What is the integral from 1 to 3 of 6/(x^2) ?

The integral from 1 to 3 of 6/(x^2) is 4

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integral from 1 to 3 of 6/(x^2)

\int_{1}^{3} \frac{6}{x ^{2}} d x

4

Solution steps

\int_{1}^{3} \frac{6}{x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{3} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{3}

= 6 [- \frac{1}{x}]_{1}^{3}

Compute the boundaries:

\frac{2}{3}

= 6 \cdot \frac{2}{3}

Simplify

= 4

\int 121 - x^{2} d x

\frac{\partial}{\partial w} (2 x^{2} w^{3} (z - 1))

\frac{d}{d x} (\frac{4 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial x} (12 x y)

\frac{d}{d x} (\frac{x + 4}{x - 4})

What is the integral from 1 to 3 of 6/(x^2) ?
The integral from 1 to 3 of 6/(x^2) is 4