What is the integral from 0 to 2 of (z^4-(z^5)/2) ?

The integral from 0 to 2 of (z^4-(z^5)/2) is 16/15

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integral from 0 to 2 of (z^4-(z^5)/2)

\int_{0}^{2} (z^{4} - \frac{z ^{5}}{2}) d z

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

\int_{0}^{2} z^{4} - \frac{z ^{5}}{2} d z

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}^{2} z^{4} d z - \int_{0}^{2} \frac{z ^{5}}{2} d z

\int_{0}^{2} z^{4} d z = \frac{32}{5}

\int_{0}^{2} \frac{z ^{5}}{2} d z = \frac{16}{3}

= \frac{32}{5} - \frac{16}{3}

Simplify

\frac{32}{5} - \frac{16}{3} : \frac{16}{15}

= \frac{16}{15}

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\int_{- 3}^{2} (12 - x^{2} - x) d x

What is the integral from 0 to 2 of (z^4-(z^5)/2) ?
The integral from 0 to 2 of (z^4-(z^5)/2) is 16/15