What is the integral from 2 to 5 of e^{-x} ?

The integral from 2 to 5 of e^{-x} is 1/(e^2)-1/(e^5)

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integral from 2 to 5 of e^{-x}

\int_{2}^{5} e^{- x} d x

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

Decimal Notation

0.12859 \dots

Solution steps

\int_{2}^{5} e^{- x} d x

Apply u-substitution

: \int_{- 2}^{- 5} - e^{u} d u

= \int_{- 2}^{- 5} - e^{u} d u

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

= - \int_{- 5}^{- 2} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 5}^{- 2} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 5}^{- 2})

Simplify

= [e^{u}]_{- 5}^{- 2}

Compute the boundaries:

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

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

\frac{\partial}{\partial y} (3 y^{2} cos (4 x))

(4 + x) \frac{d y}{d x} + y = 4 + x

\frac{\partial}{\partial y} (- 6 x y)

\frac{5}{x}

f (x) = 3 x - ln (x^{4})

What is the integral from 2 to 5 of e^{-x} ?
The integral from 2 to 5 of e^{-x} is 1/(e^2)-1/(e^5)