What is the integral from-1 to 2 of (x^4-x^5) ?

The integral from-1 to 2 of (x^4-x^5) is -39/10

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integral from-1 to 2 of (x^4-x^5)

\int_{- 1}^{2} (x^{4} - x^{5}) d x

- \frac{39}{10}

Decimal

- 3.9

Solution steps

\int_{- 1}^{2} x^{4} - x^{5} 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_{- 1}^{2} x^{4} d x - \int_{- 1}^{2} x^{5} d x

\int_{- 1}^{2} x^{4} d x = \frac{33}{5}

\int_{- 1}^{2} x^{5} d x = \frac{21}{2}

= \frac{33}{5} - \frac{21}{2}

Simplify

\frac{33}{5} - \frac{21}{2} : - \frac{39}{10}

= - \frac{39}{10}

What is the integral from-1 to 2 of (x^4-x^5) ?
The integral from-1 to 2 of (x^4-x^5) is -39/10