What is the integral from 0 to 2 of (3t-1)^{68} ?

The integral from 0 to 2 of (3t-1)^{68} is (5^{69}+1)/(207)

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integral from 0 to 2 of (3t-1)^{68}

\int_{0}^{2} (3 t - 1)^{68} d t

\frac{5 ^{69} + 1}{207}

Decimal

8.18389 E 45

Solution steps

\int_{0}^{2} (3 t - 1)^{68} d t

Apply u-substitution

: \int_{- 1}^{5} \frac{u ^{68}}{3} d u

= \int_{- 1}^{5} \frac{u ^{68}}{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^{68} d u

Apply the Power Rule

: [\frac{u ^{69}}{69}]_{- 1}^{5}

= \frac{1}{3} [\frac{u ^{69}}{69}]_{- 1}^{5}

Compute the boundaries:

\frac{5 ^{69}}{69} + \frac{1}{69}

= \frac{1}{3} (\frac{5 ^{69}}{69} + \frac{1}{69})

Simplify

= \frac{5 ^{69} + 1}{207}

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What is the integral from 0 to 2 of (3t-1)^{68} ?
The integral from 0 to 2 of (3t-1)^{68} is (5^{69}+1)/(207)