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

The integral from 1 to x^2 of 2x^2y(x+y) is x^7-x^3+2x^2(x^6-1)/3

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

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

x^{7} - x^{3} + 2 x^{2} \frac{x ^{6} - 1}{3}

Solution steps

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

Take the constant out:

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

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

Expand

(x + y) y : x y + y^{2}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{x^{2}} x y d y = \frac{x ( x ^{4} - 1 )}{2}

\int_{1}^{x^{2}} y^{2} d y = \frac{x ^{6} - 1}{3}

= 2 x^{2} (\frac{x ( x ^{4} - 1 )}{2} + \frac{x ^{6} - 1}{3})

Simplify

2 x^{2} (\frac{x ( x ^{4} - 1 )}{2} + \frac{x ^{6} - 1}{3}) : x^{7} - x^{3} + 2 x^{2} \frac{x ^{6} - 1}{3}

= x^{7} - x^{3} + 2 x^{2} \frac{x ^{6} - 1}{3}

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What is the integral from 1 to x^2 of 2x^2y(x+y) ?
The integral from 1 to x^2 of 2x^2y(x+y) is x^7-x^3+2x^2(x^6-1)/3