Question

let $f\left(x\right)=(x+1{)}^2+\frac{1}{x}$ then the value of ${\int }_{-2}^{1}f\left(x\right)(-x)dx$

• A. is equal to $\frac{-15}{4}$
• B. is equal to $-\frac{81}{10}$
• C. is equal to $\frac{81}{10}$
• D. does not exist

Answer 1

The correct option is A is equal to $\frac{-15}{4}$

We have

$f\left(x\right)={(x+1)}^2+\frac{1}{x}$

Then the value of

${\int }_{-2}^{1}f\left(x\right)(-x)dx$

Solve it:-

${\int }_{-2}^{1}f\left(x\right)(-x)dx$

$={\int }_{-2}^1[{(x+1)}^2+\frac{1}{x}](-x)dx$

$={\int }_{-2}^1[(-x){(x+1)}^2+\frac{-x}{x}]dx$

$={\int }_{-2}^1[(-x)(x^2+1+2x)-1]dx$

$={\int }_{-2}^1[(-x^3-x-2x^2)-1]dx$

$={\int }_{-2}^1[-x^3-x-2x^2-1]dx$

$={\int }_{-2}^1-x^{3}dx-{\int }_{-2}^{1}xdx-{\int }_{-2}^{1}2x^{2}dx-{\int }_{-2}^{1}1dx$

$={{\left[\frac{-x^4}{4}\right]}_{-2}}^1-{{\left[\frac{x^2}{2}\right]}_{-2}}^1-2{{\left[\frac{x^3}{3}\right]}_{-2}}^1-{{\left[x\right]}_{-2}}^1$

$=-\left[\frac{1^4-{(-2)}^4}{4}\right]-\left[\frac{1^2-{(-2)}^2}{2}\right]-2\left[\frac{1^3-{(-2)}^3}{3}\right]-[1-(-2)]$

$=\frac{15}{4}+\frac{3}{2}-2\times \frac{9}{3}-3$

$=\frac{15}{4}+\frac{3}{2}-6-3$

$=\frac{15}{4}+\frac{3}{2}-9$

$=\frac{15+6-36}{4}$

$=\frac{21-36}{4}$

$=\frac{-15}{4}$

Hence, this is the answer.