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Evaluate the integral by making an appropriate change of variables. \iint_{R} 9e^{10x + 10y} dA,...

Question:

Evaluate the integral by making an appropriate change of variables.

{eq}\displaystyle \iint_{R}9e^{10x + 10y}\,dA {/eq},

where {eq}R {/eq} is given by the inequality {eq}10|x| + 10|y| \leq 10 {/eq}.

The Integral Defined Using Substitution:

The substitution method, where we properly change one variable to another, is used to facilitate the calculation of the integral. We apply this method when we derive a part of the integral so that the other part of it is reproduced. We can also obtain this result through algebraic procedures. This method can be applied to indefinite integrals and definite integrals. In indefinite integrals, only the replacement in the function that is integrated is applied. However, when it is a definite integral, the change in the integration limits is also made, placing these limits in terms of the new variable.

This method is very important because it facilitates the solution for a wide variety of integrals. It can be used as the sole solution method or be combined with other methods.

Answer and Explanation:

{eq}\eqalign{ & {\text{First, we find the integration limits by solving the given inequality:}} \cr & 10\left| x \right| + 10\left| y \right| \leqslant 10 \cr & {\text{Dividing by 10 on both sides:}} \cr & \left| x \right| + \left| y \right| \leqslant 1 \cr & {\text{Since }}x{\text{ and }}y{\text{ are absolute values}}{\text{, the inequality becomes two inequalities}}{\text{, }} \cr & {\text{one with positive values and the other with negative values:}} \cr & x + y \leqslant 1 \cr & - x - y \leqslant 1 \cr & {\text{We find }}y{\text{ in both inequalities:}} \cr & y \leqslant 1 - x \cr & - y \leqslant 1 + x{\text{ }} \Rightarrow {\text{ }}y \geqslant - 1 - x \cr & {\text{Therefore}}{\text{, the integration limits for }}y{\text{ are:}} \cr & - 1 - x \leqslant y \leqslant 1 - x \cr & {y_1} = - 1 - x \cr & {y_2} = 1 - x \cr & {\text{Now}}{\text{, we find }}x{\text{ in both inequalities:}} \cr & x \leqslant 1 - y \cr & - x \leqslant 1 + y{\text{ }} \Rightarrow {\text{ }}x \geqslant - 1 - y \cr & {\text{The integration limits for }}x{\text{ are:}} \cr & - 1 - y \leqslant x \leqslant 1 - y \cr & {x_1} = - 1 - y \cr & {x_2} = 1 - y \cr & {\text{We write and solve the integral:}} \cr & \int_{ - 1 - x}^{1 - x} {\int_{ - 1 - y}^{1 - y} {\left( {9{e^{10x + 10y}}} \right)} } dxdy = \int_{ - 1 - x}^{1 - x} {\left[ {9\int_{ - 1 - y}^{1 - y} {{e^{10x + 10y}}dx} } \right] dy} \cr & {\text{We make an appropriate change of variables in the integral and its limits:}} \cr & u = 10x + 10y{\text{ }} \Rightarrow {\text{ }}du = 10dx{\text{ }} \Rightarrow {\text{ }}\frac{1}{{10}}du = dx \cr & {\text{The change of variables in the limits are }}u({x_1}{\text{) and }}u({x_2}){\text{:}} \cr & u({x_1}) = u( - 1 - y) = 10( - 1 - y) + 10y = - 10 - 10y + 10y = - 10 \cr & u({x_2}) = u(1 - y) = 10(1 - y) + 10y = 10 - 10y + 10y = 10 \cr & {\text{We replace the new variables and solve:}} \cr & \int_{ - 1 - x}^{1 - x} {\int_{ - 1 - y}^{1 - y} {\left( {9{e^{10x + 10y}}} \right)} } dxdy = \int_{ - 1 - x}^{1 - x} {\left[ {9\int_{ - 1 - y}^{1 - y} {{e^{10x + 10y}}dx} } \right]dy} \cr & = \int_{ - 1 - x}^{1 - x} {\left[ {9\int_{ - 10}^{10} {{e^u}\frac{1}{{10}}du} } \right]} dy = \int_{ - 1 - x}^{1 - x} {\left[ {\frac{9}{{10}}\int_{ - 10}^{10} {{e^u}du} } \right]} dy \cr & = \int_{ - 1 - x}^{1 - x} {\left[ {\frac{9}{{10}}{e^u} + k} \right]} _{ - 10}^{10}dy = \int_{ - 1 - x}^{1 - x} {\frac{9}{{10}}\left[ {{e^u} + k} \right]} _{ - 10}^{10}dy \cr & = \int_{ - 1 - x}^{1 - x} {\frac{9}{{10}} \left[ {({e^{10}} + k) - ({e^{ - 10}} + k)} \right]} dy = \int_{ - 1 - x}^{1 - x} {\frac{9}{{10}} \left[ {{e^{10}} + k - {e^{ - 10}} - k} \right]} dy \cr & = \frac{9}{{10}}\int_{ - 1 - x}^{1 - x} {\left[ {{e^{10}} - {e^{ - 10}}} \right]} dy = \frac{9}{{10}}\int_{ - 1 - x}^{1 - x} {\left[ {{e^{10}} - \frac{1}{{{e^{10}}}}} \right]} dy = \frac{9}{{10}}\int_{ - 1 - x}^{1 - x} {\left[ {\frac{{{e^{20}} - 1}}{{{e^{10}}}}} \right]} dy \cr & = \frac{9}{{10}} \left[ {\frac{{{e^{20}} - 1}}{{{e^{10}}}}} \right] \left[ {y + k} \right]_{ - 1 - x}^{1 - x} = \frac{{9 ({e^{20}} - 1)}}{{10{e^{10}}}} \left[ {(1 - x + k) - ( - 1 - x + k)} \right] \cr & = \frac{{9({e^{20}} - 1)}}{{10{e^{10}}}} \left[ {1 - x + k + 1 + x - k} \right] = \frac{{9({e^{20}} - 1)}}{{10{e^{10}}}} \left[ {1 + 1} \right] \cr & = \frac{{9({e^{20}} - 1)}}{{10{e^{10}}}} \left[ 2 \right] = \frac{{9 ({e^{20}} - 1)}}{{5{e^{10}}}} \cr} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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