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

Let $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$,where $n$ is a non-negative integer. Then,$\sum_{n=2}^{\infty} \left( \frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} \right)$ equals

If $\int_0^{10} f(x) d x=5$,then $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x=$

$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to

If $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,then the value of $k$ equals

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module