What is the integral from 0.25 to 0.6 of 5(1-y)^4 ?

The integral from 0.25 to 0.6 of 5(1-y)^4 is 0.2270646875

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integral from 0.25 to 0.6 of 5(1-y)^4

\int_{0.25}^{0.6} 5 (1 - y)^{4} d y

0.2270646875

Solution steps

\int_{0.25}^{0.6} 5 (1 - y)^{4} d y

Take the constant out:

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

= 5 \cdot \int_{0.25}^{0.6} (1 - y)^{4} d y

Apply u-substitution

: \int_{0.75}^{0.4} - u^{4} d u

= 5 \cdot \int_{0.75}^{0.4} - u^{4} d u

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

= 5 (- \int_{0.4}^{0.75} - u^{4} d u)

Take the constant out:

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

= 5 (- (- \int_{0.4}^{0.75} u^{4} d u))

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0.4}^{0.75}

= 5 (- (- [\frac{u ^{5}}{5}]_{0.4}^{0.75}))

Simplify

= 5 [0.2 u^{5}]_{0.4}^{0.75}

Compute the boundaries:

0.0454129375

= 5 \cdot 0.0454129375

Simplify

= 0.2270646875

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What is the integral from 0.25 to 0.6 of 5(1-y)^4 ?
The integral from 0.25 to 0.6 of 5(1-y)^4 is 0.2270646875