## Calculus: Early Transcendentals 8th Edition

$\frac{55}{12}$
Step 1 of 3: Area: Find the area bounded by the curves $=\sqrt[]{x}$ , $y=-\sqrt[3]{x}$ ,$y=x-2$ Step 2 of 3: Find the point of intersections of two curves. $\sqrt[]{x}=x-2 \rightarrow x=4$ $-\sqrt[3]{x}=x-2 \rightarrow x=1$ Thus the curves intersect at $=1,x=4 \rightarrow y=-1,y=2$ . Step 3 of 3: Area of the shaded region is $\int _{-1}^{0} \left( y+2+y^{3} \right) dy+ \int _{0}^{2} \left( y+2-y^{2} \right) dy$ . $\int _{-1}^{0} \left( y+2+y^{3} \right) dy+ \int _{0}^{2} \left( y+2-y^{2} \right) dy= \left[ \frac{y^{2}}{2}+2y+\frac{y^{4}}{4} \right] _{-1}^{0}+ \left[ \frac{y^{2}}{2}+2y-\frac{y^{3}}{3} \right] _{0}^{2}$ $=-\frac{1}{2}+2-\frac{1}{4}+\frac{2^{2}}{2}+4-\frac{2^{3}}{3}$ $=\frac{55}{12}$ Hence, the required area is $\frac{55}{12}$