Problem 34 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t$$ Answer: $$4\ln(3) - 4\ln(2) - 2$$

Step 1: Break down the integrand

Split the fraction in the integrand into two separate fractions to simplify the integral. $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} dt = \int_{4}^{9} \left(\frac{2}{t} + \frac{\sqrt{t}}{t}\right) dt$$

Step 2: Simplify the fractions

Simplify the fractions in the integrand. $$\int_{4}^{9} \left(\frac{2}{t} + \frac{\sqrt{t}}{t}\right) dt = \int_{4}^{9} \left(\frac{2}{t} + \frac{t^{\frac{1}{2}}}{t}\right) dt$$ $$\int_{4}^{9} \left(\frac{2}{t} + \frac{t^{\frac{1}{2}}}{t}\right) dt = \int_{4}^{9} \left(\frac{2}{t} + t^{-\frac{1}{2}}\right) dt$$

Step 3: Find the antiderivative

Find the antiderivative of the simplified integrand. $$\int \left(\frac{2}{t} + t^{-\frac{1}{2}}\right) dt = 2\ln|t| - 2\sqrt{t} + C$$

Step 4: Apply the Fundamental Theorem of Calculus

Evaluate the definite integral using the antiderivative found in Step 3. $$\int_{4}^{9} \left(\frac{2}{t} + t^{-\frac{1}{2}}\right) dt = \left[2\ln|t| - 2\sqrt{t} \right]_{4}^{9}$$

Step 5: Evaluate the limits

Substitute the limits of integration into the antiderivative and compute the difference. $$\begin{aligned} \left[2\ln|t| - 2\sqrt{t} \right]_{4}^{9} &= \left(2\ln(9) - 2\sqrt{9}\right) - \left(2\ln(4) - 2\sqrt{4}\right) \\ &= \left(2\ln(3^2) - 6\right) - \left(2\ln(2^2) - 4\right) \\ &= \left(4\ln(3) - 6\right) - \left(4\ln(2) - 4\right) \end{aligned}$$

Step 6: Combine terms and simplify

Combine like terms and simplify the result. $$\begin{aligned} \left(4\ln(3) - 6\right) - \left(4\ln(2) - 4\right) &= 4\ln(3) - 6 - 4\ln(2) + 4 \\ &= 4\ln(3) - 4\ln(2) - 2 \end{aligned}$$

Key Concepts

Definite Integral

When working with integrals, it's essential to understand the concept of a **definite integral**. This kind of integral allows us to calculate the "net area" between the x-axis and the curve defined by a function, from one point to another.
In essence, a definite integral accumulates the total value of the function over a specific interval:

• It is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• The lower limit, \( a \), signifies the beginning of the interval.
• Conversely, the upper limit, \( b \), signifies the end.

The result of a definite integral is a number that provides meaningful information about the total accumulation of the function over the interval \([a, b]\).
For example, during our exercise, we computed \( \int_{4}^{9} \left(\frac{2}{t} + t^{-\frac{1}{2}}\right) dt \), and it provided a net area between the interval [4, 9] using the antiderivative.

Antiderivative

Central to integrals is the idea of finding an **antiderivative**. Think of this as the opposite operation of deriving a function. When we derive a function, we find its rate of change; finding an antiderivative sweeps us backward, helping us determine the original function from its derivative.
There are several key points to understand:

• An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).
• To find a definite integral, we first determine its antiderivative.
• The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) dx = F(b) - F(a) \).

In our exercise, we calculated the antiderivative of the integrand \( \frac{2}{t} + t^{-\frac{1}{2}} \), resulting in \( 2\ln|t| - 2\sqrt{t} + C \).
The "+ C" represents any constant, showing there's an entire family of functions with identical derivatives, but with different vertical positions.

Logarithmic Function

**Logarithmic functions** are core to solving many integrals, especially those involving division by a variable like \( t \).
When you encounter terms like \( \frac{1}{t} \), it signals that a logarithmic function could be involved. Why is that?

• The derivative of the natural logarithm function, \( \ln(t) \), is \( \frac{1}{t} \).
• This relationship allows us to easily find the antiderivative of \( \frac{1}{t} \), denoted as \( \ln|t| \).

In calculus, when we solve integrals featuring terms like \( \frac{1}{t} \), we look for opportunities to leverage this connection to the natural logarithm. This tactic simplifies complicated expressions into manageable forms.
In our original problem, we saw this principal in action: our integration required us to identify \( 2\ln|t| \) as part of finding the antiderivative.