What is the integral from 0 to 15 of 1000e^{-0.05t} ?

The integral from 0 to 15 of 1000e^{-0.05t} is 10552.66894…

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integral from 0 to 15 of 1000e^{-0.05t}

\int_{0}^{15} 1000 e^{- 0.05 t} d t

10552.66894 \dots

Solution steps

\int_{0}^{15} 1000 e^{- 0.05 t} d t

Take the constant out:

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

= 1000 \cdot \int_{0}^{15} e^{- 0.05 t} d t

Apply u-substitution

: \int_{0}^{- 0.75} - \frac{1}{0.05} e^{u} d u

= 1000 \cdot \int_{0}^{- 0.75} - \frac{1}{0.05} e^{u} d u

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

= 1000 (- \int_{- 0.75}^{0} - \frac{1}{0.05} e^{u} d u)

Take the constant out:

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

= 1000 (- (- \frac{1}{0.05} \cdot \int_{- 0.75}^{0} e^{u} d u))

Simplify

= 1000 (- (- 20 \cdot \int_{- 0.75}^{0} e^{u} d u))

Use the common integral:

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

= 1000 (- (- 20 [e^{u}]_{- 0.75}^{0}))

Simplify

= 20000 [e^{u}]_{- 0.75}^{0}

Compute the boundaries:

0.52763 \dots

= 20000 \cdot 0.52763 \dots

Simplify

= 10552.66894 \dots

x \to 1 lim (3 - 3 x)

y^{^{'}} + \frac{1}{6} y = 40

\frac{\partial}{\partial y} ((x + y + z)^{2})

\int e^{- x} - e^{- 3 x} d x

f (x) = 5 e^{x} cos (x), (0, 5)

What is the integral from 0 to 15 of 1000e^{-0.05t} ?
The integral from 0 to 15 of 1000e^{-0.05t} is 10552.66894…