What is the integral from 1 to 2.5 of 3/(2x^2) ?

The integral from 1 to 2.5 of 3/(2x^2) is 0.9

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integral from 1 to 2.5 of 3/(2x^2)

\int_{1}^{2.5} \frac{3}{2 x ^{2}} d x

0.9

Solution steps

\int_{1}^{2.5} \frac{3}{2 x ^{2}} d x

Take the constant out:

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

= \frac{3}{2} \cdot \int_{1}^{2.5} \frac{1}{x ^{2}} d x

Apply the Power Rule

= \frac{3}{2} [- \frac{1}{x}]_{1}^{2.5}

Simplify

= 1.5 [- \frac{1}{x}]_{1}^{2.5}

Compute the boundaries:

0.6

= 1.5 \cdot 0.6

Simplify

= 0.9

What is the integral from 1 to 2.5 of 3/(2x^2) ?
The integral from 1 to 2.5 of 3/(2x^2) is 0.9