Let f(x) be an even function with period 2 an | vedclass

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

Question

(B) Given that $f(x)$ is an even function with period $2$. Since $f(x)$ is periodic with period $2$, we have $\int_0^2 f(t) dt = \int_a^{a+2} f(t) dt$

Vedclass · Accepted Answer

$g(x) + g(2)$

Vedclass · Answer

(B) Given that $f(x)$ is an even function with period $2$. Since $f(x)$ is periodic with period $2$, we have $\int_0^2 f(t) dt = \int_a^{a+2} f(t) dt$ for any $a$. Also, for an even function, $\int_{-a}^a f(t) dt = 2 \int_0^a f(t) dt$. Specifically, $\int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$. Since the period is $2$, $\int_0^2 f(t) dt = \int_{-1}^1 f(t) dt = 2 \int_0^1 f(t) dt$. Now, $g(x+2) = \int_0^{x+2} f(t) dt = \int_0^2 f(t) dt + \int_2^{x+2} f(t) dt$. Using the periodicity property $\int_2^{x+2} f(t) dt = \int_0^x f(t) dt = g(x)$. Thus, $g(x+2) = g(2) + g(x)$.

Let $f(x)$ be an even function with period $2$ and $f(x)$ be integrable on every interval. If $g(x) = \int_0^x f(t) dt$, then $g(x+2) =$

Similar Questions

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is

If the value of $\int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx = \frac{\pi}{32}$,then the value of $\int_{0}^{\pi/2} \cos^{4}(x) \cdot \sin^{2}(x) dx$ is:

Let $a$ be a positive real number such that $\int_{0}^{a} e^{x-[x]} dx = 10e - 9$,where $[x]$ is the greatest integer less than or equal to $x$. Then $a$ is equal to:

$I = \int\limits_0^1 \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module