Answer:
Option A
Explanation:
Plan Intergration by parts
$\int_{}^{} f(x)g(x)dx=f(x)\int_{}^{} g(x) dx-\int_{}^{}\left(\frac{d}{dx} [f(x)]\int_{}^{} g(x) dx\right) dx$
Given $I= \int_{0}^{1} 4x^{3}\frac{d^{2}}{dx^{2}}(1-x^{2})^{5}dx$
$=\left[4x^{3}\frac{d}{dx}(1-x^{2})^{5}\right]^{1}_{0}-\int_{0}^{1}12x^{2} \frac{d}{dx}(1-x^{2})^{5}dx$
$=\left[4x^{3}\times 5(1-x^{2})^{4}(-2x)\right]^{1}_{0}$
$-12 \left[(x^{2}(1-x^{2})^{5}\right]_0^1-\int_{0}^{1}2x(1-x^{2})^{5} dx]$
=0-0-12(0-0)
+ $12\int_{0}^{1} 2x(1-x^{2})^{5}dx$
= $12\times\left[-\frac{(1-x^{2})^{6}}{6}\right]_0^1$
= $12\left[0+\frac{1}{6}\right]=2$
