Evaluate the following iterated integrals. int_0^2int_0^1frac{8xy}{1+x^4}dxdy

Evaluate the following iterated integrals. int_0^2int_0^1frac{8xy}{1+x^4}dxdy

Question
Integrals
asked 2021-01-31
Evaluate the following iterated integrals.
\(\int_0^2\int_0^1\frac{8xy}{1+x^4}dxdy\)

Answers (1)

2021-02-01
\(\text{The given integral is }\int_0^2\int_0^1\frac{8xy}{1+x^4}dxdy\)
\(\text{Evaluate the given integral as shown below:}\)
\(\int_0^2\int_0^1\frac{8xy}{1+x^4}=\int_0^2y[\int_0^1\frac{8x}{1+x^4}dx]dy\)
\(\text{Substitute }u=x^2,du=2xdx\)
\(\int_0^2\int_0^1\frac{8xy}{1+x^4}dxdy=\int_0^2y[\int_0^1\frac{8x}{1+x^4}dx]dy\)
\(=\int_0^2y[4\int_0^1\frac{2x}{1+(x^2)^2}dx]dy\)
\(=\int_0^2y[4\int_0^1\frac{du}{1+u^2}dx]dy\)
\(=\int_0^2y[4(\tan^{-1}u)_0^1]dy\)
\(=\int_0^2y[4(\tan^{-1}1-\tan^{-1}0)]dy\)
\(=\int_0^2y[4(\pi/4-0)]dy\)
\(=\int_0^2y\pi dy\)
\(=\pi[\frac{y^2}{2}]_0^2\)
\(=\pi[\frac{2^2}{2}-0]_0^2\)
\(=2\pi\)
0

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