What is the integral from 0 to 5 of e^{-x} ?

The integral from 0 to 5 of e^{-x} is 1-1/(e^5)

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integral from 0 to 5 of e^{-x}

\int_{0}^{5} e^{- x} d x

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

Decimal Notation

0.99326 \dots

Solution steps

\int_{0}^{5} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 5} - e^{u} d u

= \int_{0}^{- 5} - e^{u} d u

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

= - \int_{- 5}^{0} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 5}^{0} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- [e^{u}]_{- 5}^{0})

Simplify

= [e^{u}]_{- 5}^{0}

Compute the boundaries:

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

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

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 3

\frac{\partial}{\partial x} (x^{2} + y^{2} + \frac{2}{x y})

r^{2} - 16 x^{2}

\frac{\partial}{\partial x} (y - x^{2})

\frac{\partial}{\partial x} (r cos (x))

What is the integral from 0 to 5 of e^{-x} ?
The integral from 0 to 5 of e^{-x} is 1-1/(e^5)