\text{Untuk fungsi ganjil }}{\text{ }}(f(–x) = -f(x)) {\text{ }}{\text{berlaku: }}$
${\text{ }}{\text{ }}{\text{ }}\displaystyle\int\limits_{ - a}^0 {f(x)\,dx + } \,\int\limits_0^a {f(x)\,dx} = \int\limits_{ - a}^a {f(x)\,dx} = 0$
$7){\text{ }}{\text{Untuk fungsi periodik }}{\text{ }}(f(–x) = f(x+p)) {\text{ }}{\text{berlaku: }}$
${\text{ }}{\text{ }}{\text{ }}\displaystyle\int\limits_a^b {f(x)\,dx = \int\limits_{a + p}^{b + p} {f(x)\,dx} }$

Contoh Soal

Soal 1

$\displaystyle\int\limits_{ - 1}^2 {3{x^2} + x - 1\;dx=...}$

Pembahasan:

$\begin{align} &= \left. {\frac{3}{{\left( {2 + 1} \right)}}{x^{\left( {2 + 1} \right)}} + \frac{1}{{\left( {1 + 1} \right)}}{x^{\left( {1 + 1} \right)}} - x} \right|_{ - 1}^2 \hfill \\ &= \left. {{x^3} + \frac{1}{2}{x^2} - x} \right|_{ - 1}^2 \hfill \\ &= \left( {{2^3} + \frac{1}{2} \cdot {2^2} - 2} \right) - \left( {{{\left( { - 1} \right)}^3} + \frac{1}{2} \cdot {{\left( { - 1} \right)}^2} - \left( { - 1} \right)} \right) \hfill \\ &= 8 - \frac{1}{2} \hfill \\ &= \boxed{\frac{{15}}{2}} \hfill \\ \end{align}$

Soal 2

$\displaystyle\int\limits_0^2 {{{{\text{(}}2x - 1{\text{)}}}^3}} dx =...$

Pembahasan:

$\begin{align} &= \left. {\frac{1}{{2 \cdot \left( {3 + 1} \right)}}{{\left( {2x - 1} \right)}^{\left( {3 + 1} \right)}}} \right|_0^2 \hfill \\ &= \left. {\frac{1}{8}{{\left( {2x - 1} \right)}^4}} \right|_0^2 \hfill \\ &= \frac{1}{8}\left[ {{{\left( {2 \cdot 2 - 1} \right)}^4} - {{\left( {2 \cdot 0 - 1} \right)}^4}} \right] \hfill \\ &= \frac{1}{8}\left[ {81 - 1} \right] \hfill \\ &= \boxed{10} \hfill \\ \end{align}$

Soal 3

$\displaystyle\int\limits_0^2 {\left| {2x - 1} \right|} \;dx =...$

Pembahasan:

Karena fungsi yang diintegralkan merupakan bentuk mutlak maka kita buat garis bilangannya terlebih dahulu.

$\begin{align} 2x - 1 &= 0 \hfill \\ x &= \frac{1}{2} \hfill \\ \end{align}$

Setelah mendapat daerah penyelesaian (daerah positif dan negatifnya) maka dilanjutkan dengan menuliskan batas integral yang diminta (0 sampai dengan 2). Perhatikan bahwa dari $x = 0$ kita harus berhenti di batas nilai $x = \tfrac{1}{2}$ terlebih dahulu baru dilanjutkan ke $x = 2$. Sehingga perlu dilakukan dua kali perhitungan nilai integral. Perlu diingat bahwa tanda di daerah himpunan penyelesaian perlu disertakan.
$\begin{align} &-\int\limits_0^{\frac{1}{2}} {2x - 1\;dx = - \left[ {{x^2} - x} \right]} _0^{\frac{1}{2}} \\ &= - \left[ {\left( {{{\left( {\frac{1}{2}} \right)}^2} - \frac{1}{2}} \right) - \left( {{0^2} - 0} \right)} \right] \\ &= \frac{1}{4} \\\ \end{align}$
$\begin{align} &\int\limits_{\frac{1}{2}}^2 {2x - 1\;dx = \left. {{x^2} - x} \right|} _{\frac{1}{2}}^2 \hfill \\ &= \left( {{2^2} - 2} \right) - \left( {{{\left( {\frac{1}{2}} \right)}^2} - \frac{1}{2}} \right) \hfill \\ &= \frac{9}{4} \hfill \\\\ \end{align}$
${\text{Sehingga}}$
$\begin{align} &\int\limits_0^2 {\left| {2x - 1} \right|} \;dx = \frac{9}{4} + \frac{1}{4} \hfill \\ &= \frac{{10}}{4} \hfill \\ &= \boxed{\frac{5}{2}} \hfill \\ \end{align}$

