To solve the integral \( \int_0^9 2.9e^{0.05x} \, dx \) using the **U-substitution method**, follow these steps:

### Step 1: Identify the substitution
Let:
\[
u = 0.05x
\]
Then, differentiate \( u \) with respect to \( x \):
\[
\frac{du}{dx} = 0.05 \quad \Rightarrow \quad dx = \frac{du}{0.05}
\]

### Step 2: Rewrite the integral
Now substitute \( u = 0.05x \) and \( dx = \frac{du}{0.05} \):
\[
\int_0^9 2.9e^{0.05x} \, dx = \int_0^{0.05 \cdot 9} 2.9e^u \cdot \frac{du}{0.05}
\]
Simplify the constants:
\[
\int_0^{0.45} \frac{2.9}{0.05} e^u \, du = \int_0^{0.45} 58e^u \, du
\]

### Step 3: Integrate
The integral of \( e^u \) is \( e^u \), so:
\[
58 \int_0^{0.45} e^u \, du = 58 \left[ e^u \right]_0^{0.45}
\]
Evaluating at the limits:
\[
58 \left( e^{0.45} - e^0 \right) = 58 \left( e^{0.45} - 1 \right)
\]

### Step 4: Final answer
Now, calculate the values of \( e^{0.45} \):
\[
e^{0.45} \approx 1.5683
\]
Thus:
\[
58 \left( 1.5683 - 1 \right) = 58 \times 0.5683 \approx 32.961
\]

So, the value of the integral is approximately:
\[
\boxed{32.96}
\]

Would you like to see more details or clarifications on any of the steps?

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Here are 5 questions related to this process:

1. How does substitution work in integrals, and why is it useful?
2. What is the significance of the exponential function \( e^x \) in calculus?
3. Can you explain why we change the limits of integration after substitution?
4. How would this method change if the exponent were more complex, such as \( e^{x^2} \)?
5. How do you compute the value of \( e^x \) manually or using a calculator?

**Tip:** Always simplify constants outside the integral before proceeding with the actual integration to reduce complexity!

Learn how to solve the integral ∫_0^9 2.9e^{0.05x} using the U-substitution method. This step-by-step guide simplifies the integral, introduces substitution, and evaluates it for the limits 0 to 9. Suitable for students in Grades 10-12 studying calculus.

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