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

$\int_0^{\pi / 2} \frac{\sin ^3 x \cos x \, dx}{\sin ^4 x+\cos ^4 x} = ?$

If $f(x) = \cos(\tan^{-1}x)$,then the value of the integral $\int_{0}^{1} x f''(x) dx$ is

The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is equal to

Let $[\bullet]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$, then $\frac{1}{\pi} \int_{0}^{\alpha\pi} \left( \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \right) d\theta$ is equal to . . . . . . .

$\int_{-1/2}^{1/2} (\cos x) \left[ \log \left( \frac{1-x}{1+x} \right) \right] dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module