What is the integral from 0 to 1 of (y^4+y^8) ?

The integral from 0 to 1 of (y^4+y^8) is 14/45

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integral from 0 to 1 of (y^4+y^8)

\int_{0}^{1} (y^{4} + y^{8}) d y

\frac{14}{45}

Decimal

0.31111 \dots

Solution steps

\int_{0}^{1} y^{4} + y^{8} 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_{0}^{1} y^{4} d y + \int_{0}^{1} y^{8} d y

\int_{0}^{1} y^{4} d y = \frac{1}{5}

\int_{0}^{1} y^{8} d y = \frac{1}{9}

= \frac{1}{5} + \frac{1}{9}

Simplify

\frac{1}{5} + \frac{1}{9} : \frac{14}{45}

= \frac{14}{45}

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What is the integral from 0 to 1 of (y^4+y^8) ?
The integral from 0 to 1 of (y^4+y^8) is 14/45