The value of $\int_{0}^{1} 4x^{3}\left\{\frac{d^{2}}{dx^{2}}\left(1-x^{2}\right)^{5}\right\}dx$ is

The value of ∫₀¹ 4x³{d²/dx²(1-x²)⁵}dx is

$I = \displaystyle\int_{0}^{1} 4x^{3}\;\dfrac{d^{2}}{dx^{2}}\!\left[(1-x^{2})^{5}\right]dx$

$\text{To find: } I$

$g(x) = (1-x^2)^5, \quad g(1)=0,\ g(0)=1$

$g'(x) = -10x(1-x^2)^4, \quad g'(1)=0,\ g'(0)=0$

$I = \Big[4x^3 g'\Big]_{0}^{1} - \int_{0}^{1} 12x^2 g'\,dx$

$I = \Big[4x^3 g'\Big]_{0}^{1} - \Big[12x^2 g\Big]_{0}^{1} + \int_{0}^{1} 24x\,g\,dx$

$g(1)=g'(1)=0; \quad x^3=x^2=0 \text{ at } x=0$

$\Rightarrow \Big[4x^3 g'\Big]_{0}^{1} = \Big[12x^2 g\Big]_{0}^{1} = 0$

$I = \int_{0}^{1} 24x(1-x^2)^5\,dx$

$u = 1-x^2 \;\Rightarrow\; x\,dx = -\dfrac{du}{2}$

$x=0\Rightarrow u=1, \quad x=1\Rightarrow u=0$

$I = 24\cdot\dfrac{1}{2}\int_{0}^{1} u^{5}\,du = 12\left[\dfrac{u^{6}}{6}\right]_{0}^{1}$

$I = 12\cdot\dfrac{1}{6} = 2$

$\boxed{I = 2}$