Question

The value of ${\int }_0^{1}4x^3\left\{\frac{d^2}{d{x}^2}\right(1-x^2{)}^5\}dx$ is?

Answer 1

Concept: Integration by parts

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx=\int \left(\frac{d}{dx}\right[f\left(x\right)]\int g(x\left)dx\right)dx$

${\int }_0^{1}4\underset{I}{x^3}\frac{d^2}{d{x}^2}(1-\underset{II}{x^2}{)}^{5}dx$

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

$={[4x^3\times 5(1-x^2{)}^4(-2x)]}_0^1-12[{\left[x^2\right(1-x^2{)}^5]}_0^1-{\int }_0^{1}2x(1-x^2{)}^{5}dx]$

$=0-0-12[0-0]+12{\int }_0^{1}2x(1-x^2{)}^{5}dx$

$=12\times {[-\frac{(1-x^2{)}^6}{6}]}_0^1$

$=12[0+\frac{1}{6}]=2$