What is the integral from 0 to 60 of 65e^{-0.02t} ?

The integral from 0 to 60 of 65e^{-0.02t} is 2271.11881…

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integral from 0 to 60 of 65e^{-0.02t}

\int_{0}^{60} 65 e^{- 0.02 t} d t

2271.11881 \dots

Solution steps

\int_{0}^{60} 65 e^{- 0.02 t} d t

Take the constant out:

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

= 65 \cdot \int_{0}^{60} e^{- 0.02 t} d t

Apply u-substitution

: \int_{0}^{- 1.2} - \frac{1}{0.02} e^{u} d u

= 65 \cdot \int_{0}^{- 1.2} - \frac{1}{0.02} e^{u} d u

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

= 65 (- \int_{- 1.2}^{0} - \frac{1}{0.02} e^{u} d u)

Take the constant out:

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

= 65 (- (- \frac{1}{0.02} \cdot \int_{- 1.2}^{0} e^{u} d u))

Simplify

= 65 (- (- 50 \cdot \int_{- 1.2}^{0} e^{u} d u))

Use the common integral:

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

= 65 (- (- 50 [e^{u}]_{- 1.2}^{0}))

Simplify

= 3250 [e^{u}]_{- 1.2}^{0}

Compute the boundaries:

0.69880 \dots

= 3250 \cdot 0.69880 \dots

Simplify

= 2271.11881 \dots

x \to 0 lim (sin (\frac{1}{x}) x)

\frac{d}{d x} (- 0.4 x^{2} + 1000)

\frac{\partial}{\partial y} (x y - 3 x z + yz)

\int \frac{x ^{5}}{x ^{2} + 8} d x

x \to 3 lim (\frac{- x}{x - 3})

What is the integral from 0 to 60 of 65e^{-0.02t} ?
The integral from 0 to 60 of 65e^{-0.02t} is 2271.11881…