Question

${\int }_0^{10}|x(x-1)(x-2\left)\right|dx=?$

• A. $160.05$
• B. $1600.5$
• C. $16.005$
• D. None of these

Answer 1

The correct option is B

$1600.5$

Explanation for the Correct Option:

Evaluating the finite integral:

${\int }_0^{10}|x(x-1)(x-2\left)\right|dx$

Expanding the integrand $\left|\right(x-2\left)\right(x-1)x|$ gives $|x²-3x²+2x|$

$\Rightarrow {\int }_0^{10}|x²-3x²+2x|dx$

Simplify $|x^3-3x^2+2x|$ assuming $x>0$

$\Rightarrow {\int }_0^{10}x|x^2-3x+2|dx$

When $0<x<1\text{or}2<x<10$, then $x^2-3x+2$ is positive.

This means$|2-3x+x^2|=x|2-3x+x^2|$

And when $1<x<2,x^2-3x+2$ is negative.

This means $x|2-3x+x^2|=x|-2+3x-x^2|$

$\Rightarrow {\int }_0^{1}x(x²-3x+2)dx+{\int }_1^{2}x(-x²+3x-2)dx+{\int }_2^{10}x(x²-3x+2)dx$ $\Rightarrow {\int }_0^1(x³-3x²+2x)dx+{\int }_1^{2}x(-x²+3x-2)dx+{\int }_2^{10}x(x²-3x+2)dx$

Integrate the sum term by term and factor out constants

$\Rightarrow {\int }_0^{1}x³dx-3{\int }_0^{1}x²dx+2{\int }_0^{1}xdx+{\int }_1^{2}x(-x²+3x-2)dx+{\int }_2^{10}x(x²-3x+2)dx$

Evaluate the antiderivative at the limits and subtract

$\Rightarrow {\left[\frac{x^4}{4}\right]}_0^1-3{\left[\frac{x^3}{3}\right]}_0^1+2{\left[\frac{x^2}{2}\right]}_0^1+{\left[\left(\frac{-x^3}{3}+\frac{3x^2}{2}-2\frac{x^2}{2}\right)\right]}_1^2+{\left[\left(\frac{x^3}{3}+\frac{3x^2}{2}-2\frac{x^2}{2}\right)\right]}_2^{10}$

Solving it we get

$\Rightarrow \frac{1}{4}-1+1-\frac{15}{4}+7-3+2496-992+96$

$\Rightarrow \frac{3201}{2}=1600.5$

Hence, option (B) is the answer.