What is the integral from 0 to 2 of (3x-1)^4 ?

The integral from 0 to 2 of (3x-1)^4 is 1042/5

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integral from 0 to 2 of (3x-1)^4

\int_{0}^{2} (3 x - 1)^{4} d x

\frac{1042}{5}

Solution steps

\int_{0}^{2} (3 x - 1)^{4} d x

Apply u-substitution

: \int_{- 1}^{5} \frac{u ^{4}}{3} d u

= \int_{- 1}^{5} \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3126}{5}

= \frac{1}{3} \cdot \frac{3126}{5}

Simplify

= \frac{1042}{5}

\int_{0}^{4 - x^{2}} (x^{2} y) d y

\int_{1}^{10} \frac{2 ln ( x )}{x} d x

\int_{1}^{3} 2^{2 x} d x

\int_{0}^{x^{2}} \frac{y}{x ^{5} + 1} d y

\int_{2}^{4} x^{2} + 2 x - 3 d x

What is the integral from 0 to 2 of (3x-1)^4 ?
The integral from 0 to 2 of (3x-1)^4 is 1042/5