What is the integral of 9e^{0.01x} ?

The integral of 9e^{0.01x} is 900e^{0.01x}+C

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integral of 9e^{0.01x}

\int 9 e^{0.01 x} d x

900 e^{0.01 x} + C

Solution steps

\int 9 e^{0.01 x} d x

Take the constant out:

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

= 9 \cdot \int e^{0.01 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.01} d u

= 9 \cdot \int e^{u} \frac{1}{0.01} d u

Take the constant out:

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

= 9 \cdot \frac{1}{0.01} \cdot \int e^{u} d u

Simplify

= 9 \cdot 100 \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot 100 e^{u}

Substitute back

u = 0.01 x

= 9 \cdot 100 e^{0.01 x}

Simplify

= 900 e^{0.01 x}

Add a constant to the solution

= 900 e^{0.01 x} + C

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What is the integral of 9e^{0.01x} ?
The integral of 9e^{0.01x} is 900e^{0.01x}+C