What is the integral from 0 to 1 of 16(2+4x)^3 ?

The integral from 0 to 1 of 16(2+4x)^3 is 1280

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integral from 0 to 1 of 16(2+4x)^3

\int_{0}^{1} 16 (2 + 4 x)^{3} d x

1280

Solution steps

\int_{0}^{1} 16 (2 + 4 x)^{3} d x

Take the constant out:

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

= 16 \cdot \int_{0}^{1} (2 + 4 x)^{3} d x

Apply u-substitution

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

= 16 \cdot \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

= 16 \cdot \frac{1}{4} \cdot \int_{2}^{6} u^{3} d u

Apply the Power Rule

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

= 16 \cdot \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{6}

Simplify

16 \cdot \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{6} : 4 [\frac{u ^{4}}{4}]_{2}^{6}

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

Compute the boundaries:

320

= 4 \cdot 320

Simplify

= 1280

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What is the integral from 0 to 1 of 16(2+4x)^3 ?
The integral from 0 to 1 of 16(2+4x)^3 is 1280