What is the integral of 0.06e^{0.02x} ?

The integral of 0.06e^{0.02x} is 3e^{0.02x}+C

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integral of 0.06e^{0.02x}

\int 0.06 e^{0.02 x} d x

3 e^{0.02 x} + C

Solution steps

\int 0.06 e^{0.02 x} d x

Take the constant out:

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

= 0.06 \cdot \int e^{0.02 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.02} d u

= 0.06 \cdot \int 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

= 0.06 \cdot \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= 0.06 \cdot 50 \cdot \int e^{u} d u

Use the common integral:

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

= 0.06 \cdot 50 e^{u}

Substitute back

u = 0.02 x

= 0.06 \cdot 50 e^{0.02 x}

Simplify

= 3 e^{0.02 x}

Add a constant to the solution

= 3 e^{0.02 x} + C

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What is the integral of 0.06e^{0.02x} ?
The integral of 0.06e^{0.02x} is 3e^{0.02x}+C