What is the integral from 0 to 1 of (8-12x)^2 ?

The integral from 0 to 1 of (8-12x)^2 is 16

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integral from 0 to 1 of (8-12x)^2

\int_{0}^{1} (8 - 12 x)^{2} d x

16

Solution steps

\int_{0}^{1} (8 - 12 x)^{2} d x

Apply u-substitution

= \int_{8}^{- 4} - \frac{u ^{2}}{12} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{8} - \frac{u ^{2}}{12} d u

Take the constant out:

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

= - (- \frac{1}{12} \cdot \int_{- 4}^{8} u^{2} d u)

Apply the Power Rule

= - (- \frac{1}{12} [\frac{u ^{3}}{3}]_{- 4}^{8})

Simplify

= \frac{1}{12} [\frac{u ^{3}}{3}]_{- 4}^{8}

Compute the boundaries:

192

= \frac{1}{12} \cdot 192

Simplify

= 16

What is the integral from 0 to 1 of (8-12x)^2 ?
The integral from 0 to 1 of (8-12x)^2 is 16