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Let's solve the given problem by finding the antiderivatives for each function, assuming \( C = 0 \).

### a. \( f(x) = e^{9x} \)

The antiderivative of \( e^{9x} \) is found using the formula:

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

For \( f(x) = e^{9x} \), \( k = 9 \):

\[
\int e^{9x} \, dx = \frac{1}{9} e^{9x}
\]

Since \( C = 0 \), the final answer is:

\[
F(x) = \frac{1}{9} e^{9x}
\]

---

### b. \( g(x) = e^{-6x} \)

Similarly, we use the same formula for \( e^{kx} \). Here, \( k = -6 \):

\[
\int e^{-6x} \, dx = \frac{1}{-6} e^{-6x}
\]

With \( C = 0 \), the answer is:

\[
G(x) = -\frac{1}{6} e^{-6x}
\]

---

### c. \( h(x) = e^{\frac{x}{11}} \)

For this function, \( k = \frac{1}{11} \). Using the same rule:

\[
\int e^{\frac{x}{11}} \, dx = \frac{1}{\frac{1}{11}} e^{\frac{x}{11}} = 11 e^{\frac{x}{11}}
\]

With \( C = 0 \), the answer is:

\[
H(x) = 11 e^{\frac{x}{11}}
\]

---

### Summary of Antiderivatives:

- **a.** \( F(x) = \frac{1}{9} e^{9x} \)
- **b.** \( G(x) = -\frac{1}{6} e^{-6x} \)
- **c.** \( H(x) = 11 e^{\frac{x}{11}} \)

Would you like further details on any of these solutions?

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Here are five questions to expand your understanding:

1. What is the general formula for the antiderivative of an exponential function \( e^{kx} \)?
2. How does the value of \( k \) in \( e^{kx} \) affect the antiderivative?
3. What happens to the constant \( C \) if it is not zero in these solutions?
4. How would you differentiate the results to verify they are correct?
5. How can you apply this concept to non-exponential functions?

**Tip:** When taking the antiderivative, always check for any constants multiplying the variable in the exponent to adjust your result accordingly.

Find the antiderivative for each function when C equals 0: a. f(x) = e^9x, b. g(x) = e^-6x, c. h(x) = e^(x/11).

Explore the detailed solutions for finding antiderivatives of exponential functions such as e^9x, e^-6x, and e^(x/11) with C=0. This page includes step-by-step methods using fundamental calculus concepts ideal for college students.

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