What is the integral from 0 to 3 of (3t-1)^{62} ?

The integral from 0 to 3 of (3t-1)^{62} is (8^{63}+1)/(189)

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integral from 0 to 3 of (3t-1)^{62}

\int_{0}^{3} (3 t - 1)^{62} d t

\frac{8 ^{63} + 1}{189}

Decimal

4.15152 E 54

Solution steps

\int_{0}^{3} (3 t - 1)^{62} d t

Apply u-substitution

: \int_{- 1}^{8} \frac{u ^{62}}{3} d u

= \int_{- 1}^{8} \frac{u ^{62}}{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^{62} d u

Apply the Power Rule

: [\frac{u ^{63}}{63}]_{- 1}^{8}

= \frac{1}{3} [\frac{u ^{63}}{63}]_{- 1}^{8}

Compute the boundaries:

\frac{8 ^{63}}{63} + \frac{1}{63}

= \frac{1}{3} (\frac{8 ^{63}}{63} + \frac{1}{63})

Simplify

= \frac{8 ^{63} + 1}{189}

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What is the integral from 0 to 3 of (3t-1)^{62} ?
The integral from 0 to 3 of (3t-1)^{62} is (8^{63}+1)/(189)