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

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

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

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

- \frac{8}{5}

Decimal Notation

- 1.6

Solution steps

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

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

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

= \frac{32}{5} - 8

Simplify

\frac{32}{5} - 8 : - \frac{8}{5}

= - \frac{8}{5}

\int cos^{5} (3 x) sin (3 x) d x

\frac{\partial}{\partial x} (\frac{4 x}{x ^{2} + y ^{2}})

x \to - 7 lim (2 x + 5)

\frac{\partial}{\partial x} (ln (2 x^{3} + 3 y^{2}))

\frac{5}{( 2 x - 3 ) ^{4}}

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