What is the integral from 1 to 2 of (6-5x)^4 ?

The integral from 1 to 2 of (6-5x)^4 is 41

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integral from 1 to 2 of (6-5x)^4

\int_{1}^{2} (6 - 5 x)^{4} d x

41

Solution steps

\int_{1}^{2} (6 - 5 x)^{4} d x

Apply u-substitution

= \int_{1}^{- 4} - \frac{u ^{4}}{5} d u

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

= - \int_{- 4}^{1} - \frac{u ^{4}}{5} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{- 4}^{1} u^{4} d u)

Apply the Power Rule

= - (- \frac{1}{5} [\frac{u ^{5}}{5}]_{- 4}^{1})

Simplify

= \frac{1}{5} [\frac{u ^{5}}{5}]_{- 4}^{1}

Compute the boundaries:

205

= \frac{1}{5} \cdot 205

Simplify

= 41

What is the integral from 1 to 2 of (6-5x)^4 ?
The integral from 1 to 2 of (6-5x)^4 is 41