Let f: R rightarrow R be a continuous functio | vedclass

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

Question

(B) Given $f(x)+f(x+1)=2$. Integrating both sides from $0$ to $1$,we get $\int_{0}^{1} f(x) d x + \int_{0}^{1} f(x+1) d x = \int_{0}^{1} 2 d x$. Using

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Given $f(x)+f(x+1)=2$. Integrating both sides from $0$ to $1$,we get $\int_{0}^{1} f(x) d x + \int_{0}^{1} f(x+1) d x = \int_{0}^{1} 2 d x$. Using the substitution $t=x+1$ in the second integral,we have $\int_{0}^{1} f(x) d x + \int_{1}^{2} f(t) d t = 2$,which implies $\int_{0}^{2} f(x) d x = 2$. Since $f(x)+f(x+1)=2$,$f(x+2) = 2-f(x+1) = 2-(2-f(x)) = f(x)$,so $f(x)$ is periodic with period $T=2$. $I_{1} = \int_{0}^{8} f(x) d x = \frac{8}{2} \int_{0}^{2} f(x) d x = 4 	imes 2 = 8$. $I_{2} = \int_{-1}^{3} f(x) d x$. Since $f(x)$ has period $2$,$\int_{a}^{a+T} f(x) d x$ is independent of $a$. Thus,$I_{2} = \int_{0}^{4} f(x) d x = \frac{4}{2} \int_{0}^{2} f(x) d x = 2 	imes 2 = 4$. Therefore,$I_{1}+2 I_{2} = 8 + 2(4) = 8 + 8 = 16$.

Let $f: R \rightarrow R$ be a continuous function such that $f(x)+f(x+1)=2$ for all $x \in R$. If $I_{1}=\int_{0}^{8} f(x) d x$ and $I_{2}=\int_{-1}^{3} f(x) d x$,then the value of $I_{1}+2 I_{2}$ is equal to:

Similar Questions

For each positive integer $n$,define $f_n(x) = \min\left(\frac{x^n}{n!}, \frac{(1-x)^n}{n!}\right)$ for $0 \leq x \leq 1$. Let $I_n = \int_{0}^{1} f_n(x) dx$ for $n \geq 1$. Then,$\sum_{n=1}^{\infty} I_n$ is equal to

The absolute value of $\int_{10}^{19} \frac{\sin x}{1 + x^8} dx$ is less than:

If $\int_0^\pi {x\,f({{\cos }^2}x + {{\tan }^4}x)\,dx} = k\int_0^{\pi /2} {f({{\cos }^2}x + {{\tan }^4}x)\,dx,}$ then the value of $k$ is

If $f(x) = \int\limits_0^\pi {\frac{t \sin t \, dt}{\sqrt{1 + \tan^2 x \sin^2 t}}}$ for $0 < x < \frac{\pi}{2}$,then which of the following is true?

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx, a > 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module