What is the integral from 0 to 12 of e^{0.1x} ?

The integral from 0 to 12 of e^{0.1x} is 23.20116…

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integral from 0 to 12 of e^{0.1x}

\int_{0}^{12} e^{0.1 x} d x

23.20116 \dots

Solution steps

\int_{0}^{12} e^{0.1 x} d x

Apply u-substitution

= \int_{0}^{1.2} e^{u} \frac{1}{0.1} d u

Take the constant out:

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

= \frac{1}{0.1} \cdot \int_{0}^{1.2} e^{u} d u

Simplify

= 10 \cdot \int_{0}^{1.2} e^{u} d u

Use the common integral:

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

= 10 [e^{u}]_{0}^{1.2}

Compute the boundaries:

2.32011 \dots

= 10 \cdot 2.32011 \dots

Simplify

= 23.20116 \dots

What is the integral from 0 to 12 of e^{0.1x} ?
The integral from 0 to 12 of e^{0.1x} is 23.20116…