Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 42

$$\frac{{124}}{{15}}\left( {{2^{3/2}} - 1} \right)$$

$$\eqalign{ & \int_1^4 {\int_0^x {\sqrt {x + y} } dydx} \cr & = \int_1^4 {\left[ {\int_0^x {\sqrt {x + y} } dy} \right]dx} \cr & {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr & = \int_0^x {\sqrt {x + y} } dy \cr & = \int_0^x {{{\left( {x + y} \right)}^{1/2}}} dy \cr & {\text{integrate by using the power rule }} \cr & = \left[ {\frac{{{{\left( {x + y} \right)}^{3/2}}}}{{3/2}}} \right]_0^x \cr & = \frac{2}{3}\left[ {{{\left( {x + y} \right)}^{3/2}}} \right]_0^x \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{2}{3}\left( {{{\left( {x + x} \right)}^{3/2}} - {{\left( {x + 0} \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{3}\left( {{{\left( {2x} \right)}^{3/2}} - {x^{3/2}}} \right) \cr & = \frac{2}{3}\left( {{2^{3/2}} - 1} \right){x^{3/2}} \cr & \cr & \int_1^4 {\left[ {\int_0^x {\sqrt {x + y} } dy} \right]dx} = \int_0^4 {\left( {\frac{2}{3}\left( {{2^{3/2}} - 1} \right){x^{3/2}}} \right)dx} \cr & = \frac{2}{3}\left( {{2^{3/2}} - 1} \right)\int_0^4 {{x^{3/2}}dx} \cr & {\text{integrating by the power rule}} \cr & = \frac{2}{3}\left( {{2^{3/2}} - 1} \right)\left( {\frac{{{x^{5/2}}}}{{5/2}}} \right)_1^4 \cr & = \frac{4}{{15}}\left( {{2^{3/2}} - 1} \right)\left( {{x^{5/2}}} \right)_1^4 \cr & {\text{evaluate}} \cr & = \frac{4}{{15}}\left( {{2^{3/2}} - 1} \right)\left( {{4^{5/2}} - 1} \right) \cr & {\text{simplifying}} \cr & = \frac{4}{{15}}\left( {{2^{3/2}} - 1} \right)\left( {32 - 1} \right) \cr & = \frac{{124}}{{15}}\left( {{2^{3/2}} - 1} \right) \cr} $$