What is the integral from 5 to 10 of ((5-x)^4)/(625) ?

The integral from 5 to 10 of ((5-x)^4)/(625) is 1

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integral from 5 to 10 of ((5-x)^4)/(625)

\int_{5}^{10} \frac{( 5 - x ) ^{4}}{625} d x

1

Solution steps

\int_{5}^{10} \frac{( 5 - x ) ^{4}}{625} d x

Take the constant out:

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

= \frac{1}{625} \cdot \int_{5}^{10} (5 - x)^{4} d x

Apply u-substitution

: \int_{0}^{- 5} - u^{4} d u

= \frac{1}{625} \cdot \int_{0}^{- 5} - u^{4} d u

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

= \frac{1}{625} (- \int_{- 5}^{0} - u^{4} d u)

Take the constant out:

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

= \frac{1}{625} (- (- \int_{- 5}^{0} u^{4} d u))

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 5}^{0}

= \frac{1}{625} (- (- [\frac{u ^{5}}{5}]_{- 5}^{0}))

Simplify

= \frac{1}{625} [\frac{u ^{5}}{5}]_{- 5}^{0}

Compute the boundaries:

625

= \frac{1}{625} \cdot 625

Simplify

= 1

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What is the integral from 5 to 10 of ((5-x)^4)/(625) ?
The integral from 5 to 10 of ((5-x)^4)/(625) is 1