Evaluate the integral of e^(6x) with respect to x.

Step-by-step solution

1. 1

   Recall why a factor appears at all. Differentiating $e^{6x}$ brings out the inner derivative:

   $\frac{d}{dx}e^{6x}=6e^{6x}$

   so integrating has to undo that extra $6$.

2. 2

   State the rule. For any nonzero constant $k$:

   $\int e^{kx}\,dx=\frac{1}{k}e^{kx}+C$

3. 3

   Apply it with $k=6$.

   $\int e^{6x}\,dx=\frac{1}{6}e^{6x}+C$

4. 4

   Derive the same thing by substitution, as a cross-check. Let $u=6x$, so $du=6\,dx$ and $dx=\tfrac{du}{6}$:

   $\int e^{u}\frac{du}{6}=\frac{1}{6}e^{u}+C=\frac{1}{6}e^{6x}+C$

5. 5

   Verify by differentiating. $\tfrac{d}{dx}\left(\tfrac16 e^{6x}\right)=\tfrac16\cdot 6e^{6x}=e^{6x}$ $\checkmark$. Numerically at $x=0.3$: the integrand is $e^{1.8}\approx 6.0496$, and a difference quotient of $\tfrac16 e^{6x}$ gives the same value $\checkmark$.