What is the integral from 0 to 2 of (3t-1)^{60} ?

The integral from 0 to 2 of (3t-1)^{60} is (5^{61}+1)/(183)

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integral from 0 to 2 of (3t-1)^{60}

\int_{0}^{2} (3 t - 1)^{60} d t

\frac{5 ^{61} + 1}{183}

Decimal

2.36984 E 40

Solution steps

\int_{0}^{2} (3 t - 1)^{60} d t

Apply u-substitution

: \int_{- 1}^{5} \frac{u ^{60}}{3} d u

= \int_{- 1}^{5} \frac{u ^{60}}{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^{60} d u

Apply the Power Rule

: [\frac{u ^{61}}{61}]_{- 1}^{5}

= \frac{1}{3} [\frac{u ^{61}}{61}]_{- 1}^{5}

Compute the boundaries:

\frac{5 ^{61}}{61} + \frac{1}{61}

= \frac{1}{3} (\frac{5 ^{61}}{61} + \frac{1}{61})

Simplify

= \frac{5 ^{61} + 1}{183}

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What is the integral from 0 to 2 of (3t-1)^{60} ?
The integral from 0 to 2 of (3t-1)^{60} is (5^{61}+1)/(183)