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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

= - (- \int_{1}^{2} \frac{1}{u ^{2}} d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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