What is the integral from 1 to 2 of (x-2)^{-2} ?

The integral from 1 to 2 of (x-2)^{-2} is diverges

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integral from 1 to 2 of (x-2)^{-2}

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

diverges

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{u ^{2}} d u

= \int_{- 1}^{0} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{- 1}^{0}

= [- \frac{1}{u}]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\frac{2}{x ^{6}}

\int_{0}^{1} sin (π x cos (π) x) d x

\int 4 x^{3} + 1 d x

f (x) = x^{3} - 9 x (3)

What is the integral from 1 to 2 of (x-2)^{-2} ?
The integral from 1 to 2 of (x-2)^{-2} is diverges