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

The integral from 1 to 4 of (6/(x^2)) is 9/2

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

\int_{1}^{4} (\frac{6}{x ^{2}}) d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{1}^{4} \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}^{4} \frac{1}{x ^{2}} d x

Apply the Power Rule

= 6 [- \frac{1}{x}]_{1}^{4}

Compute the boundaries:

\frac{3}{4}

= 6 \cdot \frac{3}{4}

Simplify

= \frac{9}{2}

What is the integral from 1 to 4 of (6/(x^2)) ?
The integral from 1 to 4 of (6/(x^2)) is 9/2