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

The integral from 1 to 2 of (y^2+1)/(y^2) is 3/2

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

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

\frac{y ^{2} + 1}{y ^{2}} = \frac{1}{y ^{2}} + 1

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{y ^{2}} = y^{- 2}

= \int_{1}^{2} y^{- 2} + 1 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^{- 2} d y + \int_{1}^{2} 1 d y

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

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

= \frac{1}{2} + 1

Simplify

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

= \frac{3}{2}

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What is the integral from 1 to 2 of (y^2+1)/(y^2) ?
The integral from 1 to 2 of (y^2+1)/(y^2) is 3/2