What is the integral from 0 to 4 of (1-e^{-x}) ?

The integral from 0 to 4 of (1-e^{-x}) is 1/(e^4)+3

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integral from 0 to 4 of (1-e^{-x})

\int_{0}^{4} (1 - e^{- x}) d x

\frac{1}{e ^{4}} + 3

Decimal

3.01831 \dots

Solution steps

\int_{0}^{4} 1 - e^{- x} d x

Apply u-substitution

= \int_{0}^{- 4} - 1 + e^{u} d u

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

= - \int_{- 4}^{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_{- 4}^{0} 1 d u + \int_{- 4}^{0} e^{u} d u)

\int_{- 4}^{0} 1 d u = 4

\int_{- 4}^{0} e^{u} d u = 1 - \frac{1}{e ^{4}}

= - (- 4 + 1 - \frac{1}{e ^{4}})

Simplify

= \frac{1}{e ^{4}} + 3

What is the integral from 0 to 4 of (1-e^{-x}) ?
The integral from 0 to 4 of (1-e^{-x}) is 1/(e^4)+3