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

Popular >

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

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

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

Decimal Notation

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_{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

\int_{- 3 x}^{3 x} z^{3} d z

\int_{0}^{c} sin (3 x) cos (n x) d x

\int_{10}^{14} \frac{1}{12} d x

\int_{- 3}^{3} (2 x^{2} - 18) d x

\int_{2}^{6} \frac{4}{x} d x

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