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

The integral from 1 to 2 of 1/(4x^2) is 1/8

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

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

\frac{1}{8}

Decimal

0.125

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{4} [- \frac{1}{x}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{4} \cdot \frac{1}{2}

Simplify

= \frac{1}{8}

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What is the integral from 1 to 2 of 1/(4x^2) ?
The integral from 1 to 2 of 1/(4x^2) is 1/8