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

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

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

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

x^{2} + \frac{7}{3}

Solution steps

\int_{1}^{2} x^{2} + 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

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

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

\int_{1}^{2} y^{2} d y = \frac{7}{3}

= x^{2} + \frac{7}{3}

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