What is the integral from 1 to 2 of (y-1/(y^2)) ?

The integral from 1 to 2 of (y-1/(y^2)) is 1

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integral from 1 to 2 of (y-1/(y^2))

\int_{1}^{2} (y - \frac{1}{y ^{2}}) d y

1

Solution steps

\int_{1}^{2} y - \frac{1}{y ^{2}} d y

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}^{2} y d y - \int_{1}^{2} \frac{1}{y ^{2}} d y

\int_{1}^{2} y d y = \frac{3}{2}

\int_{1}^{2} \frac{1}{y ^{2}} d y = \frac{1}{2}

= \frac{3}{2} - \frac{1}{2}

Simplify

\frac{3}{2} - \frac{1}{2} : 1

= 1

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What is the integral from 1 to 2 of (y-1/(y^2)) ?
The integral from 1 to 2 of (y-1/(y^2)) is 1