What is the integral from 1 to 2.5 of e^{-x} ?

The integral from 1 to 2.5 of e^{-x} is 1/e-0.08208…

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integral from 1 to 2.5 of e^{-x}

\int_{1}^{2.5} e^{- x} d x

\frac{1}{e} - 0.08208 \dots

Decimal

0.28579 \dots

Solution steps

\int_{1}^{2.5} e^{- x} d x

Apply u-substitution

: \int_{- 1}^{- 2.5} - e^{u} d u

= \int_{- 1}^{- 2.5} - e^{u} d u

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

= - \int_{- 2.5}^{- 1} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 2.5}^{- 1} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 2.5}^{- 1})

Simplify

= [e^{u}]_{- 2.5}^{- 1}

Compute the boundaries:

\frac{1}{e} - 0.08208 \dots

= \frac{1}{e} - 0.08208 \dots

\frac{d}{d x} (\frac{x ^{2} - 2 x + 1}{x + 1})

\int_{- \infty}^{- 3} x e^{x} d x

\int \frac{y}{y + 2} d y

\frac{d}{d x} (\frac{e ^{x}}{6 + e ^{x}})

\frac{d}{d x} (\frac{64 x}{x ^{2} + 64})

What is the integral from 1 to 2.5 of e^{-x} ?
The integral from 1 to 2.5 of e^{-x} is 1/e-0.08208…