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

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

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

\int_{1}^{2} \frac{2}{x ^{2}} d x

1

Solution steps

\int_{1}^{2} \frac{2}{x ^{2}} 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}{x ^{2}} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2}

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

Simplify

= 1

\frac{d}{d x} (\frac{3 x - x ^{2}}{x} - 4 cot (x))

f (x) = 3 cos (cos (x^{6}))

\int_{- \infty}^{0} \frac{1}{2 - 10 x} d x

(x^{2} sin (x) + 4 y) d x + x d y = 0

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

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