What is the integral from 0 to 2 of (3t-1)^{56} ?

The integral from 0 to 2 of (3t-1)^{56} is (5^{57}+1)/(171)

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integral from 0 to 2 of (3t-1)^{56}

\int_{0}^{2} (3 t - 1)^{56} d t

\frac{5 ^{57} + 1}{171}

Decimal

4.05783 E 37

Solution steps

\int_{0}^{2} (3 t - 1)^{56} d t

Apply u-substitution

: \int_{- 1}^{5} \frac{u ^{56}}{3} d u

= \int_{- 1}^{5} \frac{u ^{56}}{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}^{5} u^{56} d u

Apply the Power Rule

: [\frac{u ^{57}}{57}]_{- 1}^{5}

= \frac{1}{3} [\frac{u ^{57}}{57}]_{- 1}^{5}

Compute the boundaries:

\frac{5 ^{57}}{57} + \frac{1}{57}

= \frac{1}{3} (\frac{5 ^{57}}{57} + \frac{1}{57})

Simplify

= \frac{5 ^{57} + 1}{171}

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What is the integral from 0 to 2 of (3t-1)^{56} ?
The integral from 0 to 2 of (3t-1)^{56} is (5^{57}+1)/(171)