Calculate the double integral.

$\int {\int}_{R}\frac{1}{1+x+y}dA,\text{}R=[1,3]\times [1,2]$

Ramsey
2021-10-15
Answered

Calculate the double integral.

$\int {\int}_{R}\frac{1}{1+x+y}dA,\text{}R=[1,3]\times [1,2]$

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broliY

Answered 2021-10-16
Author has **97** answers

Once again use the fact that:

$\int \frac{{f}^{\prime}\left(x\right)}{f\left(x\right)}dx=\mathrm{ln}\left|f\left(x\right)\right|+C$

It doesn't matter do you opt to solve the first integral with respect to x or with respect to y, Here, we opt to solve the integral with respect to y first:

${\int}_{1}^{3}{\int}_{1}^{2}\frac{1}{1+x+y}dydx={\int}_{1}^{3}\mathrm{ln}(x+3)-\mathrm{ln}(x+2)$

Now using the integration by parts we obtain that the integral is equal to:

$((x+3)\mathrm{ln}(x+3)-(x+3))-((x+2)\mathrm{ln}(x+2)-(x+2)){\mid}_{1}^{3}=6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3$

Result:$6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3$

It doesn't matter do you opt to solve the first integral with respect to x or with respect to y, Here, we opt to solve the integral with respect to y first:

Now using the integration by parts we obtain that the integral is equal to:

Result:

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