What is the integral from 0 to 1 of (6-9x)^2 ?

The integral from 0 to 1 of (6-9x)^2 is 9

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integral from 0 to 1 of (6-9x)^2

\int_{0}^{1} (6 - 9 x)^{2} d x

9

Solution steps

\int_{0}^{1} (6 - 9 x)^{2} d x

Apply u-substitution

: \int_{6}^{- 3} - \frac{u ^{2}}{9} d u

= \int_{6}^{- 3} - \frac{u ^{2}}{9} d u

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

= - \int_{- 3}^{6} - \frac{u ^{2}}{9} d u

Take the constant out:

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

= - (- \frac{1}{9} \cdot \int_{- 3}^{6} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 3}^{6}

= - (- \frac{1}{9} [\frac{u ^{3}}{3}]_{- 3}^{6})

Simplify

= \frac{1}{9} [\frac{u ^{3}}{3}]_{- 3}^{6}

Compute the boundaries:

81

= \frac{1}{9} \cdot 81

Simplify

= 9

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What is the integral from 0 to 1 of (6-9x)^2 ?
The integral from 0 to 1 of (6-9x)^2 is 9