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

The integral from 1 to 2 of (3x-1)^3 is 203/4

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

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

\frac{203}{4}

Decimal

50.75

Solution steps

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

Apply u-substitution

: \int_{2}^{5} \frac{u ^{3}}{3} d u

= \int_{2}^{5} \frac{u ^{3}}{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_{2}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{5}

= \frac{1}{3} [\frac{u ^{4}}{4}]_{2}^{5}

Compute the boundaries:

\frac{609}{4}

= \frac{1}{3} \cdot \frac{609}{4}

Simplify

= \frac{203}{4}

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What is the integral from 1 to 2 of (3x-1)^3 ?
The integral from 1 to 2 of (3x-1)^3 is 203/4