The value of int_{-2}^2(x cos x+sin x+1) d x | vedclass

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

Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $I = \int_{-2}^2(x \cos x+\sin x+1) d x$. We can split the integral as $I = \int_{-2}^2 x \cos x d x + \int_{-2}^2 \sin x d x + \int_{-2}^2 1 d x$. Recall the property of definite integrals: $\int_{-a}^a f(x) d x = 0$ if $f(x)$ is an odd function, and $2 \int_0^a f(x) d x$ if $f(x)$ is an even function. Let $f_1(x) = x \cos x$. Since $f_1(-x) = (-x) \cos(-x) = -x \cos x = -f_1(x)$, $f_1(x)$ is an odd function. Thus, $\int_{-2}^2 x \cos x d x = 0$. Let $f_2(x) = \sin x$. Since $f_2(-x) = \sin(-x) = -\sin x = -f_2(x)$, $f_2(x)$ is an odd function. Thus, $\int_{-2}^2 \sin x d x = 0$. Therefore, $I = 0 + 0 + \int_{-2}^2 1 d x = [x]_{-2}^2 = 2 - (-2) = 4$.

The value of $\int_{-2}^2(x \cos x+\sin x+1) d x$ is

Similar Questions

If $f(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt$ and $g(x) = \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$, then the value of $f(x) + g(x)$ is

If $[.]$ denotes the greatest integer function,then $\int_0^{1000} e^{x-[x]} dx=$

Evaluate the definite integral: $\int_0^{2a} f(x) dx$

The value of the integral $\int_0^{\infty} \frac{dx}{(1+x^2)(1+x)^2}$ is

$\int_{-1}^{1} \sin^3 x \cos^2 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module