Question

Evaluate the definite integral:

Answer 1

\[\int\limits_{1}^{7}\]\[(2x^2+8)dx \over \sqrt{x}\]

Answer 2

Let's remember that roots are really fractional exponents:\[\int_{1}^{7}\frac{2x^2+8}{\sqrt x}dx=\int_{1}^{7}\frac{2x^2+8}{x^{1/2}}dx=\int_{1}^{7}2x^{3/2}+8x^{-1/2}dx\]Now remembering the basic inverse power rule\[\int_{a}^{b} x^ndx=\frac{x^{n+1}}{n+1}\large{|}_{a}^{b}\]this should be a snap :D