Question

The value of ${\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx$ is

• A. $2$
• B. $0$
• C. $-2$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx$

$$\begin{gathered} \Rightarrow {\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx \\ \Rightarrow {\int }_{-2}^2\left[x\cos \left(x\right)\right]dx+{\int }_{-2}^2\left[\sin \left(x\right)\right]dx+{\int }_{-2}^2\left[1\right]dx \end{gathered}$$

using integration by parts $\int f\left(x\right)·g\left(x\right)·dx=f\left(x\right)·g\left(x\right)\text{'}-\int f\left(x\right)\text{'}\left(\int g\left(x\right)·dx\right)·dx$

put $f\left(x\right)=x,g\left(x\right)=\cos \left(x\right)$

$$\begin{gathered} {\int }_{-2}^2(x·\cos (x\left)\right)dx=-x·\sin \left(x\right)+\int \left(\sin \right(x)·dx \\ {\int }_{-2}^2(x·\cos \left(x\right))dx=-x·\sin (x)+(\cos \left(x\right) \end{gathered}$$

$$\begin{gathered} {\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx={\left|\cos \left(x\right)-x·\sin \left(x\right)\right|}_{-2}^2-{\left|\cos \left(x\right)\right|}_{-2}^2+{\left|x\right|}_{-2}^2 \\ {\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx=\cos \left(2\right)-\cos (-2)-2\sin \left(2\right)-2\sin (-2)-\cos \left(2\right)+\cos (-2)+2-(-2) \\ {\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx=\cos \left(2\right)-\cos \left(2\right)-2\sin \left(2\right)+2\sin \left(2\right)-\cos \left(2\right)+\cos (-2)+4 \\ {\int }_{-2}^2\left[x\cos \left(x\right)+\sin \left(x\right)+1\right]dx=4 \end{gathered}$$

Hence, option $D$ is the correct answer.