What is the integral from 1 to 2 of (4x-5)^3 ?

The integral from 1 to 2 of (4x-5)^3 is 5

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integral from 1 to 2 of (4x-5)^3

\int_{1}^{2} (4 x - 5)^{3} d x

5

Solution steps

\int_{1}^{2} (4 x - 5)^{3} d x

Apply u-substitution

: \int_{- 1}^{3} \frac{u ^{3}}{4} d u

= \int_{- 1}^{3} \frac{u ^{3}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 1}^{3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 1}^{3}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{- 1}^{3}

Compute the boundaries:

20

= \frac{1}{4} \cdot 20

Simplify

= 5

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

\int_{1}^{4} 6 x + x d x

\int_{- 3}^{0} x^{3} - 9 x d x

\int_{- 2}^{2} (x^{2} + 3)^{2} d x

\int_{- 1}^{2} (x^{3} + x^{2}) d x

What is the integral from 1 to 2 of (4x-5)^3 ?
The integral from 1 to 2 of (4x-5)^3 is 5