What is the integral from 0 to 1 of (2-3x)^4 ?

The integral from 0 to 1 of (2-3x)^4 is 11/5

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integral from 0 to 1 of (2-3x)^4

\int_{0}^{1} (2 - 3 x)^{4} d x

\frac{11}{5}

Decimal

2.2

Solution steps

\int_{0}^{1} (2 - 3 x)^{4} d x

Apply u-substitution

: \int_{2}^{- 1} - \frac{u ^{4}}{3} d u

= \int_{2}^{- 1} - \frac{u ^{4}}{3} d u

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

= - \int_{- 1}^{2} - \frac{u ^{4}}{3} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 1}^{2} u^{4} d u)

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 1}^{2}

= - (- \frac{1}{3} [\frac{u ^{5}}{5}]_{- 1}^{2})

Simplify

= \frac{1}{3} [\frac{u ^{5}}{5}]_{- 1}^{2}

Compute the boundaries:

\frac{33}{5}

= \frac{1}{3} \cdot \frac{33}{5}

Simplify

= \frac{11}{5}

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What is the integral from 0 to 1 of (2-3x)^4 ?
The integral from 0 to 1 of (2-3x)^4 is 11/5