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

The integral from 1 to 2 of (4x-2)^3 is 80

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

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

80

Solution steps

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

Apply u-substitution

: \int_{2}^{6} \frac{u ^{3}}{4} d u

= \int_{2}^{6} \frac{u ^{3}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{2}^{6} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2}^{6}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{6}

Compute the boundaries:

320

= \frac{1}{4} \cdot 320

Simplify

= 80

\int_{0}^{4} t (4 - t) d t

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

\int_{- 1}^{4} 2 x + 1 d x

\int_{0}^{1} x^{2} + x + 1 d x

\int_{0}^{3} e^{\frac{x}{3}} d x

What is the integral from 1 to 2 of (4x-2)^3 ?
The integral from 1 to 2 of (4x-2)^3 is 80