Question

Area of the figure bounded by the lines $y=\sqrt{x},x\in [0,1],y=x^2,x\in [1,2]$ and $y=-x^2+2x+4,x\in 0,2]$ is:

• A. $\frac{10}{7}$
• B. $\frac{3}{5}$
• C. $\frac{4}{3}$
• D. $\frac{19}{3}$

Answer 1

Answer: (D)

$\frac{19}{3}$

As shown in the figure, the area bounded
$={\int }_0^2(-x^2+2x+4)dx-{\int }_0^1\sqrt{x}dx-{\int }_1^2{x}^{2}dx$
$=\frac{19}{3}$