Question

The area of the region bounded by the curve $y=|x^3–4x^2+3x|$ and the $x$-axis, $0\le x\le 3$, is

• A. $\frac{37}{6}$
• B. $\frac{9}{4}$
• C. $\frac{37}{12}$
• D. $0$

Answer 1

The correct option is C $\frac{37}{12}$

$y=|x^3–4x^2+3x|=|x(x-1)(x-3)|$
Area under curve,
$A=\underset{0}{\overset{1}{\int }}{x}^3–4x^2+3xdx-\underset{1}{\overset{3}{\int }}{x}^3–4x^2+3xdx\Rightarrow A=\frac{1}{4}-\frac{4}{3}+\frac{3}{2}-{(\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2})}_1^3\Rightarrow A=\frac{5}{12}-(\frac{80}{4}-\frac{4\times 26}{3}+\frac{3\times 8}{2})\Rightarrow A=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}$