Soal 4

$\displaystyle\int_{-2}^{3}\left|x^{2}-x-2\right| d x=...$

Pembahasan:

Karena fungsi yang diintegralkan merupakan bentuk mutlak maka kita buat garis bilangannya terlebih dahulu.

$\begin{aligned} x^{2}-x-2 &=0 \\ (x+1)(x-2) &=0 \\ x=-1 \vee x=2 \end{aligned}$

$\begin{aligned} &\int_{-2}^{-1} x^{2}-x-2 d x=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-\left.2 x\right|_{-2} ^{-1}\\ &=\left(\frac{1}{3}(-1)^{3}-\frac{1}{2}(-1)^{2}-2(-1)\right)-\left(\frac{1}{3}(-2)^{3}-\frac{1}{2}(-2)^{2}-2(-2)\right) \\ &=\frac{11}{6}\\ \end{aligned}$

$\begin{aligned} &-\int_{-1}^{2} x^{2}-x-2 d x =-\left[\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-2 x\right]_{-1}^{2} \\ &=-\left[\left(\frac{1}{3} \cdot 2^{3}-\frac{1}{2} \cdot 2^{2}-2\cdot 2\right)-\left(\frac{1}{3}(-1)^{3}-\frac{1}{2}(-1)^{2}-2(-1)\right)\right] \\ &=\frac{9}{2} \end{aligned}$

$\begin{aligned} &\int_{2}^{3} x^{2}-x-2 d x=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-\left.2 x\right|_{2} ^{3}\\ &=\left(\frac{1}{3}(3)^{3}-\frac{1}{2}(3)^{2}-2(3)\right)-\left(\frac{1}{3}(2)^{3}-\frac{1}{2}(2)^{2}-2(2)\right) \\ &=\frac{11}{6}\\ \end{aligned}$

${\text{Sehingga}}$
$\begin{aligned} \int_{-2}^{3}\left|x^{2}-x-2\right| d x &=\frac{11}{6}+\frac{9}{2}+\frac{11}{6} \\ &=\boxed{\frac{49}{6}} \end{aligned}$

Soal 5

Diketahui $\displaystyle\int_{a}^{3}\left(3 x^{2}+2 x+1\right) d x=25$. Nilai $\frac{1}{2} a=\ldots$

Pembahasan:
$\begin{align} x^{3}+x^{2}+\left.x\right|_{a} ^{3}&=25 \\ \left(3^{3}+3^{2}+2\right)-\left(a^{3}+a^{2}+a\right)&=25 \\ a^{3}+a^{2}+a&=14\\ \end{align}$

Dengan coba-coba atau dengan menggunakan teorema horner didapat $a = 2$ sehingga $\boxed{\frac{1}{2}a = 1}$.

