Question

If $I=\stackrel{1}{\underset{0}{\int }}x(1-x{)}^{1/2}dx$ and $60I+k=25$ then $k=$ _________. $(k\in R)$.

• A. $9$
• B. $25$
• C. $60$
• D. $41$

Answer 1

Answer: (A)

$9$

$I=\stackrel{1}{\underset{0}{\int }}x(1-x{)}^{\frac{1}{2}}dx$
Taking $=(1-x{)}^{\frac{1}{2}}=t$

$⟹1-x=t^2$

$⟹x=1-t^2$
$\therefore dx=-2t$ dt
For x$=0$ we have $t=1$
And for $x=1$ we have $t=0$

$={\int }_1^0(1-t^2)\left(t\right)(-2t)dt$

$=-{\int }_1^{0}2t^2(1-t^2)dt$

$={\int }_0^{1}2t^2(1-t^2)dt\left({\int }_a^{b}f\right(x)=-{\int }_b^{a}f(x\left)\right)$

$={\int }_0^1(2t^2-2t^4)dt$

$={[\frac{2}{3}{t}^3-\frac{2}{5}{t}^5]}_0^1$

$=\frac{2}{3}-\frac{2}{5}=\frac{4}{15}$

$\therefore I=\frac{4}{15}$

and,$60I+k=25⟹60\times \frac{4}{15}+k=25$

$16+k=25⟹k=9$