Q. The area bounded by $f (x) =x^{2}, 0 \le x \le 1, g\left(x\right)=x+2,1\le x\le2$ and $x$- axis is
VITEEEVITEEE 2017
Solution:
Required area $=$ Area of $OAB +$ Area of $A B C$
Now, Area of $OAB =\int\limits_{0}^{1} f(x) d x+\int\limits_{1}^{2} g(x) d x$
$=\int\limits_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x$
$=\frac{x^{3}}{3} 1_{0}^{1}+\left[\frac{-x^{2}}{2}+2 x\right]_{1}^{2}$
$=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}\right)+2\right]$
$=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]$
$=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$ sq unit
