What is the integral of 105e^{0.05t} ?

The integral of 105e^{0.05t} is 2100e^{0.05t}+C

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integral of 105e^{0.05t}

\int 105 e^{0.05 t} d t

2100 e^{0.05 t} + C

Solution steps

\int 105 e^{0.05 t} d t

Take the constant out:

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

= 105 \cdot \int e^{0.05 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{0.05} d u

= 105 \cdot \int e^{u} \frac{1}{0.05} d u

Take the constant out:

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

= 105 \cdot \frac{1}{0.05} \cdot \int e^{u} d u

Simplify

= 105 \cdot 20 \cdot \int e^{u} d u

Use the common integral:

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

= 105 \cdot 20 e^{u}

Substitute back

u = 0.05 t

= 105 \cdot 20 e^{0.05 t}

Simplify

= 2100 e^{0.05 t}

Add a constant to the solution

= 2100 e^{0.05 t} + C

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What is the integral of 105e^{0.05t} ?
The integral of 105e^{0.05t} is 2100e^{0.05t}+C