JEE Advanced 2014 · Paper 1, Mathematics question 55

Integrate by parts twice. Let $y=(1-x^2)^5$, so $y'=-10x(1-x^2)^4$. Both $y$ and $y'$ vanish at $x=1$, and $x^3$, $x^2$ vanish at 0.
$$\int_0^14x^3y''\,dx=\Big[4x^3y'\Big]_0^1-\int_0^112x^2y'\,dx=-\Big[12x^2y\Big]_0^1+\int_0^124xy\,dx.$$

Last integral.
$$24\int_0^1x(1-x^2)^5dx=24\cdot\frac1{12}=2.$$

Answer: 2