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

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

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

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

- \frac{32}{5}

Decimal Notation

- 6.4

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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)

Join

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

= - \frac{32}{5}

\int_{- 1}^{1} ∣ e^{x} - 1 ∣ d x

\int_{2}^{4} 3 x^{2} d x

\int_{- 1}^{1} x + 2 x^{2} - x^{3} + 5 x^{4} d x

\int_{0}^{1} (x + 3) d x

\int_{0}^{c o s (x)} e^{- t^{3}} d t

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