What is the integral from 0 to 1 of 1/((x-2)^2) ?

The integral from 0 to 1 of 1/((x-2)^2) is 1/2

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integral from 0 to 1 of 1/((x-2)^2)

\int_{0}^{1} \frac{1}{( x - 2 ) ^{2}} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{1} \frac{1}{( x - 2 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

x \to 0 lim (\frac{e ^{x} cos ( x )}{4})

\frac{d}{d x} (\frac{e ^{x}}{4 + e ^{x}})

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2} - 36 x}{x ^{2} + 4 x})

\frac{25}{x ( x ^{2} + 6 x + 25 )}

\frac{d x}{d t} = x - 12

What is the integral from 0 to 1 of 1/((x-2)^2) ?
The integral from 0 to 1 of 1/((x-2)^2) is 1/2