int_{0}^{4/pi} left( 3x^2 sin frac{1}{x} - x | vedclass

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

Question

(C) Let $I = \int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$. Consider the integral $\int 3x^2 \sin \frac{1}{x} dx$. Us

Vedclass · Accepted Answer

$\frac{32\sqrt{2}}{\pi^3}$

Vedclass · Answer

(C) Let $I = \int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$. Consider the integral $\int 3x^2 \sin \frac{1}{x} dx$. Using integration by parts,let $u = \sin \frac{1}{x}$ and $dv = 3x^2 dx$. Then $du = -\cos \frac{1}{x} \cdot (-\frac{1}{x^2}) dx = \frac{1}{x^2} \cos \frac{1}{x} dx$ and $v = x^3$. Thus,$\int 3x^2 \sin \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x^3 \cdot \frac{1}{x^2} \cos \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx$. Substituting this into the original integral: $I = \left[ x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} (-x \cos \frac{1}{x}) dx$. $I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} x \cos \frac{1}{x} dx + \int_{0}^{4/\pi} x \cos \frac{1}{x} dx$. $I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} = (4/\pi)^3 \sin(\pi/4) - 0 = \frac{64}{\pi^3} \cdot \frac{1}{\sqrt{2}} = \frac{64}{\sqrt{2} \pi^3} = \frac{32 \cdot 2}{\sqrt{2} \pi^3} = \frac{32\sqrt{2}}{\pi^3}$.

$\int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$ has the value:

Similar Questions

$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

Consider the integral $I = \int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} dx$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule,$\int_0^2 f(x) dx$ is equal to

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) d x=$,where $[.]$ denotes the Greatest Integer Function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module