Problem 11 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{1}^{9} \frac{x-1}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{40}{3} \).

Step 1: Rewrite the integrand

To integrate the function, start by rewriting the integrand \( \frac{x-1}{\sqrt{x}} \) as \( x^{1/2} - x^{-1/2} \). This can be achieved by breaking it down as follows: \( \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}} = x^{1/2} - x^{-1/2} \).

Step 2: Integrate term by term

Integrate each term separately. The integral of \( x^{1/2} \) becomes \( \frac{2}{3}x^{3/2} \), and the integral of \( x^{-1/2} \) becomes \( 2x^{1/2} \).

Step 3: Apply the limits of integration

Evaluate the integrals from 1 to 9. First substitute \( x = 9 \) into the result from step 2: \( \frac{2}{3}(9)^{3/2} - 2(9)^{1/2} \). Then substitute \( x = 1 \) into the result: \( \frac{2}{3}(1)^{3/2} - 2(1)^{1/2} \).

Step 4: Calculate the definite integral

Perform the calculations for \( \frac{2}{3}(9\sqrt{9}) - 2\sqrt{9} = \frac{2}{3}(27) - 6 \) which simplifies to \( 18 - 6 = 12 \). For \( x = 1 \), compute \( \frac{2}{3}(1) - 2(1) = \frac{2}{3} - 2 = -\frac{4}{3} \).

Step 5: Find the final answer

Subtract the value calculated at \( x = 1 \) from that at \( x = 9 \): \( 12 - \left(-\frac{4}{3}\right) \). Simplify to get \( 12 + \frac{4}{3} = \frac{36}{3} + \frac{4}{3} = \frac{40}{3} \).

Key Concepts

Definite Integrals

When we talk about definite integrals, we are noting a process to find the net area under a curve between two points on the x-axis. This specific interval in this exercise is from 1 to 9 on the x-axis. Essentially, this is a limit of a sum of areas of rectangles under the curve. It's like adding up slices of a cake, where each slice has a thickness given by a tiny change in x.

One important aspect of definite integrals is they give us a number. This number represents the total accumulation of quantities, whether that’s space (like area), distance, or even something abstract like work. They are the end results of integrating a function over a set interval, providing insights into cumulative quantities.

• The bounds or limits of integration, in this case, 1 and 9, indicate the precise stretch along the x-axis we're focusing on.
• The notation lntegral symbol + function + limits all together define the structure of a definite integral.

It's crucial to emphasize that definite integrals result in a finite value and solve for the net area. Positive parts of the area add up, while negative ones can subtract, leading to the concept of 'net' area.

Integration Techniques

Integrating functions often requires finesse in breaking down expressions into friendlier parts. In this exercise, we saw the method of separating a more complex expression into simpler terms. This is a classic technique in integration.

• Decomposing: For our problem, breaking down \( \frac{x-1}{\sqrt{x}} \) into \( x^{1/2} - x^{-1/2} \) makes the function manageable to integrate.
• Integrating basic power functions: Recall the power rule for integration, \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \, n eq -1 \). This rule lets us integrate each term separately.

Recognizing standard forms and applying straightforward rules allow us to simplify the process. Integration becomes less about blind application and more about strategic decomposition and recognizing familiar patterns.

Furthermore, the principle of integrating term by term is highly productive as it breaks what could be one tricky problem into more straightforward, bite-sized calculations.

Mathematical Problem Solving

Mathematical problem-solving, especially in integral calculus, is centered around logical progression and precise calculations.

• Start by fully understanding the problem, reading through the integral expression and determining which technique fits best for simplification.
• Be diligent with calculations: small errors in these steps can lead to incorrect results.

In this particular solution, each transformation step was deliberate. From rewriting the integrand to applying limits to the integral, each phase was handled methodically.

It's essential to check each step for accuracy. Finally, understanding how each step builds upon the last can clarify any concept that's puzzling. Encourage critical thinking, taking it slow, and understanding the 'why' behind each operation.