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Popular >

integral from 0 to 1 of integral from-1 to |x|^{-1 of}(x+y)^2

Solution

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

Solution

diverges

Solution steps

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

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

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

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

diverges

= diverges

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of integral from-1 to |x|^{-1 of}(x+y)^2 ?
The integral from 0 to 1 of integral from-1 to |x|^{-1 of}(x+y)^2 is diverges