Evaluate the definite integral: int_0^{2a} f( | vedclass

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

Question

(B) We are given the integral $I = \int_0^{2a} f(x) dx$. Using the property of definite integrals, we can split the interval: $I = \int_0^a f(x) dx +

Vedclass · Accepted Answer

$\int_0^a (f(x) + f(2a - x)) dx$

Vedclass · Answer

(B) We are given the integral $I = \int_0^{2a} f(x) dx$. Using the property of definite integrals, we can split the interval: $I = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$. In the second integral, let $x = 2a - t$. Then $dx = -dt$. When $x = a$, $t = a$, and when $x = 2a$, $t = 0$. Substituting these into the second integral: $\int_a^{2a} f(x) dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) dt = \int_0^a f(2a - x) dx$. Thus, $I = \int_0^a f(x) dx + \int_0^a f(2a - x) dx = \int_0^a (f(x) + f(2a - x)) dx$. Comparing this with the given options, option $(b)$ is the correct representation.

Evaluate the definite integral: $\int_0^{2a} f(x) dx$

Similar Questions

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\pi} \log (1+\cos x) d x$.

The value of $\int_0^1 \tan^{-1}(1-x+x^2) dx$ is

If $f(x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g(x) = \ln x$ for $x > 0$,then the value of the integral $\int_{-\pi/4}^{\pi/4} g(f(x)) dx$ is

$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x=$

If $a > 0$, then $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module