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

If $\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in R$,where $[x]$ is the greatest integer less than or equal to $x$,then the value of $\alpha$ is :

The value of the integral $\sum\limits_{k = 1}^n {\int_0^1 {f(k - 1 + x)\,dx} } $ is

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

If $f(x) = \frac{e^x}{1+e^x}$, $l_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) dx$ and $l_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx$, then the value of $\frac{l_2}{l_1}$ is

The value of the integral $\int_0^\infty \frac{\log_e(x)}{x^2+4} dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module