Let T 0 be a fixed number. f: R rightarrow | vedclass

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

Question

(C) Given,$I = \int_0^T f(x) dx$. Since $f(x+T) = f(x)$,$f$ is a periodic function with period $T$. We need to evaluate $J = \int_0^{5T} f(2x) dx$. Le

Vedclass · Accepted Answer

$5I$

Vedclass · Answer

(C) Given,$I = \int_0^T f(x) dx$. Since $f(x+T) = f(x)$,$f$ is a periodic function with period $T$. We need to evaluate $J = \int_0^{5T} f(2x) dx$. Let $2x = y$,then $dx = \frac{1}{2} dy$. When $x = 0$,$y = 0$. When $x = 5T$,$y = 10T$. Substituting these into the integral: $J = \int_0^{10T} f(y) \cdot \frac{1}{2} dy = \frac{1}{2} \int_0^{10T} f(y) dy$. Using the property of periodic functions $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$: $J = \frac{1}{2} 	imes 10 \int_0^T f(y) dy = 5 \int_0^T f(x) dx = 5I$.

Let $T > 0$ be a fixed number. $f: R \rightarrow R$ is a continuous function such that $f(x+T) = f(x)$ for all $x \in R$. If $I = \int_0^T f(x) dx$,then find the value of $\int_0^{5T} f(2x) dx$.

Similar Questions

If $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$,then:

$\int_{ - 1/2}^{1/2} {\cos x \ln \left( \frac{1 + x}{1 - x} \right) dx}$ is equal to

$\int_{1}^{6\pi}([\sec^{-1}x]+[\cot^{-1}x])dx$ is equal to (where $[.]$ denotes the greatest integer function).

If ${I_n} = \int_{0}^{\pi /4} {\tan^n x} \,dx$,then $\lim_{n \to \infty} n[{I_n} + {I_{n - 2}}]$ equals

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module