What is the integral of 12e^{2t} ?

The integral of 12e^{2t} is 6e^{2t}+C

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integral of 12e^{2t}

\int 12 e^{2 t} d t

6 e^{2 t} + C

Solution steps

\int 12 e^{2 t} d t

Take the constant out:

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

= 12 \cdot \int e^{2 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= 12 \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= 12 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 12 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 t

= 12 \cdot \frac{1}{2} e^{2 t}

Simplify

12 \cdot \frac{1}{2} e^{2 t} : 6 e^{2 t}

= 6 e^{2 t}

Add a constant to the solution

= 6 e^{2 t} + C

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