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

The integral from 0 to 3 of (3t-1)^{68} is (8^{69}+1)/(207)

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

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

\frac{8 ^{69} + 1}{207}

Decimal

9.93662 E 59

Solution steps

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

Apply u-substitution

= \int_{- 1}^{8} \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}^{8} u^{68} d u

Apply the Power Rule

= \frac{1}{3} [\frac{u ^{69}}{69}]_{- 1}^{8}

Compute the boundaries:

\frac{8 ^{69}}{69} + \frac{1}{69}

= \frac{1}{3} (\frac{8 ^{69}}{69} + \frac{1}{69})

Simplify

= \frac{8 ^{69} + 1}{207}

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What is the integral from 0 to 3 of (3t-1)^{68} ?
The integral from 0 to 3 of (3t-1)^{68} is (8^{69}+1)/(207)