What is the integral from 0 to 1 of 7xysqrt(x^2+y^2) ?

The integral from 0 to 1 of 7xysqrt(x^2+y^2) is (7x((x^2+1)^{3/2}-x^3))/3

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integral from 0 to 1 of 7xysqrt(x^2+y^2)

\int_{0}^{1} 7 x y x^{2} + y^{2} d y

\frac{7 x ( ( x ^{2} + 1 ) ^{\frac{3}{2}} - x ^{3} )}{3}

Solution steps

\int_{0}^{1} 7 x y x^{2} + y^{2} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 x \cdot \int_{0}^{1} x^{2} + y^{2} y d y

Apply u-substitution

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

= 7 x \cdot \int_{x^{2}}^{x^{2} + 1} u^{2} d u

Apply the Power Rule

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

= 7 x [\frac{u ^{3}}{3}]_{x^{2}}^{x^{2} + 1}

Apply radical rule:

assuming

a \geq 0

= 7 x [\frac{u ^{3}}{3}]_{x}^{x^{2} + 1}

Compute the boundaries:

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

= 7 x \frac{( x ^{2} + 1 ) ^{\frac{3}{2}}}{3} - \frac{x ^{3}}{3}

Simplify

= \frac{7 x ( ( x ^{2} + 1 ) ^{\frac{3}{2}} - x ^{3} )}{3}

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What is the integral from 0 to 1 of 7xysqrt(x^2+y^2) ?
The integral from 0 to 1 of 7xysqrt(x^2+y^2) is (7x((x^2+1)^{3/2}-x^3))/3