\int \:0.9^{2-x}\left(0.00022+0.0000027\cdot 1.124^{47+x}\right)dx

We are tasked with solving the following integral:

\[
I = \int 0.9^{2-x} \left(0.00022 + 0.0000027 \cdot 1.124^{47+x} \right) dx
\]

This is a combination of two terms inside the integral. Let’s break it down into two parts and solve it term by term:

\[
I = \int 0.9^{2-x} \cdot 0.00022 \, dx + \int 0.9^{2-x} \cdot 0.0000027 \cdot 1.124^{47+x} \, dx
\]

### First Integral:
\[
I_1 = \int 0.9^{2-x} \cdot 0.00022 \, dx
\]

This can be simplified to:
\[
I_1 = 0.00022 \int 0.9^{2-x} \, dx
\]

Now, rewrite \(0.9^{2-x} = 0.9^2 \cdot 0.9^{-x}\). So we have:
\[
I_1 = 0.00022 \cdot 0.9^2 \int 0.9^{-x} \, dx
\]

The integral of \(a^{-x}\) is \(\frac{a^{-x}}{-\ln(a)}\). Thus:
\[
I_1 = 0.00022 \cdot 0.9^2 \cdot \frac{0.9^{-x}}{-\ln(0.9)}
\]

### Second Integral:
\[
I_2 = \int 0.9^{2-x} \cdot 0.0000027 \cdot 1.124^{47+x} \, dx
\]

This can be rewritten as:
\[
I_2 = 0.0000027 \cdot 0.9^2 \cdot 1.124^{47} \int 0.9^{-x} \cdot 1.124^x \, dx
\]

Now, combine the terms inside the integral:
\[
0.9^{-x} \cdot 1.124^x = \left(\frac{1.124}{0.9}\right)^x
\]

Thus, we get:
\[
I_2 = 0.0000027 \cdot 0.9^2 \cdot 1.124^{47} \int \left(\frac{1.124}{0.9}\right)^x \, dx
\]

The integral of \(a^x\) is \(\frac{a^x}{\ln(a)}\), so we have:
\[
I_2 = 0.0000027 \cdot 0.9^2 \cdot 1.124^{47} \cdot \frac{\left(\frac{1.124}{0.9}\right)^x}{\ln\left(\frac{1.124}{0.9}\right)}
\]

### Final Answer:

Summing both integrals, the solution is:

\[
I = 0.00022 \cdot 0.9^2 \cdot \frac{0.9^{-x}}{-\ln(0.9)} + 0.0000027 \cdot 0.9^2 \cdot 1.124^{47} \cdot \frac{\left(\frac{1.124}{0.9}\right)^x}{\ln\left(\frac{1.124}{0.9}\right)}
\]

Would you like me to evaluate this numerically for specific limits or further simplify it?

---

**Relative Questions:**
1. How does the logarithmic function behave in integrals involving exponential terms?
2. Can you explain how to deal with products of exponentials in integration?
3. What is the process of simplifying constants in integrals?
4. What happens if the base of the exponential functions were negative?
5. How would changing the limits of integration affect the outcome of this problem?

**Tip:** When dealing with exponentials in integrals, factoring constants and recognizing basic patterns like \(a^x\) can simplify the process significantly.

\int 0.9^{2-x}\left(0.00022+0.0000027\cdot 1.124^{47+x}\right)dx

Explore the step-by-step solution for the integral \int 0.9^{2-x}\left(0.00022 + 0.0000027 \cdot 1.124^{47+x}\right)dx, breaking down exponential functions, constants, and logarithms. This problem is ideal for calculus students, covering integration techniques and exponential products.

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