What is the integral from 0 to 2 of (-4x^4+8x^3) ?

The integral from 0 to 2 of (-4x^4+8x^3) is 32/5

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integral from 0 to 2 of (-4x^4+8x^3)

\int_{0}^{2} (- 4 x^{4} + 8 x^{3}) d x

\frac{32}{5}

Decimal

6.4

Solution steps

\int_{0}^{2} - 4 x^{4} + 8 x^{3} 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}^{2} 4 x^{4} d x + \int_{0}^{2} 8 x^{3} d x

\int_{0}^{2} 4 x^{4} d x = \frac{128}{5}

\int_{0}^{2} 8 x^{3} d x = 32

= - \frac{128}{5} + 32

Simplify

- \frac{128}{5} + 32 : \frac{32}{5}

= \frac{32}{5}

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What is the integral from 0 to 2 of (-4x^4+8x^3) ?
The integral from 0 to 2 of (-4x^4+8x^3) is 32/5