JEE Advanced 2014 Paper 1Question 1
The value of $$\int_0^1 4x^3\left\{\frac{d^2}{dx^2}(1-x^2)^5\right\}dx$$ is
Definite Integration
Answer$2$
Solution
Concept. Integrate by parts twice, moving both derivatives from $(1-x^2)^5$ onto $4x^3$. The boundary terms vanish because $x=0$ kills the power of $x$ and $x=1$ kills $(1-x^2)$. Formulas used. (1) $\int_0^1u\,v'\,dx=\big[uv\big]_0^1-\int_0^1u'v\,dx$(2) $\int_0^12x(1-x^2)^5dx=\Big[-\frac{(1-x^2)^6}{6}\Big]_0^1=\frac16$Step 1 - First by parts. Let $g=(1-x^2)^5$, $g'=-10x(1-x^2)^4$. By (1), $$I=\big[4x^3g'\big]_0^1-\int_0^112x^2g'\,dx=-\int_0^112x^2g'\,dx,$$ since $g'(1)=0$ and $x^3=0$ at $0$. Step 2 - Second by parts. $$-\int_0^112x^2g'\,dx=-\big[12x^2g\big]_0^1+\int_0^124x\,g\,dx=\int_0^124x(1-x^2)^5dx.$$ Step 3 - Value. By (2), $I=12\cdot\frac16=2$. Answer: 2