MCQ
$\int_0^a {{x^2}{{({a^2} - {x^2})}^{3/2}}dx = } $
  • $\frac{{\pi {a^6}}}{{32}}$
  • B
    $\frac{{2{a^5}}}{{15}}$
  • C
    $\frac{{{a^6}}}{{32}}$
  • D
    None of these

Answer

Correct option: A.
$\frac{{\pi {a^6}}}{{32}}$
a
(a) $I = \int_0^a {{x^2}{{({a^2} - {x^2})}^{3/2}}dx} $

Put $x = a\sin \theta \Rightarrow dx = a\cos \theta \,d\theta $

$I = \int_0^{\pi /2} {{a^2}{{\sin }^2}\theta .{a^3}{{\cos }^3}\theta .a\cos \theta \,d\theta } $

$ = {a^6}\int_0^{\pi /2} {{{\sin }^2}\theta {{\cos }^4}\theta \,d\theta= {a^6}\frac{{\Gamma \frac{3}{2}.\,\Gamma \frac{5}{2}}}{{2.\Gamma \frac{8}{2}}}} $

$= {a^6}\frac{{\frac{1}{2}.\sqrt \pi .\frac{3}{2}.\frac{1}{2}.\sqrt \pi }}{{2.3.2.1}} $

$= \frac{{\pi {a^6}}}{{32}}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$ is equal to:
$\frac{\text{d}}{\text{dx}}\Big\{\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)\Big\}$ equals:
The maximum possible area bounded by the parabola $y=x^2+x+10$ and a chord of the parabola of length $1$ is
If $ D, E, F$  be the middle points of the sides $BC, CA $ and  $ AB$  of the triangle $\overrightarrow {AD}  + \overrightarrow {BE}  + \overrightarrow {CF} $ is
If $\mathrm{a, b, c},$ are in $\mathrm{A.P}$, then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is

The coordinates of the point on the ellipse $16 x^2+9 y^2=400$ where the ordinate decreases at the same rate at which the abscissa increases, are :
The degree of the differential equation $\text{y}_3^\frac{2}{3}+2+3\text{y}_2+\text{y}_1=0$ is:
If A is a singular matrix, then adj A is.
The solution of the differential equation $\cos y\log (\sec x + \tan x)dx = \cos x\log (\sec y + \tan y)dy$ is
Let $|M|$ denote the determinant of a square matrix $M$. Let $g:\left[0, \frac{\pi}{2}\right] \rightarrow R$ be the function defined by

$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$

where

$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.

Let $p (x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?

$(A)$ $p \left(\frac{3+\sqrt{2}}{4}\right)<0$

$(B)$ $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$

$(C)$ $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$

$(D)$ $p \left(\frac{5-\sqrt{2}}{4}\right)<0$