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

The integral from 1 to 2 of 2/((2x-5)^3) is -4/9

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

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

- \frac{4}{9}

Decimal Notation

- 0.44444 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{1}^{2} \frac{1}{( 2 x - 5 ) ^{3}} d x

Apply u-substitution

: \int_{- 3}^{- 1} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{- 3}^{- 1} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{- 3}^{- 1} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{- 3}^{- 1}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{- 3}^{- 1}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{- 3}^{- 1} : [- \frac{1}{2 u ^{2}}]_{- 3}^{- 1}

= [- \frac{1}{2 u ^{2}}]_{- 3}^{- 1}

Compute the boundaries:

- \frac{4}{9}

= - \frac{4}{9}

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What is the integral from 1 to 2 of 2/((2x-5)^3) ?
The integral from 1 to 2 of 2/((2x-5)^3) is -4/9