Check the solution to "Evaluate the following iterated integrals. ∫02∫018xy1+x4dxdy"

Kye

Answered question

2021-01-31

Evaluate the following iterated integrals.

\int_{0}^{2} \int_{0}^{1} \frac{8 x y}{1 + x^{4}} d x d y

StrycharzT

Skilled2021-02-01Added 102 answers

The given integral is \int_{0}^{2} \int_{0}^{1} \frac{8 x y}{1 + x^{4}} d x d y

Evaluate the given integral as shown below:

\int_{0}^{2} \int_{0}^{1} \frac{8 x y}{1 + x^{4}} = \int_{0}^{2} y [\int_{0}^{1} \frac{8 x}{1 + x^{4}} d x] d y

Substitute u = x^{2}, d u = 2 x d x

\int_{0}^{2} \int_{0}^{1} \frac{8 x y}{1 + x^{4}} d x d y = \int_{0}^{2} y [\int_{0}^{1} \frac{8 x}{1 + x^{4}} d x] d y

= \int_{0}^{2} y [4 \int_{0}^{1} \frac{2 x}{1 + (x^{2})^{2}} d x] d y

= \int_{0}^{2} y [4 \int_{0}^{1} \frac{d u}{1 + u^{2}} d x] d y

= \int_{0}^{2} y [4 (\tan^{- 1} u)_{0}^{1}] d y

= \int_{0}^{2} y [4 (\tan^{- 1} 1 - \tan^{- 1} 0)] d y

= \int_{0}^{2} y [4 (π / 4 - 0)] d y

= \int_{0}^{2} y π d y

= π [\frac{y^{2}}{2}]_{0}^{2}

= π [\frac{2^{2}}{2} - 0]_{0}^{2}

= 2 π

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?