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

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

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

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

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

Solution steps

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

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

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

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

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

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

\int t^{2} e^{- 6 t} d t

n = 1 \sum \infty n!

\frac{d}{d x} (12 x^{3} + 6 x^{2} + 2 x)

y = cos (a^{5} + x^{5})

f (x) = 3 x^{2} + 2

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