What is the integral from 1 to 1.5 of x^2+1/(4x^2) ?

The integral from 1 to 1.5 of x^2+1/(4x^2) is 0.875

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integral from 1 to 1.5 of x^2+1/(4x^2)

\int_{1}^{1.5} x^{2} + \frac{1}{4 x ^{2}} d x

0.875

Solution steps

\int_{1}^{1.5} x^{2} + \frac{1}{4 x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{1.5} x^{2} d x + \int_{1}^{1.5} \frac{1}{4 x ^{2}} d x

\int_{1}^{1.5} x^{2} d x = 0.79166 \dots

\int_{1}^{1.5} \frac{1}{4 x ^{2}} d x = 0.08333 \dots

= 0.79166 \dots + 0.08333 \dots

Simplify

= 0.875

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What is the integral from 1 to 1.5 of x^2+1/(4x^2) ?
The integral from 1 to 1.5 of x^2+1/(4x^2) is 0.875