What is the integral from-2 to 0 of (u^5-u^3+u^2) ?

The integral from-2 to 0 of (u^5-u^3+u^2) is -4

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integral from-2 to 0 of (u^5-u^3+u^2)

\int_{- 2}^{0} (u^{5} - u^{3} + u^{2}) d u

- 4

Solution steps

\int_{- 2}^{0} u^{5} - u^{3} + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 2}^{0} u^{5} d u - \int_{- 2}^{0} u^{3} d u + \int_{- 2}^{0} u^{2} d u

\int_{- 2}^{0} u^{5} d u = - \frac{32}{3}

\int_{- 2}^{0} u^{3} d u = - 4

\int_{- 2}^{0} u^{2} d u = \frac{8}{3}

= - \frac{32}{3} - (- 4) + \frac{8}{3}

Simplify

- \frac{32}{3} - (- 4) + \frac{8}{3} : - 4

= - 4

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\frac{d ^{10}}{d x ^{10}} ((2 x + 5)^{7})

\int \frac{x ^{2}}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (e^{- x} - e^{- 2 x})

What is the integral from-2 to 0 of (u^5-u^3+u^2) ?
The integral from-2 to 0 of (u^5-u^3+u^2) is -4