Question

Evaluate the integral

Answer 1

Topic:

Integration

Solution:

We have to evaluate the given definite integral.

$\displaystyle\sf I = \int_0^2x^2\sqrt{4 - x^2}\, dx$

$\rm Put\ x = 2\sin(\theta)\implies dx = 2\cos(\theta)\ d\theta$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}(2\sin\theta)^2\sqrt{4 - (2\sin\theta)^2} \ 2\cos\theta\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}4\sin^2\theta\sqrt{4(1- \sin^2\theta)} \ 2\cos\theta\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}4\sin^2\theta\sqrt{4(\cos^2\theta)} \ 2\cos\theta\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}4\sin^2(\theta)\ 2\cos(\theta) \ 2\cos(\theta)\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}16\Big(\sin\theta\cos\theta\Big)^2\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}16\left(\dfrac{2\sin\theta\cos\theta}{2}\right)^2\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}16\left(\dfrac{\sin2\theta}{2}\right)^2\ d\theta\qquad(\because\ 2\sin x\cos x = \sin2x)}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}4\sin^2(2\theta)\ d\theta}$

${\displaystyle\implies \sf I = \int_0^{\frac\pi2}2(1 - \cos4\theta)\ d\theta\qquad\Big(\because \ 2\sin^2(x) = 1 - \cos(2x)\Big)}$

${\displaystyle\implies \sf I = 2\int_0^{\frac\pi2}d\theta - 2\int_0^{\frac\pi2} \cos4\theta\ d\theta}$

${\displaystyle\implies \sf I = 2\theta\bigg|^{\frac\pi2}_0 - \dfrac12\sin4\theta\bigg|^{\frac\pi2}_0}$

$\displaystyle\implies \sf I = \pi -0$

${\displaystyle\implies \sf I = \pi }$

Hence the required answer is,

$\blue{\boxed{\tt \int_0^2x^2\sqrt{4 - x^2}\, dx = \pi}}$

More to know:

$\sf\circ\quad\int x^n\ dx = \begin{cases}\sf\dfrac{x^{n+1}}{n+1} + C, if\ n\not = -1\\\\\sf\log|x| + C, if \ n = -1\end{cases}$

$\sf\circ\quad\int e^x \ dx = e^x + C$

$\sf\circ\quad\int \log(x) \ dx = x\log(x) -x +C$

$\sf\circ\quad\int \cos(x) \ dx = \sin(x) + C$

$\sf\circ\quad\int \sin(x) \ dx = -\cos(x) + C$

$\sf\circ\quad\int \sin(ax + b) \ dx = -\dfrac{\cos(x)}{a} + C$

$\sf\circ\quad\int \cos(ax + b) \ dx = \dfrac{\sin(x)}{a} + C$