If p(x) = (4e) 2x , then what is ∫ p(x) dx equal to?
Understanding the Integration Problem
The problem asks us to find the indefinite integral of the function \(p(x) = (4e)^{2x}\). This is an example of integrating an exponential function where the base is a constant raised to a linear expression involving the variable \(x\).
The given function \(p(x)\) can be written in the form \(a^{kx}\), where:
- The base \(a\) is \(4e\).
- The exponent is \(2x\), so the coefficient \(k\) of \(x\) in the exponent is \(2\).
We need to find \(\int p(x) dx\), which is \(\int (4e)^{2x} dx\).
General Formula for Integrating Exponential Functions
The general formula for the integral of an exponential function of the form \(a^{kx}\) is given by:
\[ \int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C \]
where \(a\) is the base (\(a > 0, a \neq 1\)), \(k\) is the coefficient of \(x\), and \(C\) is the constant of integration.
Applying the Formula to p(x) = (4e) 2x
In our case, \(a = 4e\) and \(k = 2\). We can substitute these values into the general formula:
\[ \int (4e)^{2x} dx = \frac{(4e)^{2x}}{2 \ln(4e)} + C \]
Simplifying the Expression
Now, let's simplify the denominator, which involves the term \(\ln(4e)\). Using the properties of logarithms, \(\ln(uv) = \ln u + \ln v\):
\[ \ln(4e) = \ln(4) + \ln(e) \]
We know that \(\ln(e) = 1\). Also, we can write \(4\) as \(2^2\). Using the property \(\ln(u^v) = v \ln u\):
\[ \ln(4) = \ln(2^2) = 2 \ln(2) \]
So, \(\ln(4e) = 2 \ln(2) + 1\). We can rearrange this as \(1 + 2 \ln 2\).
Substitute this simplified expression back into the integral result:
\[ \int (4e)^{2x} dx = \frac{(4e)^{2x}}{2 (1 + 2 \ln 2)} + C \]
Since \(p(x) = (4e)^{2x}\), we can replace \((4e)^{2x}\) with \(p(x)\) in the numerator:
\[ \int p(x) dx = \frac{p(x)}{2 (1 + 2 \ln 2)} + C \]
Comparing with Options
Let's compare our derived result with the given options:
\[ \frac{p(x)}{2 (1 + 2 \ln 2)} + C \]
This matches the second option provided.
| Step | Description | Calculation |
|---|---|---|
| 1 | Identify \(p(x)\) form \(a^{kx}\) | \(p(x) = (4e)^{2x}\), \(a = 4e\), \(k = 2\) |
| 2 | Apply integral formula \(\int a^{kx} dx\) | \(\int (4e)^{2x} dx = \frac{(4e)^{2x}}{2 \ln(4e)} + C\) |
| 3 | Simplify \(\ln(4e)\) | \(\ln(4e) = \ln(4) + \ln(e)\) |
| 4 | Evaluate \(\ln(4)\) and \(\ln(e)\) | \(\ln(4) = \ln(2^2) = 2 \ln 2\), \(\ln(e) = 1\) |
| 5 | Combine terms for \(\ln(4e)\) | \(\ln(4e) = 2 \ln 2 + 1 = 1 + 2 \ln 2\) |
| 6 | Substitute back into integral | \(\frac{(4e)^{2x}}{2 (1 + 2 \ln 2)} + C\) |
| 7 | Replace \((4e)^{2x}\) with \(p(x)\) | \(\frac{p(x)}{2 (1 + 2 \ln 2)} + C\) |
Conclusion
By identifying the function \(p(x) = (4e)^{2x}\) as an exponential function of the form \(a^{kx}\) and applying the standard integration formula \(\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C\), we found the integral to be \(\frac{(4e)^{2x}}{2 \ln(4e)} + C\). Simplifying the logarithm term \(\ln(4e)\) to \(1 + 2 \ln 2\) led to the final result \(\frac{(4e)^{2x}}{2 (1 + 2 \ln 2)} + C\), which is equal to \(\frac{p(x)}{2 (1 + 2 \ln 2)} + C\).
Revision Table: Key Integration Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Indefinite Integral | The reverse process of differentiation; finding the antiderivative of a function. Includes the constant of integration, \(C\). | \(\int f(x) dx = F(x) + C\), where \(F'(x) = f(x)\) |
| Integral of \(a^x\) | Integral of a basic exponential function. | \(\int a^x dx = \frac{a^x}{\ln a} + C\) (for \(a > 0, a \neq 1\)) |
| Integral of \(e^x\) | Special case where the base is \(e\). | \(\int e^x dx = e^x + C\) (since \(\ln e = 1\)) |
| Integral of \(a^{kx}\) | Integral of an exponential function with a linear term in the exponent. | \(\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C\) (for \(a > 0, a \neq 1\), \(k \neq 0\)) |
| Logarithm Property \(\ln(uv)\) | Used to expand logarithms of products. | \(\ln(uv) = \ln u + \ln v\) (for \(u, v > 0\)) |
| Logarithm Property \(\ln(u^v)\) | Used to simplify logarithms of powers. | \(\ln(u^v) = v \ln u\) (for \(u > 0\)) |
Additional Information: Exponential Function Integration
Integrating exponential functions is a fundamental skill in calculus. The key is to recognize the form of the function and apply the appropriate standard integral formula.
- Functions like \(e^{ax}\) integrate to \(\frac{1}{a} e^{ax} + C\).
- Functions like \(a^{bx}\) integrate to \(\frac{1}{b \ln a} a^{bx} + C\).
- Our problem \(p(x) = (4e)^{2x}\) fits the \(a^{kx}\) form where the base itself is a product \((4e)\).
Understanding logarithm properties is crucial for simplifying the denominator involving \(\ln(\text{base})\). The natural logarithm \(\ln x\) is the logarithm with base \(e\).
Always remember to include the constant of integration, \(C\), when performing indefinite integration, as it represents the family of all antiderivatives.
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