Question

The value of the integral $\underset{-2}{\overset{2}{\int }}\left(a{x}^5+b{x}^3+cx+d\right)dx,$ where a, b, c, d are constants, depends only on ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{-2}{\overset{2}{\int }}\left(a{x}^5+b{x}^3+cx+d\right)dx \\ =\underset{-2}{\overset{2}{\int }}\left(a{x}^5+b{x}^3+cx\right)dx+\underset{-2}{\overset{2}{\int }}\left(d\right)dx \\ \mathrm{Let}, \\ f\left(x\right)=a{x}^5+b{x}^3+cx \\ f\left(-x\right)=a{\left(-x\right)}^5+b{\left(-x\right)}^3+c\left(-x\right) \\ =-a{x}^5-b{x}^3-cx \\ =-\left(a{x}^5+b{x}^3+cx\right) \\ =-f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\\ 2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=\underset{-2}{\overset{2}{\int }}\left(a{x}^5+b{x}^3+cx\right)dx+\underset{-2}{\overset{2}{\int }}\left(d\right)dx \\ =0+\underset{-2}{\overset{2}{\int }}\left(d\right)dx \\ ={\left(dx\right)}_{-2}^2 \\ =\left(2d-\left(-2d\right)\right) \\ =2d+2d \\ =4d \end{gathered}$$

​Hence, the value of the integral $\underset{-2}{\overset{2}{\int }}\left(a{x}^5+b{x}^3+cx+d\right)dx,$ where a, b, c, d are constants, depends only on d.