What is the integral from-1 to 0 of e^{-x} ?

The integral from-1 to 0 of e^{-x} is e-1

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integral from-1 to 0 of e^{-x}

\int_{- 1}^{0} e^{- x} d x

e - 1

Decimal Notation

1.71828 \dots

Solution steps

\int_{- 1}^{0} e^{- x} 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{d}{d x} (5 tan (3 x))

1, \frac{1}{x}, 1, 2

\frac{d}{d x} (tan (g (x)) x)

x \to 0 - lim (x)

\frac{d y}{d x} = - y - 3

What is the integral from-1 to 0 of e^{-x} ?
The integral from-1 to 0 of e^{-x} is e-1