What is the integral from 0 to 1 of 3/2 (x^2+y^2) ?

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

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integral from 0 to 1 of 3/2 (x^2+y^2)

\int_{0}^{1} \frac{3}{2} (x^{2} + y^{2}) d x

\frac{3}{2} y^{2} + \frac{1}{2}

Solution steps

\int_{0}^{1} \frac{3}{2} (x^{2} + y^{2}) d x

Take the constant out:

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

= \frac{3}{2} \cdot \int_{0}^{1} y^{2} + x^{2} d x

Apply the Sum Rule:

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

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

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

Simplify

\frac{3}{2} (y^{2} + \frac{1}{3}) : \frac{3}{2} y^{2} + \frac{1}{2}

= \frac{3}{2} y^{2} + \frac{1}{2}

\frac{d}{d x} (7^{x} (- 6 x + \frac{2}{x}))

\frac{\partial}{\partial x} (2 x \cdot sin (9 x^{2} y))

\int \frac{1}{25 x ^{2} + 1} d x

n = 1 \sum \infty \frac{3}{4 + 6 n}

n = 1 \sum \infty \frac{2}{3 n + 5}

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