What is the integral from 0 to 1 of 3/2+e^{-y} ?

The integral from 0 to 1 of 3/2+e^{-y} is 5/2-1/e

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integral from 0 to 1 of 3/2+e^{-y}

\int_{0}^{1} \frac{3}{2} + e^{- y} d y

\frac{5}{2} - \frac{1}{e}

Decimal Notation

2.13212 \dots

Solution steps

\int_{0}^{1} \frac{3}{2} + e^{- y} 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_{0}^{1} \frac{3}{2} d y + \int_{0}^{1} e^{- y} d y

\int_{0}^{1} \frac{3}{2} d y = \frac{3}{2}

\int_{0}^{1} e^{- y} d y = 1 - \frac{1}{e}

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

Simplify

\frac{3}{2} + 1 - \frac{1}{e} : \frac{5}{2} - \frac{1}{e}

= \frac{5}{2} - \frac{1}{e}

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What is the integral from 0 to 1 of 3/2+e^{-y} ?
The integral from 0 to 1 of 3/2+e^{-y} is 5/2-1/e