$$ Page $663$ : Problem $1-20($problem set $10-2)$

${\displaystyle \underset{2}{\overset{5}{}}}2dx$

${\displaystyle \underset{1}{\overset{3}{}}}3dx$

${\displaystyle \underset{-1}{\overset{4}{}}}2xdx$

${\displaystyle \underset{-3}{\overset{9}{}}}3xdx$

${\displaystyle \underset{2}{\overset{6}{}}}(x+1)dx.$

${\displaystyle \underset{-1}{\overset{3}{}}}(2x-1)dx$

${\displaystyle \underset{0}{\overset{8}{}}}{x}^{\frac{2}{3}}dx.$

${\displaystyle \underset{0}{\overset{9}{}}}{x}^{\frac{1}{2}}dx.$

${\displaystyle \underset{1}{\overset{2}{}}}(x^2-3x+5)dx.$

${\displaystyle \underset{0}{\overset{3}{}}}(x^2+5x-2)dx.$

${\displaystyle \underset{1}{\overset{9}{}}}(5+y^{-\frac{1}{2}})dy.$

${\displaystyle \underset{1}{\overset{8}{}}}(1+2y^{-\frac{1}{3}})dy.$

${\displaystyle \underset{2}{\overset{6}{}}}(2x-3{)}^{\frac{1}{2}}dx.$

${\displaystyle \underset{1}{\overset{8}{}}}(5x-4{)}^{-\frac{1}{2}}dx.$

${\displaystyle \underset{1}{\overset{6}{}}}\frac{60dx}{(3x+2{)}^2}.$

${\displaystyle \underset{0}{\overset{6}{}}}\frac{20}{(4x+1{)}^{\frac{3}{2}}}dx.$

${\displaystyle \underset{a}{\overset{b}{}}}3x^{2}dx.$

${\displaystyle \underset{c}{\overset{d}{}}}4x^{3}dx.$

${\displaystyle \underset{1}{\overset{n}{}}}(x+1)dx.$

${\displaystyle \underset{n}{\overset{1}{}}}(2x-1)dx.$