Question

${\int }_0^3\left|x^3+x^2+3x\right|dx=$

• A. $\frac{171}{2}$
• B. $\frac{171}{4}$
• C. $\frac{170}{4}$
• D. $\frac{170}{3}$

Answer 1

The correct option is B

$\frac{171}{4}$

Explanation for correct option:

Integrating using modulus function:

Let $I={\int }_0^3\left|x^3+x^2+3x\right|dx$.

Here we will use the property of modulus function.

Let $f\left(x\right)=\left|x^3+x^2+3x\right|$

$f\left(x\right)=\left|x(x^2+x+3)\right|$

Now if the discriminant of the quadratic equation is less than zero, the roots of the equation will be greater than zero. Now the discriminant will be,

$$\begin{gathered} D\text{=}{b}^2-4ac \\ =1-4\times 1\times 3 \\ =1-12 \\ =-11 \end{gathered}$$

Therefore, $f\left(x\right)>0$ Now we have,

$f\left(x\right)=\left\{\begin{array}{l}-x(x^2+x+3)x<0\\ x(x^2+x+3)x\ge 0\end{array}\right.$

Therefore,

$\begin{array}{rcl}I& =& {\int }_0^3\left|x^3+x^2+3x\right|dx\\ & =& {\int }_0^3(x^3+x^2+3x)dx\\ & =& {\left[\frac{x^4}{4}+\frac{x^3}{3}+\frac{3x^2}{2}\right]}_0^3\\ & =& \frac{81}{4}+\frac{27}{3}+\frac{27}{2}\\ & =& \frac{81}{4}+\frac{45}{2}\\ I& =& \frac{171}{4}\end{array}$

Therefore the value of given integral is equal to $\frac{171}{4}$

Hence, option (B) is the correct answer.