What is the integral from 3 to 7 of 4(x-5)^3 ?

The integral from 3 to 7 of 4(x-5)^3 is 0

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integral from 3 to 7 of 4(x-5)^3

\int_{3}^{7} 4 (x - 5)^{3} d x

0

Solution steps

\int_{3}^{7} 4 (x - 5)^{3} d x

Take the constant out:

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

= 4 \cdot \int_{3}^{7} (x - 5)^{3} d x

Apply u-substitution

: \int_{- 2}^{2} u^{3} d u

= 4 \cdot \int_{- 2}^{2} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 2}^{2}

= 4 [\frac{u ^{4}}{4}]_{- 2}^{2}

Compute the boundaries:

0

= 4 \cdot 0

Simplify

= 0

\int_{5}^{\infty} x e^{- \frac{x}{2}} d x

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What is the integral from 3 to 7 of 4(x-5)^3 ?
The integral from 3 to 7 of 4(x-5)^3 is 0