If int_0^{2a} f(x) , dx = 2 int_0^a f(x) , dx | vedclass

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

Question

(B) We are given the property $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$. By substituting $x = 2a - t$ in the second inte

Vedclass · Accepted Answer

$f(2a - x) = f(x)$

Vedclass · Answer

(B) We are given the property $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$. By substituting $x = 2a - t$ in the second integral $\int_a^{2a} f(x) \, dx$,we get $dx = -dt$. When $x = a$,$t = a$,and when $x = 2a$,$t = 0$. Thus,$\int_a^{2a} f(x) \, dx = \int_a^0 f(2a - t) (-dt) = \int_0^a f(2a - t) \, dt = \int_0^a f(2a - x) \, dx$. Substituting this back,we have $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx = \int_0^a [f(x) + f(2a - x)] \, dx$. Given that $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$,it follows that $\int_0^a [f(x) + f(2a - x)] \, dx = \int_0^a 2f(x) \, dx$. This equality holds if $f(2a - x) = f(x)$.

If $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$,then:

Similar Questions

$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$

The value of $\int_{0}^{\pi /2} \frac{2^{\sin x}}{2^{\sin x} + 2^{\cos x}} dx$ is

Suppose $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Then,the value of $(M - N)$ equals

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module