The area bounded by $f(x)=x^{2}, 0 \leq x \leq 1$, $\mathrm{g}(\mathrm{x})=-\mathrm{x}+2,1 \leq \mathrm{x} \leq 2$ and $\mathrm{x}$-axis is

• A

  $\frac{3}{2}$
• B

  $\frac{4}{3}$
• C

  $\frac{8}{3}$
• Correct answer
  D

  None of these

Solution

Required area $=$ Area of OAB + Area of ABC
Now, Area of $\mathrm{OAB}=\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{1}^{2} \mathrm{~g}(\mathrm{x}) \mathrm{d} \mathrm{x}$ $=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x$ $=\left.\frac{x^{3}}{3}\right|_{0} ^{1}+\left[\frac{-x^{2}}{2}+2 x\right]_{1}^{2}$ $=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]$ $=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]$ $=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$ sq unit

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