What is the integral of 2e^{r/2} ?

The integral of 2e^{r/2} is 4e^{r/2}+C

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integral of 2e^{r/2}

\int 2 e^{\frac{r}{2}} d r

4 e^{\frac{r}{2}} + C

Solution steps

\int 2 e^{\frac{r}{2}} d r

Take the constant out:

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

= 2 \cdot \int e^{\frac{r}{2}} d r

Apply u-substitution

: \int e^{u} \cdot 2 d u

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot 2 e^{u}

Substitute back

u = \frac{r}{2}

= 2 \cdot 2 e^{\frac{r}{2}}

Simplify

= 4 e^{\frac{r}{2}}

Add a constant to the solution

= 4 e^{\frac{r}{2}} + C

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\frac{d y}{d x} = sin (6 x)

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