MCQ
$\int_0^a {x{{(2ax - {x^2})}^{\frac{3}{2}}}\,dx = } $
- A${a^5}\left[ {\frac{{3\pi }}{{16}} - 1} \right]$
- B${a^5}\left[ {\frac{{3\pi }}{{16}} + 1} \right]$
- ✓${a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$
- Dએકપણ નહીં.
$\Rightarrow dx = 2a\sin 2\theta \,d\theta $
Therefore, $\int_0^a {x{{(2ax - {x^2})}^{3/2}}dx} $
$ = \int_0^{\pi /4} {2{a^5}(1 - \cos 2\theta ){{\sin }^4}2\theta \,\,d\theta } $
Now again, put $2\theta = \phi $
$ = {a^5}\left[ {\int_0^{\pi /2} {{{\sin }^4}\phi \,d\phi } - \int_0^{\pi /2} {{{\sin }^4}\phi \cos \phi \,d\phi } } \right]$
$ = {a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
કારણ $(R) : \vec b $ એ $\vec c $ ને લંબ હોય , તો $\vec b \, \times \,\vec c \,\, = \,\,0$