If $\frac{d}{dx}\{f(x)\} = g(x)$, then $\int_a^b f(x) g(x) dx$ is equal to
- A$\frac{1}{2}\left[f^2(b) - f^2(a)\right]$
- B$\frac{1}{2}\left[g^2(b) - g^2(a)\right]$
- C$f(b) - f(a)$
- D$\frac{1}{2}\left[f(b^2) - f(a^2)\right]$
Explore More
Similar Questions
Let $y = \{x\}^{[x]}$ where $\{x\}$ denotes the fractional part of $x$ and $[x]$ denotes the greatest integer $\le x$. Then $\int_{0}^{3} y \, dx = $
$\int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $
Easy
View Solution$\int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx =$
If $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$,then find $f'(9)$.
$\int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx = } $
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo