What is the integral from 0 to-8 of (-4\sqrt[3]{x}) ?

The integral from 0 to-8 of (-4\sqrt[3]{x}) is 48\sqrt[3]{-1}

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integral from 0 to-8 of (-4\sqrt[3]{x})

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: [\frac{3}{4} x^{\frac{4}{3}}]_{- 8}^{0}

= - (- 4 [\frac{3}{4} x^{\frac{4}{3}}]_{- 8}^{0})

Simplify

= 4 [\frac{3}{4} x^{\frac{4}{3}}]_{- 8}^{0}

Compute the boundaries:

Simplify

\int \frac{5 x}{( x + 3 ) ( x - 2 )} d x

\int u^{- 5} (5 - u^{5} + u^{10}) d u

\int x^{18} d x

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

\int_{0}^{2 π} 5 d x

What is the integral from 0 to-8 of (-4\sqrt[3]{x}) ?
The integral from 0 to-8 of (-4\sqrt[3]{x}) is 48\sqrt[3]{-1}