The value of the integral int_{-2}^{2}(1+2 si | vedclass

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

Question

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$. We can split the integral as: $I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x

Vedclass · Accepted Answer

$2(e^{2}-1)$

Vedclass · Answer

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$. We can split the integral as: $I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x$. Consider the first part: $f(x) = e^{|x|}$. Since $f(-x) = e^{|-x|} = e^{|x|} = f(x)$, $f(x)$ is an even function. Thus, $\int_{-2}^{2} e^{|x|} d x = 2 \int_{0}^{2} e^{x} d x = 2[e^{x}]_{0}^{2} = 2(e^{2}-1)$. Consider the second part: $g(x) = \sin x e^{|x|}$. Since $g(-x) = \sin(-x) e^{|-x|} = -\sin x e^{|x|} = -g(x)$, $g(x)$ is an odd function. Thus, $\int_{-2}^{2} \sin x e^{|x|} d x = 0$. Therefore, $I = 2(e^{2}-1) + 2(0) = 2(e^{2}-1)$.

The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is equal to

Similar Questions

Let $f, f', f''$ be continuous in $[0, \ln 2]$ and $f(0) = 0, f'(0) = 3, f(\ln 2) = 6, f'(\ln 2) = 4$ and $\int_{0}^{\ln 2} e^{-2x} f(x) dx = 3$,then $\int_{0}^{\ln 2} e^{-2x} f''(x) dx$ is

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

$\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx$ is equal to

$\int_{2 - \log 3}^{3 + \log 3} \frac{\log (4 + x)}{\log (4 + x) + \log (9 - x)} \, dx = $

If $f(x) = f(a-x)$, then $\int_0^a x f(x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module