What is the integral from 0 to 5 of 20e^{0.02x} ?

The integral from 0 to 5 of 20e^{0.02x} is 105.17091…

integral from 0 to 5 of 20e^{0.02x}

\int_{0}^{5} 20 e^{0.02 x} d x

105.17091 \dots

Solution steps

\int_{0}^{5} 20 e^{0.02 x} d x

Take the constant out:

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

= 20 \cdot \int_{0}^{5} e^{0.02 x} d x

Apply u-substitution

= 20 \cdot \int_{0}^{0.1} e^{u} \frac{1}{0.02} d u

Take the constant out:

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

= 20 \cdot \frac{1}{0.02} \cdot \int_{0}^{0.1} e^{u} d u

Simplify

= 20 \cdot 50 \cdot \int_{0}^{0.1} e^{u} d u

Use the common integral:

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

= 20 \cdot 50 [e^{u}]_{0}^{0.1}

Simplify

= 1000 [e^{u}]_{0}^{0.1}

Compute the boundaries:

0.10517 \dots

= 1000 \cdot 0.10517 \dots

Simplify

= 105.17091 \dots

d er i v a t i v e f (x) = \frac{x ^{3}}{12} + \frac{1}{x}

\frac{\partial}{\partial y} (z y^{2} (t + 1))

\int e^{- s i n (x)} cos (x) d x

x \to 0 - lim (1 - \frac{3}{x} + \frac{1}{x ^{2}})

d er i v a t i v e y = (1 + 4 x)^{\frac{3}{2}}

What is the integral from 0 to 5 of 20e^{0.02x} ?
The integral from 0 to 5 of 20e^{0.02x} is 105.17091…