Expert answers to "Calculate the double integral. ∫∫R4x1+xydA R=[0,4]×[0,1]"

coexpennan

Answered question

2021-05-12

Calculate the double integral.

\int \int_{R} \frac{4 x}{1 + x y} d A

R = [0, 4] \times [0, 1]

toroztatG

Skilled2021-05-13Added 98 answers

Explanations for your question:

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

Consider the double integral,

$\int \int_{R} \frac{4 x}{1 + x y} d A$

Where $R = [0, 4] \times [0, 1]$

Then $\int \int_{R} \frac{4 x}{1 + x y} d A = \int_{0}^{4} \int_{0}^{1} \frac{4 x}{1 + x y} d y d x$

$= \int_{0}^{4} 4 x [\frac{\ln (1 + x y)}{x}]_{0}^{1} d x$

$= \int_{0}^{4} 4 [\ln | 1 + x | - \ln | 1 + 0 |] d x$

$= 4 \int_{0}^{4} [\ln | 1 + x | - 0] d x$

$= 4 \int_{0}^{4} \ln (1 + x) d x$

Now consider the integral $\int_{0}^{4} \ln (1 + x) d x$

Let $u = \ln (1 + x), d v = d x$

Then $d u = \frac{1}{1 + x} d x, v = x$

Use integration by parts,

$\int u d v = u v - \int v d u$

$\int_{0}^{4} \ln (1 + x) d x = [\ln (1 + x) x - \int x \frac{1}{1 + x} d x]_{0}^{4}$

$= [x \ln (1 + x) - \int \frac{1 + x - 1}{1 + x} d x]_{0}^{4}$

$= [x \ln (1 + x) - (x - \ln | 1 + x |)]_{0}^{4}$

$= [4 \ln 5 - 3 + \ln 5] - [0 - 0 + 0]$

$= 5 \ln 5 - 4$ (2)

Use equation (2) in equation (1) to get,

$\int \int_{R} \frac{4 x}{1 + x y} d A = 4 [5 \ln 5 - 4]$

$= 20 \ln 5 - 16$

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