What is the integral from 2 to 3 of 7/((5x-5)^2) ?

The integral from 2 to 3 of 7/((5x-5)^2) is 7/50

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integral from 2 to 3 of 7/((5x-5)^2)

\int_{2}^{3} \frac{7}{( 5 x - 5 ) ^{2}} d x

\frac{7}{50}

Decimal

0.14

Solution steps

\int_{2}^{3} \frac{7}{( 5 x - 5 ) ^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{2}^{3} \frac{1}{( 5 x - 5 ) ^{2}} d x

Take the partial fraction of

\frac{1}{( 5 x - 5 ) ^{2}} : \frac{1}{25 ( x - 1 ) ^{2}}

= 7 \cdot \int_{2}^{3} \frac{1}{25 ( x - 1 ) ^{2}} d x

Take the constant out:

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

= 7 \cdot \frac{1}{25} \cdot \int_{2}^{3} \frac{1}{( x - 1 ) ^{2}} d x

Apply u-substitution

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

= 7 \cdot \frac{1}{25} \cdot \int_{1}^{2} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

= 7 \cdot \frac{1}{25} [- \frac{1}{u}]_{1}^{2}

Simplify

7 \cdot \frac{1}{25} [- \frac{1}{u}]_{1}^{2} : \frac{7}{25} [- \frac{1}{u}]_{1}^{2}

= \frac{7}{25} [- \frac{1}{u}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{7}{25} \cdot \frac{1}{2}

Simplify

= \frac{7}{50}

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What is the integral from 2 to 3 of 7/((5x-5)^2) ?
The integral from 2 to 3 of 7/((5x-5)^2) is 7/50