What is the integral from 0 to 2 of 9000e^{0.06t} ?

The integral from 0 to 2 of 9000e^{0.06t} is 19124.52773…

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integral from 0 to 2 of 9000e^{0.06t}

\int_{0}^{2} 9000 e^{0.06 t} d t

19124.52773 \dots

Solution steps

\int_{0}^{2} 9000 e^{0.06 t} d t

Take the constant out:

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

= 9000 \cdot \int_{0}^{2} e^{0.06 t} d t

Apply u-substitution

: \int_{0}^{0.12} e^{u} \frac{1}{0.06} d u

= 9000 \cdot \int_{0}^{0.12} e^{u} \frac{1}{0.06} d u

Take the constant out:

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

= 9000 \cdot \frac{1}{0.06} \cdot \int_{0}^{0.12} e^{u} d u

Simplify

= 9000 \cdot 16.66666 \dots \cdot \int_{0}^{0.12} e^{u} d u

Use the common integral:

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

= 9000 \cdot 16.66666 \dots [e^{u}]_{0}^{0.12}

Simplify

= 150000 [e^{u}]_{0}^{0.12}

Compute the boundaries:

0.12749 \dots

= 150000 \cdot 0.12749 \dots

Simplify

= 19124.52773 \dots

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What is the integral from 0 to 2 of 9000e^{0.06t} ?
The integral from 0 to 2 of 9000e^{0.06t} is 19124.52773…