What is the integral from 0 to 5 of 40e^{-0.3t} ?

The integral from 0 to 5 of 40e^{-0.3t} is 103.58264…

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integral from 0 to 5 of 40e^{-0.3t}

\int_{0}^{5} 40 e^{- 0.3 t} d t

103.58264 \dots

Solution steps

\int_{0}^{5} 40 e^{- 0.3 t} d t

Take the constant out:

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

= 40 \cdot \int_{0}^{5} e^{- 0.3 t} d t

Apply u-substitution

: \int_{0}^{- 1.5} - \frac{1}{0.3} e^{u} d u

= 40 \cdot \int_{0}^{- 1.5} - \frac{1}{0.3} e^{u} d u

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

= 40 (- \int_{- 1.5}^{0} - \frac{1}{0.3} e^{u} d u)

Take the constant out:

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

= 40 (- (- \frac{1}{0.3} \cdot \int_{- 1.5}^{0} e^{u} d u))

Simplify

= 40 (- (- 3.33333 \dots \cdot \int_{- 1.5}^{0} e^{u} d u))

Use the common integral:

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

= 40 (- (- 3.33333 \dots [e^{u}]_{- 1.5}^{0}))

Simplify

= 133.33333 \dots [e^{u}]_{- 1.5}^{0}

Compute the boundaries:

0.77686 \dots

= 133.33333 \dots \cdot 0.77686 \dots

Simplify

= 103.58264 \dots

y^{^{'}} = 7 x e^{y}, y (0) = 0

x \to 0 lim (\frac{a ^{x} - 1}{2 x})

x \to - \infty lim (4 x^{3})

x \to 0 lim (\frac{8}{x + 2})

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

What is the integral from 0 to 5 of 40e^{-0.3t} ?
The integral from 0 to 5 of 40e^{-0.3t} is 103.58264…