The area bounded by the curve $y=x^3-3 x^2+2 x$ and the $X$-axis is (in square units)

• Correct answer
  A

  $\frac{1}{2}$
• B

  $\frac{5}{2}$
• C

  1
• D

  4

Solution

Area Bounded by $$ \begin{aligned} y & =x^3-3 x^2+2 x \text { and } X \text {-axis } \\ y & =x\left(x^2-3 x+2\right) \\ \Rightarrow \quad y & =x(x-1)(x-2) \end{aligned} $$
Similarly part in 0 to 1 and 1 to 2 $$ \begin{array}{ll} \text { So, } \operatorname{area}=2 \int_0^1 y d x & \Rightarrow 2 \int_0^1 x^3-3 x^2+2 x d x \\ \Rightarrow 2\left[\frac{x^4}{4}-\frac{3 x^3}{3}+\frac{2 x^2}{2}\right]_0^1 & \Rightarrow 2\left[\frac{1}{4}-1+1-0\right]=\frac{1}{2} \end{array} $$

Hence, the correct answer is option: 'A'.