What is the integral from 0 to 3 of (1-e^{-x}) ?

The integral from 0 to 3 of (1-e^{-x}) is 1/(e^3)+2

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integral from 0 to 3 of (1-e^{-x})

\int_{0}^{3} (1 - e^{- x}) d x

\frac{1}{e ^{3}} + 2

Decimal Notation

2.04978 \dots

Solution steps

\int_{0}^{3} 1 - e^{- x} d x

Apply u-substitution

: \int_{0}^{- 3} - 1 + e^{u} d u

= \int_{0}^{- 3} - 1 + e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 3}^{0} - 1 + e^{u} d u

Apply the Sum Rule:

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

= - (- \int_{- 3}^{0} 1 d u + \int_{- 3}^{0} e^{u} d u)

\int_{- 3}^{0} 1 d u = 3

\int_{- 3}^{0} e^{u} d u = 1 - \frac{1}{e ^{3}}

= - (- 3 + 1 - \frac{1}{e ^{3}})

Simplify

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

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