int_0^pi x sin x , dx = | vedclass

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

Question

(A) ધારો કે $I = \int_0^\pi x \sin x \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $I = \int_0^\pi (\pi -

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) ધારો કે $I = \int_0^\pi x \sin x \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $I = \int_0^\pi (\pi - x) \sin(\pi - x) \, dx$ કારણ કે $\sin(\pi - x) = \sin x$,તેથી: $I = \int_0^\pi (\pi - x) \sin x \, dx = \pi \int_0^\pi \sin x \, dx - \int_0^\pi x \sin x \, dx$ $I = \pi \int_0^\pi \sin x \, dx - I$ $2I = \pi \int_0^\pi \sin x \, dx$ $2I = \pi [-\cos x]_0^\pi$ $2I = \pi [-\cos \pi - (-\cos 0)]$ $2I = \pi [-(-1) - (-1)] = \pi [1 + 1] = 2\pi$ $I = \pi$.

$\int_0^\pi x \sin x \, dx = $

Similar Questions

જો $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$ હોય,તો $27 I^2$ ની કિંમત . . . . . . . . થાય.

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

$\int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = $

$\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ ની કિંમત શોધો.

$\int_0^1 \tan^{-1}(1-x+x^2) dx$ નું મૂલ્ય શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module