Mathematics - Area Under Curves Question with Solution | TestHub

$0\le y\le x^2+1, 0\le y\le x+1, \frac{1}{2}\le x\le 2$

Drawing the graph according to the given information,

Area is, $A={\int }_{\frac{1}{2}}^1\left(x^2+1\right)dx+{\int }_1^2\left(x+1\right)dx$

$A={\left[\frac{x^3}{3}+x\right]}_{\frac{1}{2}}^1+{\left[\frac{x^2}{2}+x\right]}_1^2$

$A=\left[\frac{1}{3}+1\right]-\left[{\left(\frac{1}{2}\right)}^3·\frac{1}{3}+\frac{1}{2}\right]+\left[\frac{4}{2}+2\right]-\left[\frac{1}{2}+1\right]$

$A=\frac{4}{3}-\left(\frac{1}{24}+\frac{1}{2}\right)+\left(2+2\right)-\frac{3}{2}$

$A=\frac{4}{3}-\frac{13}{24}+4-\frac{3}{2}$

$A=\frac{32-13+96-36}{24}$

$A=\frac{79}{24}$ sq. units.