Let f(x) be a differentiable function defined | vedclass

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

Question

(B) Given $f^{\prime}(x) = f^{\prime}(2-x)$. Integrating both sides with respect to $x$,we get $f(x) = -f(2-x) + C$,which implies $f(x) + f(2-x) = C$.

Vedclass · Accepted Answer

$1+e^{2}$

Vedclass · Answer

(B) Given $f^{\prime}(x) = f^{\prime}(2-x)$. Integrating both sides with respect to $x$,we get $f(x) = -f(2-x) + C$,which implies $f(x) + f(2-x) = C$.At $x=0$,$f(0) + f(2) = C$. Given $f(0) = 1$ and $f(2) = e^{2}$,we have $C = 1 + e^{2}$.Thus,$f(x) + f(2-x) = 1 + e^{2}$.Let $I = \int_{0}^{2} f(x) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have $I = \int_{0}^{2} f(2-x) dx$.Adding the two expressions for $I$,we get $2I = \int_{0}^{2} (f(x) + f(2-x)) dx$.Substituting the sum,$2I = \int_{0}^{2} (1 + e^{2}) dx = (1 + e^{2}) [x]_{0}^{2} = 2(1 + e^{2})$.Therefore,$I = 1 + e^{2}$.

Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime}(x) = f^{\prime}(2-x)$ for all $x \in (0,2)$,$f(0) = 1$ and $f(2) = e^{2}$. Then the value of $\int_{0}^{2} f(x) dx$ is ..... .

Similar Questions

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

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

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx=$

$\int_{-1}^{1} x|x| \, dx = $

The value of the integral $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module