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

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

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

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

e - \frac{1}{e}

Decimal

2.35040 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

e - \frac{1}{e}

= e - \frac{1}{e}

f (x) = \frac{1}{( 1 + x ^{2} )}, at x = - 1

\frac{5}{x}, at m

n = 0 \sum \infty (\frac{( - 1 ) ^{n}}{2 ^{n}}) x^{2 n}

\frac{d y}{d x} = x (x - 2)

ln (6 x)

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