What is the integral from 0 to 1 of ysqrt(x^2+y^2) ?

The integral from 0 to 1 of ysqrt(x^2+y^2) is ((x^2+1)^{3/2}-x^3)/3

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integral from 0 to 1 of ysqrt(x^2+y^2)

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

\frac{( x ^{2} + 1 ) ^{\frac{3}{2}} - x ^{3}}{3}

Solution steps

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

Apply u-substitution

: \int_{x^{2}}^{x^{2} + 1} u^{2} d u

= \int_{x^{2}}^{x^{2} + 1} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{x^{2}}^{x^{2} + 1}

= [\frac{u ^{3}}{3}]_{x^{2}}^{x^{2} + 1}

Apply radical rule:

assuming

a \geq 0

= [\frac{u ^{3}}{3}]_{x}^{x^{2} + 1}

Compute the boundaries:

\frac{( x ^{2} + 1 ) ^{\frac{3}{2}}}{3} - \frac{x ^{3}}{3}

= \frac{( x ^{2} + 1 ) ^{\frac{3}{2}}}{3} - \frac{x ^{3}}{3}

Simplify

= \frac{( x ^{2} + 1 ) ^{\frac{3}{2}} - x ^{3}}{3}

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What is the integral from 0 to 1 of ysqrt(x^2+y^2) ?
The integral from 0 to 1 of ysqrt(x^2+y^2) is ((x^2+1)^{3/2}-x^3)/3