What is the integral from 1 to 2 of e^{-x} ?

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

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integral from 1 to 2 of e^{-x}

\int_{1}^{2} e^{- x} d x

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

Decimal Notation

0.23254 \dots

Solution steps

\int_{1}^{2} e^{- x} d x

Apply u-substitution

: \int_{- 1}^{- 2} - e^{u} d u

= \int_{- 1}^{- 2} - e^{u} d u

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

= - \int_{- 2}^{- 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}^{- 1} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 2}^{- 1})

Simplify

= [e^{u}]_{- 2}^{- 1}

Compute the boundaries:

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

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

4 x = y^{2} + 12, 3 \leq x \leq 11

\int_{0}^{\frac{π}{2}} ∣ 7 sin (x) - 7 cos (2 x) ∣ d x

y = \frac{2}{x ^{2}}

\int_{6}^{\infty} \frac{1}{x ^{2} + x} d x

\frac{d y}{d t} = 2 - y

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