Soal 6

Jika $\displaystyle\int_{2}^{0} 2 f(x) d x=10$ dan $\displaystyle\int_{4}^{2} f(x) d x=-4$ maka nilai dari $\displaystyle\int_{0}^{4} f(x) d x=\dots$

Pembahasan:
Kita gunakan sifat integral tentu nomor 2) dan 4)
$\begin{align} &\int\limits_2^0 {2f\left( x \right)\;dx = 10} \to \int\limits_2^0 {f\left( x \right)\;dx = 5} \to \int\limits_0^2 {f\left( x \right)\;dx = - 5} \\\ &\int\limits_4^2 {f\left( x \right)\;dx = - 4 \to } \int\limits_2^4 {f\left( x \right)\;dx} = 4\ \end{align}$ $\begin{align} \int\limits_0^4 {f\left( x \right)\;dx} &= \int\limits_0^2 {f\left( x \right)\;dx + } \int\limits_2^4 {f\left( x \right)\;dx} \hfill \\ &= - 5 + 4 \hfill \\ &= \boxed{ - 1} \hfill \\ \end{align}$

Soal 7

Diketahui $f(x)$ merupakan fungsi genap dengan $\displaystyle\int\limits_{ - 4}^4 {f\left( x \right)\;dx = 8}$ dan $\displaystyle\int\limits_6^4 {f\left( x \right)\;dx = - 2}$. Nilai dari $\displaystyle \int\limits_0^6 {f\left( x \right)\;dx = ...}$

Pembahasan:
Kita gunakan sifat integral tentu nomor 2), 4), dan 5)
$\begin{align} &\int\limits_6^4 {f\left( x \right)\;dx = - 2 \to } \int\limits_4^6 {f\left( x \right)\;dx = 2} \hfill \\ &\int\limits_{ - 4}^4 {f\left( x \right)\;dx = 8 \to } \int\limits_0^4 {f\left( x \right)\;dx = 4} \hfill \\ \end{align}$
$\begin{align} \int\limits_0^6 {f\left( x \right)\;dx} &= \int\limits_0^4 {f\left( x \right)\;dx + } \int\limits_4^6 {f\left( x \right)\;dx} \hfill \\ &= 4 + 2 \hfill \\ &= \boxed6 \hfill \\ \end{align}$

Soal 8

Diketahui $f(x)$ merupakan fungsi ganjil dengan $\displaystyle\int\limits_{ - 5}^0 {f\left( x \right)\;dx = 3}$ dan $\displaystyle\int\limits_5^{10} {f\left( x \right)\;dx = 7}$. Nilai dari $\displaystyle \int\limits_0^10 {f\left( x \right)\;dx = ...}$

Pembahasan:
Kita gunakan sifat integral tentu nomor 4) dan 6)
$\begin{align} &\int\limits_{ - 5}^0 {f\left( x \right)\;dx = 3 \to } \int\limits_0^5 {f\left( x \right)\;dx = - 3} \\\ \end{align}$ $\begin{align} \int\limits_0^{10} {f\left( x \right)\;dx} &= \int\limits_0^5 {f\left( x \right)\;dx + } \int\limits_5^{10} {f\left( x \right)\;dx} \hfill \\ &= - 3 + 7 \hfill \\ &= \boxed4 \hfill \\ \end{align}$

Soal 9

Diketahui $f\left( x \right) = f\left( {x + 3} \right)$ dan $\displaystyle\int\limits_0^3 {f\left( x \right)\;dx = 4}$. Nilai dari $\displaystyle \int\limits_9^{15} {f\left( x \right)\;dx = ...}$

Pembahasan:
Kita gunakan sifat integral tentu nomor 4) dan 7)
$\begin{align} \int\limits_0^3 {f\left( x \right)\;dx = } \int\limits_3^6 {f\left( x \right)\;dx = } \int\limits_6^9 {f\left( x \right)\;dx} = \hfill \\ = \int\limits_9^{12} {f\left( x \right)\;dx = \int\limits_{12}^{15} {f\left( x \right)\;dx = } } 4 \hfill \\ \end{align}$
$\begin{align} {\text{Sehingga}} \hfill \\ \int\limits_9^{15} {f\left( x \right)\;dx} &= \int\limits_9^{12} {f\left( x \right)\;dx} + \int\limits_{12}^{15} {f\left( x \right)\;dx} \hfill \\ &= 4 + 4 \hfill \\ &= \boxed8 \hfill \\ \end{align}$