Question
Evaluate the following integral:
(i) $\int_0^{\pi / 2} \sin x d x$
(ii) $\int_1^5 x d x$
(i) $\int_0^{\pi / 2} \sin x d x$
(ii) $\int_1^5 x d x$
✓
Answer
Using $\int_a^b f(x) d x=\left.F(x)\right|_a ^b$
$\therefore \quad \int_0^{\frac{\pi}{2}} \sin x d x=-\left.\cos x\right|_0 ^{z / 2}=-\left[\cos \left(\frac{\pi}{2}\right)-\cos 0\right]$
Since, $\cos \left(\frac{\pi}{2}\right)=0$ and $\cos 0=1$
$\int_0^{\frac{\pi}{2}} \sin x d x=-(0-1)=1$
(ii) $\int_1^5 x d x$
Solution:
Using, $\int_a^b f(x) d x=\left.F(x)\right|_a ^b$
$\int_1^5 x d x=\left.\frac{x^2}{2}\right|_1 ^5$
$=\frac{5^2}{2}-\frac{1^2}{2}$
$=\frac{25-1}{2}$
$=\frac{24}{2}$
$=12$
$\therefore \quad \int_0^{\frac{\pi}{2}} \sin x d x=-\left.\cos x\right|_0 ^{z / 2}=-\left[\cos \left(\frac{\pi}{2}\right)-\cos 0\right]$
Since, $\cos \left(\frac{\pi}{2}\right)=0$ and $\cos 0=1$
$\int_0^{\frac{\pi}{2}} \sin x d x=-(0-1)=1$
(ii) $\int_1^5 x d x$
Solution:
Using, $\int_a^b f(x) d x=\left.F(x)\right|_a ^b$
$\int_1^5 x d x=\left.\frac{x^2}{2}\right|_1 ^5$
$=\frac{5^2}{2}-\frac{1^2}{2}$
$=\frac{25-1}{2}$
$=\frac{24}{2}$
$=12$
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