What is the integral from 0 to 3 of (3t-1)^{52} ?

The integral from 0 to 3 of (3t-1)^{52} is (8^{53}+1)/(159)

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integral from 0 to 3 of (3t-1)^{52}

\int_{0}^{3} (3 t - 1)^{52} d t

\frac{8 ^{53} + 1}{159}

Decimal Notation

4.59592 E 45

Solution steps

\int_{0}^{3} (3 t - 1)^{52} d t

Apply u-substitution

: \int_{- 1}^{8} \frac{u ^{52}}{3} d u

= \int_{- 1}^{8} \frac{u ^{52}}{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^{52} d u

Apply the Power Rule

: [\frac{u ^{53}}{53}]_{- 1}^{8}

= \frac{1}{3} [\frac{u ^{53}}{53}]_{- 1}^{8}

Compute the boundaries:

\frac{8 ^{53}}{53} + \frac{1}{53}

= \frac{1}{3} (\frac{8 ^{53}}{53} + \frac{1}{53})

Simplify

= \frac{8 ^{53} + 1}{159}

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What is the integral from 0 to 3 of (3t-1)^{52} ?
The integral from 0 to 3 of (3t-1)^{52} is (8^{53}+1)/(159)