What is the integral of 2e^{x/3+4} ?

The integral of 2e^{x/3+4} is 6e^{4+x/3}+C

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integral of 2e^{x/3+4}

\int 2 e^{\frac{x}{3} + 4} d x

6 e^{4 + \frac{x}{3}} + C

Solution steps

\int 2 e^{\frac{x}{3} + 4} d x

Take the constant out:

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

= 2 \cdot \int e^{\frac{x}{3} + 4} d x

e^{\frac{x}{3} + 4} = e^{4} e^{\frac{x}{3}}

= 2 \cdot \int e^{4} e^{\frac{x}{3}} d x

Take the constant out:

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

= 2 e^{4} \cdot \int e^{\frac{x}{3}} d x

Apply u-substitution

: \int e^{u} \cdot 3 d u

= 2 e^{4} \cdot \int e^{u} \cdot 3 d u

Take the constant out:

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

= 2 e^{4} \cdot 3 \cdot \int e^{u} d u

Use the common integral:

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

= 2 e^{4} \cdot 3 e^{u}

Substitute back

u = \frac{x}{3}

= 2 e^{4} \cdot 3 e^{\frac{x}{3}}

Simplify

2 e^{4} \cdot 3 e^{\frac{x}{3}} : 6 e^{4 + \frac{x}{3}}

= 6 e^{4 + \frac{x}{3}}

Add a constant to the solution

= 6 e^{4 + \frac{x}{3}} + C

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