What is the integral of 0.06e^{0.04x} ?

The integral of 0.06e^{0.04x} is 1.5e^{0.04x}+C

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integral of 0.06e^{0.04x}

\int 0.06 e^{0.04 x} d x

1.5 e^{0.04 x} + C

Solution steps

\int 0.06 e^{0.04 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^{0.04 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.04} d u

= 0.06 \cdot \int e^{u} \frac{1}{0.04} 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}{0.04} \cdot \int e^{u} d u

Simplify

= 0.06 \cdot 25 \cdot \int e^{u} d u

Use the common integral:

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

= 0.06 \cdot 25 e^{u}

Substitute back

u = 0.04 x

= 0.06 \cdot 25 e^{0.04 x}

Simplify

= 1.5 e^{0.04 x}

Add a constant to the solution

= 1.5 e^{0.04 x} + C

\int 3 x^{2} cos (x) d x

n = 0 \sum \infty \frac{n !}{10 7 ^{n}}

\frac{d}{d x} (\frac{x ^{2}}{5 x + 2})

3 t^{2} - 30 t + 72

f (x) = \frac{8}{x}

What is the integral of 0.06e^{0.04x} ?
The integral of 0.06e^{0.04x} is 1.5e^{0.04x}+C