Question

Calculate the iterated integral.

\(\int {_1^4\int {_0^2\left( {6{x^2}y - 2x} \right)dydx} } \)

Short answer

We get , \(\int_0^2 {\int_0^4 {{y^3}{e^{2x}}dydx = 32\left( {{e^4} - 1} \right)} } \)

Step-by-step solution

Step 1:- Given and to find

\(\int_0^2 {\int_0^4 {{y^3}{e^{2x}}dydx)} } \)(Given)

The value of the given integral) (to find)

Step 2:-Find the value of the integral with respect to y

\(\begin{array}{l}\int_{\rm{0}}^{\rm{2}} {\int_{\rm{0}}^{\rm{4}} {{{\rm{y}}^{\rm{3}}}{{\rm{e}}^{{\rm{2x}}}}{\rm{dydx = }}\int_{\rm{0}}^{\rm{2}} {\left( {\left( {{{\rm{e}}^{{\rm{2x}}}}} \right)_{\rm{0}}^{\rm{4}}{{\rm{y}}^{\rm{3}}}{\rm{dy}}} \right)} } } {\rm{dx}}\\ & & {\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{{\rm{e}}^{{\rm{2x}}}}\left( {\frac{{{{\rm{4}}^{\rm{4}}}}}{{\rm{4}}}} \right)} \right){\rm{dx}}} \\ & & {\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{{\rm{e}}^{{\rm{2x}}}}\left( {\frac{{{{\rm{4}}^{\rm{4}}}}}{{\rm{4}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{4}}}}}{{\rm{4}}}} \right)} \right){\rm{dx}}} \\ & & {\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{{\rm{e}}^{{\rm{2x}}}}\left( {\frac{{{\rm{256}}}}{{\rm{4}}}} \right)} \right){\rm{dx}}} \\ & & {\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{{\rm{e}}^{{\rm{2x}}}}{\rm{64}}} \right){\rm{dx}}} \\ & & {\rm{ = 64}}\int_{\rm{0}}^{\rm{2}} {\left( {{{\rm{e}}^{{\rm{2x}}}}} \right)} {\rm{dx}}\end{array}\)

Step 3:- Find the value of final integral

\(\begin{array}{l}\int_{\rm{0}}^{\rm{2}} {\int_{\rm{0}}^{\rm{4}} {{{\rm{y}}^{\rm{3}}}{{\rm{e}}^{{\rm{2x}}}}{\rm{dydx = }}} } {\rm{64}}\int_{\rm{0}}^{\rm{2}} {{{\rm{e}}^{{\rm{2x}}}}} {\rm{dx}}\\ & & {\rm{ = 64}}{\left( {\frac{{{{\rm{e}}^{{\rm{2x}}}}}}{{\rm{2}}}} \right)^{\rm{2}}}\\ & & {\rm{ = 64}}\left( {\frac{{{{\rm{e}}^{{\rm{2 \times 2}}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{e}}^{{\rm{2 \times 0}}}}}}{{\rm{2}}}} \right)\\ & & {\rm{ = 64}}\left( {\frac{{{{\rm{e}}^{\rm{4}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{e}}^{\rm{0}}}}}{{\rm{2}}}} \right)\\ & & {\rm{ = 32}}\left( {{{\rm{e}}^{\rm{4}}}{\rm{ - 1}}} \right)\\{\rm{Therefore,}}\int_{\rm{0}}^{\rm{2}} {\int_{\rm{0}}^{\rm{4}} {{{\rm{y}}^{\rm{3}}}{{\rm{e}}^{{\rm{2x}}}}{\rm{dydx = 32}}\left( {{{\rm{e}}^{\rm{4}}}{\rm{ - 1}}} \right)} } \end{array}\)