Question

Find the area bounded by y=√x and y=x^3+2 and x=4 and Y-axis

Make a sketch, use vertical elements, show the element of area with its dimensions and corresponding differential of area.

Answer 1

$A=\int_0^4 (x^3+2-\sqrt{x})dx=(\frac{x^4}{4}+2x-\frac{2\sqrt{x^3}}{3})|_0^4 =$

$=(\frac{4^4}{4}+2\cdot 4-\frac{2\sqrt{4^3}}{3})-(\frac{0^4}{4}+2\cdot 0-\frac{2\sqrt{0^3}}{3})=$

$=4^3+8-\frac{16}{3}=64+8-5\frac{1}{3}=66\frac{2}{3}.$