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opatovaL

Answered question

2021-11-08

Calculate the iterated integral.

\int_{0}^{1} \int_{0}^{1} {(x + y)}^{2} d x d y

Alix Ortiz

Skilled2021-11-09Added 109 answers

Given

\int_{0}^{1} \int_{0}^{1} {(x + y)}^{2} d x d y

solution

\int_{0}^{1} \int_{0}^{1} {(x + y)}^{2} d x d y = \int_{0}^{1} \int_{0}^{1} (x^{2} + 2 x y + y^{2}) d x d y

integrating with respect to x,

\int_{0}^{1} \int_{0}^{1} {(x + y)}^{2} d x d y = \int_{0}^{1} {[\frac{x^{3}}{3} + \frac{2 x^{2} y}{2} + y^{2} x]}_{0}^{1} d y

= \int_{0}^{1} (\frac{1}{3} + y + y^{2} - 0) d y

= {[\frac{1}{3} y + \frac{y^{2}}{2} + \frac{y^{3}}{3}]}_{0}^{1}

= \frac{1}{3} + \frac{1}{2} + \frac{1}{3} - 0

\int_{0}^{1} \int_{0}^{1} {(x + y)}^{2} d x d y = \frac{7}{6}

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