Evaluate the definite integral int_{0}^{frac{ | vedclass

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

Question

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$. We know that the antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$. By the Second Fundamental T

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$. We know that the antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$. By the Second Fundamental Theorem of Calculus,we have: $I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$ $I = \frac{1}{2} \left[ \sin(2 \cdot \frac{\pi}{2}) - \sin(2 \cdot 0) \right]$ $I = \frac{1}{2} [\sin \pi - \sin 0]$ Since $\sin \pi = 0$ and $\sin 0 = 0$,we get: $I = \frac{1}{2} [0 - 0] = 0$.

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$.

Similar Questions

$\int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx =$

If $f(x) = \begin{cases} \sqrt{1-x} & 0 \leqslant x \leqslant 1 \\ (7x-6)^{-1/3} & 1 < x \leqslant 2 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is equal to

$\int_{0}^{\pi} \sin x \, dx = $ . . . . . . .

Evaluate the definite integral $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module