Integral Tertentu

Jika f kontinu pada interval [a, b] dan F adalah antiturunan dari f pada interval tersebut, maka :

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\mathrm{\int_{a}^{b}f(x)\;dx=\left [ F(x) \right ]_{a}^{b}=F(b)-F(a)}$
Contoh :
1.

\int_{1}^{3} (2 x - 1) d x

$\mathrm{\int_{1}^{3}(2x-1)\;dx}$

= {[x^{2} - x]}_{1}^{3}

$\mathrm{=\left [ x^{2}-x \right ]_{1}^{3}}$

= (3^{2} - 3) - (1^{2} - 1)

$\mathrm{=\left ( 3^{2}-3 \right )-\left ( 1^{2}-1 \right )}$

= 6 - 0

$=6-0$

= 6

$=6$

2.

\int_{0}^{1} (x^{2} - 3 x) d x

$\mathrm{\int_{0}^{1}\left ( x^{2}-3x \right )\;dx}$

= {[\frac{1}{3} x^{3} - \frac{3}{2} x^{2}]}_{0}^{1}

$\mathrm{=\left [ \frac{1}{3}x^{3}-\frac{3}{2}x^{2} \right ]_{0}^{1}}$

= (\frac{1}{3} (1)^{3} - \frac{3}{2} (1)^{2}) - (\frac{1}{3} (0)^{3} - \frac{3}{2} (0)^{2})

$\mathrm{=\left (\frac{1}{3}(1)^{3}-\frac{3}{2}(1)^{2} \right )-\left (\frac{1}{3}(0)^{3}-\frac{3}{2}(0)^{2} \right )}$

= \frac{1}{3} - \frac{3}{2}

$=\frac{1}{3}-\frac{3}{2}$

= - \frac{7}{6}

$=-\frac{7}{6}$

Sifat-Sifat Integral Tertentu

\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x

$\mathrm{\int_{a}^{b}k\:f(x)\;dx=k\int_{a}^{b}f(x)\:dx}$

2.

\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x

$\mathrm{\int_{a}^{b}\left [f(x)\pm g(x) \right ]\;dx=\int_{a}^{b}f(x)\:dx\pm \int_{a}^{b}g(x)\:dx}$

3.

\int_{a}^{a} f (x) d x = 0

$\mathrm{\int_{a}^{a}f(x)\;dx=0}$

4.

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x

$\mathrm{\int_{a}^{b}f(x)\;dx=-\int_{b}^{a}f(x)\;dx}$

5.

\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

$\mathrm{\int_{a}^{c}f(x)\;dx=\int_{a}^{b}f(x)\;dx+\int_{b}^{c}f(x)\;dx}$
dengan : a < b < c

Berikut contoh-contoh latihan soal integral tertentu :

Contoh 1
Jika

f (x) = 3 x^{2} - 4 x

$\mathrm{f(x)=3x^{2}-4x}$ dan

g (x) = 2 x + 1

$\mathrm{g(x)=2x+1}$ , tentukan nilai dari :

a.

\int_{1}^{3} f (x) d x

$\mathrm{\int_{1}^{3}f(x)\:dx}$

Jawab :

\Rightarrow \int_{1}^{3} (3 x^{2} - 4 x) d x

$\mathrm{\Rightarrow \int_{1}^{3}\left ( 3x^{2}-4x \right )\:dx}$

= {[x^{3} - 2 x^{2}]}_{1}^{3}

$\mathrm{= \left [x^{3}-2x^{2} \right ]_{1}^{3}}$

= ((3)^{3} - 2 (3)^{2}) - ((1)^{3} - 2 (1)^{2})

$\mathrm{= \left ((3)^{3}-2(3)^{2} \right )-\left ((1)^{3}-2(1)^{2} \right )}$

= 9 - (- 1)

$=9-(-1)$

= 10

$\mathrm{=10}$

b.

\int_{- 2}^{1} g (x) d x

$\mathrm{\int_{-2}^{1}g(x)\:dx}$

Jawab :

\Rightarrow \int_{- 2}^{1} (2 x + 1) d x

$\mathrm{\Rightarrow\int_{-2}^{1}(2x+1)\:dx}$

= {[x^{2} + x]}_{- 2}^{1}

$\mathrm{=\left [x^{2}+x \right ]_{-2}^{1}}$

= (1^{2} + 1) - ((- 2)^{2} + (- 2))

$\mathrm{=\left ( 1^{2}+1 \right )-\left ( (-2)^{2}+(-2) \right )}$

= 2 - 2

$=2-2$

= 0

$=0$

c.

\int_{0}^{1} (f (x) - g (x)) d x

$\mathrm{\int_{0}^{1}\left (f(x)-g(x) \right )\:dx}$

Jawab :

\Rightarrow \int_{0}^{1} ((3 x^{2} - 4 x) - (2 x + 1)) d x

$\mathrm{\Rightarrow \int_{0}^{1}\left (\left ( 3x^{2}-4x \right )-(2x+1) \right )\:dx}$

\Rightarrow \int_{0}^{1} (3 x^{2} - 6 x - 1) d x

$\mathrm{\Rightarrow \int_{0}^{1}\left (3x^{2}-6x-1 \right )\:dx}$

= {[x^{3} - 3 x^{2} - x]}_{0}^{1}

$\mathrm{=\left [x^{3}-3x^{2}-x \right ]_{0}^{1}}$

= (1^{3} - 3 (1)^{2} - 1) - (0^{3} - 3 (0)^{2} - 0)

$\mathrm{=\left (1^{3}-3(1)^{2}-1 \right )-\left ( 0^{3}-3(0)^{2}-0 \right )}$

= - 3 - 0

$=-3-0$

= - 3

$=-3$

Contoh 2
Tentukan nilai a dari pengintegralan berikut !

a.

