Problem 27 - Essential Calculus

Question

Find the indefinite integral and check the result by differentiation. $$ \int\left(2 x-4^{x}\right) d x $$

Step-by-Step Solution

Answer

The indefinite integral of \(2x - 4^x\) is \(x^2 - \frac{1}{\ln 4} e^{x \ln 4} + C\), and it has been confirmed by differentiation.

Step 1: Finding the Integral

To find the indefinite integral of the given function \(2x - 4^x\), we break it into two integrals based on the subtraction operation. It becomes \(\int 2x \, dx\) and \(- \int 4^x \, dx\). We apply the integral calculus rule to the first part (which follows the power rule for integration) and we get \(x^2\). The second part is a bit tricky. We can't directly apply the exponential integration rule because the base is not 'e'. However, we can utilize a change of base formula that lets us rewrite this term as \(e^{x \ln 4}\). Then we integrate \(- \int e^{x \ln 4} \, dx\) using the exponential integration rule. The result is \(-\frac{1}{\ln 4} e^{x \ln 4}\). We do not forget to add the constant of integration 'C'. Hence the integral solution becomes \(x^2 - \frac{1}{\ln 4} e^{x \ln 4} + C\).

Step 2: Checking the Result by Differentiation

In order to confirm our solution, we differentiate our result by using the power rule and chain rule. We first differentiate \(x^2\) and we get \(2x\). Then, we take the derivative of \(-\frac{1}{\ln 4} e^{x \ln 4}\) which becomes \(-4^x\). And the derivative of the constant 'C' is 0. Combining these gives us back the original function \(2x - 4^x\) which validates our solution.

Key Concepts

Integration by Parts

Integration by parts is a helpful technique for finding integrals, particularly when products of functions are involved. The method is derived from the product rule for differentiation. This rule states that the derivative of a product of two functions is given by \[ (u imes v)' = u'v + uv' \] Here, integration by parts reverses this concept. The formula used is:

• \[ \int u \, dv = uv - \int v \, du \]

Choosing which function to be "\(u\)" and "\(dv\)" can be a balancing act, often guided by the LIATE rule (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential) or a similar logical choice. This approach wasn't directly used in our exercise because the given integral involved simpler terms: a polynomial and an exponential. In this case, breaking down into separate simpler integrals for each term made the problem more manageable. Simplified problems like these can reinforce understanding of when this tool is most useful, especially in scenarios involving more complex products.

Exponential Functions

Exponential functions feature prominently in calculus and typically take the form \(a^x\) where \(a\) is a constant. One key property of exponential functions is their rapid growth, depending on the base. In calculus, it’s important to transform base \(a\) functions to base \(e\) for easier integration, as the natural logarithm base \(e\) has special properties allowing straightforward differentiation and integration.
In our exercise, the term \(4^x\) was converted to \(e^{x \ln 4}\) using the formula \(a^x = e^{x \ln a}\). This conversion exploits the fact that exponential growth can be expressed equivalently with base \(e\), making it simpler to handle mathematically.

• Exponential functions: \(e^{x}\), \(a^{x} = e^{x \ln a}\)
• Integration: Similar to \(e^x\), integrating \(e^{x \ln a}\) gives a result scaled by \(\frac{1}{\ln a}\)

Chain Rule in Differentiation

The chain rule is a vital tool for differentiation, used when differentiating a "function of a function". This rule states that to find the derivative of a composite function, you differentiate the outer function and multiply it by the derivative of the inner function. Mathematically, it's represented as:

• \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \]

The chain rule was crucial in our exercise to check the correctness of integration by differentiating back to the original function. When differentiating \(-\frac{1}{\ln 4} e^{x \ln 4}\) to retrieve \(-4^x\), the chain rule helped in managing the \(x \ln 4\) exponent as the inner function. This allowed us to multiply with \(\ln 4\), providing the correct rate of change relative to the base 4, and ensuring our derivative matches the original integrand. It's this meticulous process that illustrates the elegance and interconnectedness of calculus principles in practice.