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

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

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

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

e - 1

Decimal Notation

1.71828 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{1}^{0} - e^{u} d u

= \int_{1}^{0} - e^{u} d u

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

= - \int_{0}^{1} - e^{u} d u

Take the constant out:

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

= - (- \int_{0}^{1} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{0}^{1})

Simplify

= [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= e - 1

\frac{6}{5 x + 3}

u (x) = 2 x

\frac{d}{d x} (cos (e^{3 x}) - sin (e^{3 x}))

\int 5 sin (x) ln (sin (x)) d x

n = 0 \sum \infty n^{0.5}

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