The value of ∫014x^3(d^2)/(dx^2)1-x^25dx is ____________

Correct answer: 2. ∫014x^3⏟d^2dx^2 1-x2^5⏟dx =4x^3d/dx 1-x^2501- ∫0112x^2d/dx 1-x^25 dx =4x^3 ×51-x^24 -2x01-12x^21-x^2501- ∫012x1-x^25dx =0-0-120-0+12 ∫012x1-x^25 dx =12 ×-(1-x^2)/6601 =120+1/6=2

Q37 / 60 · JEE Advanced 2014 (Paper 1) · Mathematics

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

Correct answer: 2

Solution

\int_{0}^{1} \underset{⏟}{4 x^{3}} \underset{⏟}{\frac{d^{2}}{d x^{2}} {(1 - x^{2})}^{5}} d x

= {[4 x^{3} \frac{d}{d x} {(1 - x^{2})}^{5}]}_{0}^{1} - \int_{0}^{1} 12 x^{2} \frac{d}{d x} {(1 - x^{2})}^{5} d x

= {[4 x^{3} \times 5 {(1 - x^{2})}^{4} (- 2 x)]}_{0}^{1} - 12 [{[x^{2} {(1 - x^{2})}^{5}]}_{0}^{1} - \int_{0}^{1} 2 x {(1 - x^{2})}^{5} d x]

= 0 - 0 - 12 [0 - 0] + 12 \int_{0}^{1} 2 x {(1 - x^{2})}^{5} d x

= 12 \times {[- \frac{{(1 - x^{2})}^{6}}{6}]}_{0}^{1}

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

Hence, the correct answer is 2.

The value of ∫014x3d2dx21-x25dx is ____________

Solution

More Definite Integration Questions

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