What is the integral from 0 to 100 of e^{-0.01x} ?

The integral from 0 to 100 of e^{-0.01x} is 100(1-1/e)

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integral from 0 to 100 of e^{-0.01x}

\int_{0}^{100} e^{- 0.01 x} d x

100 (1 - \frac{1}{e})

Decimal Notation

63.21205 \dots

Solution steps

\int_{0}^{100} e^{- 0.01 x} d x

Apply u-substitution

: \int_{0}^{- 1} - \frac{1}{0.01} e^{u} d u

= \int_{0}^{- 1} - \frac{1}{0.01} e^{u} d u

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

= - \int_{- 1}^{0} - \frac{1}{0.01} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.01} \cdot \int_{- 1}^{0} e^{u} d u)

Simplify

= - (- 100 \cdot \int_{- 1}^{0} e^{u} d u)

Use the common integral:

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

= - (- 100 [e^{u}]_{- 1}^{0})

Simplify

= 100 [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 100 (1 - \frac{1}{e})

\frac{x y ^{^{'}}}{e ^{x} + 1} = 2

f (x) = 3 x^{2} - 24 x + 45

x^{2} - 6 x y + y^{2} = 6

f (x) = 2 cos (2 x)

1 + 5 e^{3 t}

What is the integral from 0 to 100 of e^{-0.01x} ?
The integral from 0 to 100 of e^{-0.01x} is 100(1-1/e)