\int_{a}^{a + 1} (2 x + 1) d x = 4

$\mathrm{\int_{a}^{a+1}\left ( 2x+1 \right )\;dx=4}$

Jawab :

\Rightarrow {[x^{2} + x]}_{a}^{a + 1} = 4

$\mathrm{\Rightarrow \left [ x^{2}+x \right ]_{a}^{a+1}=4}$

\Rightarrow [{(a + 1)}^{2} + (a + 1)] - [a^{2} + a] = 4

$\mathrm{\Rightarrow \left [ \left ( a+1 \right )^{2}+\left ( a+1 \right ) \right ]-\left [ a^{2}+a \right ]=4}$

\Rightarrow a^{2} + 2 a + 1 + a + 1 - a^{2} - a = 4

$\mathrm{\Rightarrow a^{2}+2a+1+a+1-a^{2}-a=4}$

\Rightarrow 2 a + 2 = 4

$\mathrm{\Rightarrow 2a+2=4}$

\Rightarrow 2 a = 2

$\mathrm{\Rightarrow 2a=2}$

\Rightarrow a = 1

$\mathrm{\Rightarrow a=1}$

b.

\int_{1}^{a} (2 x - 2) d x = 2 a + 1; a > 0

$\mathrm{\int_{1}^{a}\left ( 2x-2 \right )\;dx=2a+1\;\;;\,a>0}$

Jawab :

\Rightarrow {[x^{2} - 2 x]}_{1}^{a} = 2 a + 1

$\mathrm{\Rightarrow\left [ x^{2}-2x \right ]_{1}^{a}=2a+1}$

\Rightarrow (a^{2} - 2 a) - (1^{2} - 2.1) = 2 a + 1

$\mathrm{\Rightarrow \left ( a^{2}-2a \right )-\left ( 1^{2}-2.1 \right )=2a+1}$

\Rightarrow a^{2} - 2 a + 1 = 2 a + 1

$\mathrm{\Rightarrow a^{2}-2a+1=2a+1}$

\Rightarrow a^{2} - 4 a = 0

$\mathrm{\Rightarrow a^{2}-4a=0}$

\Rightarrow a (a - 4) = 0

$\mathrm{\Rightarrow a\left ( a-4 \right )=0}$

\Rightarrow a = 0 a t a u a = 4

$\mathrm{\Rightarrow a=0\;atau\;a=4}$
Karena a > 0, maka

a = 4

$\mathrm{a=4}$

c.

\int_{a}^{1} (3 x^{2} + 2 x) d x = - 10; a \in R

$\mathrm{\int_{a}^{1}\left ( 3x^{2}+2x \right )\;dx=-10\;\;;\,a\in R}$

Jawab ;

\Rightarrow {[x^{3} + x^{2}]}_{a}^{1} = - 10

$\mathrm{\Rightarrow \left [ x^{3}+x^{2} \right ]_{a}^{1}=-10}$

\Rightarrow (1^{3} + 1^{2}) - (a^{3} + a^{2}) = - 10

$\mathrm{\Rightarrow \left (1^{3}+1^{2} \right )-\left ( a^{3}+a^{2} \right )=-10}$

\Rightarrow 2 - a^{3} - a^{2} = - 10

$\mathrm{\Rightarrow 2-a^{3}-a^{2}=-10}$

\Rightarrow a^{3} + a^{2} - 12 = 0

$\mathrm{\Rightarrow a^{3}+a^{2}-12=0}$

Nilai a yang mungkin adalah faktor dari 12, yaitu : ±1, ±2, ±3, ±4, ±6, ±12.
Nilai a yang memenuhi adalah a = 2, karena :

\Rightarrow 2^{3} + 2^{2} - 12 = 0

$\mathrm{\Rightarrow 2^{3}+2^{2}-12=0}$

Contoh 3
Nilai dari

\int_{0}^{\frac{π}{2}} (2 c o s 2 x - s i n x) d x

$\mathrm{\int_{0}^{\frac{\pi }{2}}\left (2\,cos\:2x-sin\,x \right )\;dx}$ adalah...

Jawab :

\int_{0}^{\frac{π}{2}} (2 c o s 2 x - s i n x) d x

$\mathrm{\int_{0}^{\frac{\pi }{2}}\left (2\,cos\:2x-sin\,x \right )\;dx}$

= {[2 \frac{1}{2} s i n 2 x - (- c o s x)]}_{0}^{\frac{π}{2}}

$\mathrm{=\left [ 2\,\frac{1}{2}sin\,2x-(-cos\,x) \right ]_{0}^{\frac{\pi }{2}}}$

= {[s i n 2 x + c o s x]}_{0}^{\frac{π}{2}}

$\mathrm{=\left [ sin\,2x+cos\,x \right ]_{0}^{\frac{\pi }{2}}}$

= (s i n 2 (\frac{π}{2}) + c o s (\frac{π}{2})) - (s i n 2 (0) + c o s 0)

$\mathrm{=\left ( sin\,2\left ( \frac{\pi }{2} \right )+cos\,\left ( \frac{\pi }{2} \right ) \right )-\left ( sin\,2(0)+cos\,0 \right )}$

= (s i n π + c o s (\frac{π}{2})) - (s i n 0 + c o s 0)

$\mathrm{=\left ( sin\,\pi+cos\left ( \frac{\pi }{2} \right ) \right )-\left ( sin\,0+cos\,0 \right )}$

= (0 + 0) - (0 + 1)

$=\left ( 0+0 \right )-\left ( 0+1 \right )$

= - 1

$=-1$