Problem 30 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{1 / 3}^{1 / 2} \frac{10^{1 / x}}{x^{2}} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{1/3}^{1/2} \frac{10^{1/x}}{x^2} dx$$ Answer: The definite integral evaluates to \(-25\).

Step 1: Simplify the integrand

Rewrite the exponent as \(10^{1/x} = e^{\ln(10^{1/x})} = e^{(1/x)\ln(10)}\). Now our integral becomes $$\int_{1 / 3}^{1 / 2} \frac{e^{(1/x)\ln(10)}}{x^{2}} d x$$

Step 2: Substitute u

Let \(u = \frac{\ln(10)}{x}\), so \(\frac{du}{dx} = -\frac{\ln(10)}{x^2}\). Now the integral is $$\int_{\ln(30)}^{\ln(5)} e^u du$$

Step 3: Find the antiderivative

The antiderivative of \(e^u\) is simply \(e^u + C\), where C is the constant of integration.

Step 4: Apply the fundamental theorem of calculus

Now we can evaluate the definite integral by subtracting the antiderivative at the lower limit of integration from the antiderivative at the upper limit of integration: $$[e^u]_{\ln(30)}^{\ln(5)} = e^{\ln(5)} - e^{\ln(30)}$$

Step 5: Simplify the final result

Using the property of exponentials \(e^{\ln(x)} = x\), we can further simplify the final answer: $$5 - 30 = -25$$ Thus, the definite integral \(\int_{1/3}^{1/2} \frac{10^{1/x}}{x^2} dx\) is equal to \(-25\).

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent. In the context of this exercise, we deal with functions like \(10^{1/x}\). To manipulate exponential expressions, it's often useful to rewrite them in terms of natural logarithms, as this simplifies integration and differentiation. For instance, \(10^{1/x}\) can be rewritten using the identity \(e^{ ext{ln}(a)} = a\), giving us \(e^{(1/x) \ln(10)}\). Breaking down the expression like this helps us take advantage of the properties of the exponential and logarithm functions, making it easier to integrate or differentiate.

Substitution Method

The substitution method is a fundamental technique in calculus used to simplify integration. It's often utilized when dealing with complex integrals, where substitution can transform the integral into a simpler form. Here, we made a substitution by letting \(u = \frac{\ln(10)}{x}\), which led to a cleaner expression for integration.

• This method involves replacing a part of the integral with a new variable (\(u\)) so the integration process becomes more manageable.
• After substituting, you also need to adjust the differential (\(du\)) accordingly, which in this case was \(\frac{du}{dx} = -\frac{\ln(10)}{x^2}\).

This simplifies our integral to \(\int e^u du\), which is much easier to evaluate.

Antiderivative

Finding the antiderivative is the process of reverse differentiation. For an integral, it means identifying a function whose derivative returns the integrand. In our exercise, we needed the antiderivative for \(e^u\). The antiderivative of an exponential function is straightforward; it is the exponential function itself plus a constant of integration.

• For \(e^u\), the antiderivative is \(e^u + C\), where \(C\) is a constant.

Once the antiderivative is determined, we can proceed with calculating the definite integral by evaluating this expression over the limits of integration.

Fundamental Theorem of Calculus

The fundamental theorem of calculus bridges the gap between differentiation and integration, providing a practical way to compute definite integrals. It tells us that to find the integral of a function over a particular interval, we can evaluate the antiderivative at the boundaries of this interval and subtract.
For this problem, after determining the antiderivative as \(e^u\), the theorem was used to evaluate:

• \([e^u]_{\ln(30)}^{\ln(5)} = e^{\ln(5)} - e^{\ln(30)}\).

This simplifies to \(5 - 30 = -25\). Applying the fundamental theorem here allows us to easily find the area under the curve of the original function over the specified limits, leading us to our final result.