What is the integral from 1 to 2 of (10-9x)^2 ?

The integral from 1 to 2 of (10-9x)^2 is 19

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integral from 1 to 2 of (10-9x)^2

\int_{1}^{2} (10 - 9 x)^{2} d x

19

Solution steps

\int_{1}^{2} (10 - 9 x)^{2} d x

Apply u-substitution

: \int_{1}^{- 8} - \frac{u ^{2}}{9} d u

= \int_{1}^{- 8} - \frac{u ^{2}}{9} d u

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

= - \int_{- 8}^{1} - \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_{- 8}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 8}^{1}

= - (- \frac{1}{9} [\frac{u ^{3}}{3}]_{- 8}^{1})

Simplify

= \frac{1}{9} [\frac{u ^{3}}{3}]_{- 8}^{1}

Compute the boundaries:

171

= \frac{1}{9} \cdot 171

Simplify

= 19

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What is the integral from 1 to 2 of (10-9x)^2 ?
The integral from 1 to 2 of (10-9x)^2 is 19