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

On the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,the greatest value of the function $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$ is:

If $I$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$, then

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

$ \int_{-5}^{5} |x+2| \, dx $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module