What is the integral from-2 to 1 of 2\sqrt[3]{2-3x} ?

The integral from-2 to 1 of 2\sqrt[3]{2-3x} is 15/2

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integral from-2 to 1 of 2\sqrt[3]{2-3x}

\frac{15}{2}

Decimal Notation

7.5

Solution steps

Take the constant out:

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

Apply u-substitution

\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} u^{\frac{4}{3}}]_{- 1}^{8}

= 2 (- (- \frac{1}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{8}))

Simplify

2 (- (- \frac{1}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{8})) : \frac{2}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{8}

= \frac{2}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{8}

Compute the boundaries:

\frac{45}{4}

= \frac{2}{3} \cdot \frac{45}{4}

Simplify

= \frac{15}{2}

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What is the integral from-2 to 1 of 2\sqrt[3]{2-3x} ?
The integral from-2 to 1 of 2\sqrt[3]{2-3x} is 15/2