The value of int_0^1 4 x^3 left{ frac{d^2}{d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_0^1 4 x^3 \frac{d^2}{d x^2} (1-x^2)^5 d x$.Using integration by parts,let $u = 4x^3$ and $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.Then $d

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $I = \int_0^1 4 x^3 \frac{d^2}{d x^2} (1-x^2)^5 d x$.Using integration by parts,let $u = 4x^3$ and $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.Then $du = 12x^2 dx$ and $v = \frac{d}{dx}(1-x^2)^5 = 5(1-x^2)^4(-2x) = -10x(1-x^2)^4$.$I = [4x^3 \cdot (-10x(1-x^2)^4)]_0^1 - \int_0^1 (-10x(1-x^2)^4) \cdot 12x^2 dx$.The boundary term $[ -40x^4(1-x^2)^4 ]_0^1 = 0 - 0 = 0$.So,$I = 120 \int_0^1 x^3(1-x^2)^4 dx$.Let $t = 1-x^2$,then $dt = -2x dx$,so $x^2 = 1-t$ and $x dx = -\frac{1}{2} dt$.$I = 120 \int_1^0 (1-t) t^4 (-\frac{1}{2} dt) = 60 \int_0^1 (t^4 - t^5) dt$.$I = 60 [\frac{t^5}{5} - \frac{t^6}{6}]_0^1 = 60 (\frac{1}{5} - \frac{1}{6}) = 60 (\frac{6-5}{30}) = 60 (\frac{1}{30}) = 2$.

The value of $\int_0^1 4 x^3 \left\{ \frac{d^2}{d x^2} (1-x^2)^5 \right\} d x$ is

Similar Questions

If $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$,then $a$ and $b$ are respectively

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi \,d \phi$.

$\int_0^1 {{\cos }^{ - 1}}x\,dx = $

If $[x]$ is the greatest integer function not greater than $x$,then $\int_0^8 [x] dx$ is equal to

The positive value of $x$ satisfying the equation $\int_x^1(1-t) dt = \frac{1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module