The value of ∫limits_0^1 4 x^3≤ft(d^2)/(d x^2)≤ft(1-x^2)^5 d x, is

Let $I=\int\limits_0^1 4 x^3\left\{\frac{d^2}{d x^2}\left(1-x^2\right)^5\right\} d x$

$I=\left[4 x^3 \frac{d}{d x}\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 12 x^2\left\{\frac{d}{d x}\left(1-x^2\right)^5\right\} d x$

$\Rightarrow I=\left[4 x^3 \times(-10 x)\left(1-x^2\right)^4\right]_0^1 - 12\left[\left[x^2\left(1-x^2\right)^5\right]_0^1-\int\limits_0^1 2 x\left(1-x^2\right)^5 d x\right]$

$\Rightarrow I=0-12\left[0+\int\limits_0^1\left(1-x^2\right)^5(-2 x) d x\right]$

$\Rightarrow I=-12\left[\frac{\left(1-x^2\right)^6}{6}\right]_0^1=-12\left(0-\frac{1}{6}\right)=2$