$ \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x $ ની કિંમત શોધો.

  • A
    $ 10 $
  • B
    $ 00 $
  • C
    $ 08 $
  • D
    $ 03 $

Explore More

Similar Questions

સંકલન $\frac{48}{\pi^{4}} \int_{0}^{\pi} \left(\frac{3 \pi x^{2}}{2} - x^{3}\right) \frac{\sin x}{1 + \cos^{2} x} dx$ નું મૂલ્ય કેટલું થાય?

$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx = \dots$

$\int_0^\infty \frac{\log(1 + x^2)}{1 + x^2} \,dx = $

Difficult
View Solution

ધારો કે $g_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}, i=1, 2$,અને $f: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}$ એવા વિધેયો છે કે જેથી $g_1(x)=1, g_2(x)=|4x-\pi|$ અને $f(x)=\sin^2 x$,દરેક $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$ માટે.
$S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, i=1, 2$ વ્યાખ્યાયિત કરો.
$(1)$ $\frac{16S_1}{\pi}$ નું મૂલ્ય.
$(2)$ $\frac{48S_2}{\pi^2}$ નું મૂલ્ય.

જો $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$ હોય,તો $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo