What is the integral from 0 to 3 of (3x-1)^{58} ?

The integral from 0 to 3 of (3x-1)^{58} is (8^{59}+1)/(177)

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integral from 0 to 3 of (3x-1)^{58}

\int_{0}^{3} (3 x - 1)^{58} d x

\frac{8 ^{59} + 1}{177}

Decimal

1.08227 E 51

Solution steps

\int_{0}^{3} (3 x - 1)^{58} d x

Apply u-substitution

: \int_{- 1}^{8} \frac{u ^{58}}{3} d u

= \int_{- 1}^{8} \frac{u ^{58}}{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}^{8} u^{58} d u

Apply the Power Rule

: [\frac{u ^{59}}{59}]_{- 1}^{8}

= \frac{1}{3} [\frac{u ^{59}}{59}]_{- 1}^{8}

Compute the boundaries:

\frac{8 ^{59}}{59} + \frac{1}{59}

= \frac{1}{3} (\frac{8 ^{59}}{59} + \frac{1}{59})

Simplify

= \frac{8 ^{59} + 1}{177}

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What is the integral from 0 to 3 of (3x-1)^{58} ?
The integral from 0 to 3 of (3x-1)^{58} is (8^{59}+1)/(177)