Problem 8 - Calculus: Early Transcendentals

Question

Find the area bounded by the curves. \(y=\sqrt{x}\) and \(y=\sqrt{x+1}, 0 \leq x \leq 4\)

Step-by-Step Solution

Answer

The area is approximately 1.45 square units.

Step 1: Find Points of Intersection

We need to find the points where the curves intersect. Set the two expressions for \(y\) equal: \(\sqrt{x} = \sqrt{x+1}\). Squaring both sides gives \(x = x + 1\), which simplifies to a contradiction. Thus, they do not intersect within the given range. The curves do not intersect for \(x\geq 0\).

Step 2: Set Up the Integral

To find the area between the curves, we need the integral of the absolute value of the difference of the functions. The area between \(y=\sqrt{x}\) and \(y=\sqrt{x+1}\) from \(x=0\) to \(x=4\) is given by: \[\int_{0}^{4} (\sqrt{x+1} - \sqrt{x})\,dx.\]

Step 3: Integrate the Function

Integrate \(\sqrt{x+1}\) and \(\sqrt{x}\) separately. The antiderivative of \(\sqrt{x+1}\) is \(\frac{2}{3}(x+1)^{3/2}\), and the antiderivative of \(\sqrt{x}\) is \(\frac{2}{3}x^{3/2}\).

Step 4: Evaluate the Definite Integrals

Insert the bounds 0 and 4 into the antiderivatives and subtract: \[\left[\frac{2}{3}(x+1)^{3/2}\right]_{0}^{4} - \left[\frac{2}{3}x^{3/2}\right]_{0}^{4}.\]Calculate: \(\frac{2}{3}(5^{3/2}) - \frac{2}{3}(1^{3/2}) - \left(\frac{2}{3}(4^{3/2}) - \frac{2}{3}(0^{3/2})\right)\).

Step 5: Simplify the Expression

Compute each term: - For \((5^{3/2}): 5^{3/2} = 5 \times \sqrt{5} \approx 11.18\), so \(\frac{2}{3}(11.18)\approx 7.45\).- For \((1^{3/2}): 1^{3/2} = 1\), \(\frac{2}{3}(1) = \frac{2}{3}\).- For \((4^{3/2}): 4^{3/2} = 8\), \(\frac{2}{3}(8) = \frac{16}{3} \approx 5.33\).Now combine the computations:\(7.45 - \frac{2}{3} - (5.33 - 0) = 6.78 - 5.33 \approx 1.45\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, utilized to find quantities like area under a curve or between curves, volume, and more. It revolves around summing infinitesimally small quantities to find a whole. In our problem, integration helps us calculate the area between two curves, where simple geometry fails.

To understand integration, it's useful to think about the area under a curve as a series of thin vertical strips. Each strip resembles a rectangle with a very small width. Adding up all these little rectangles' areas approximates the total area.

In calculus, integration is often symbolized with an elongated "S" shape known as the integral sign (∫). To integrate a function, you'd find its antiderivative, which is a function that, when differentiated, returns the original function. The definite integral, which we'll dive into next, is specifically useful for bounded quantities.

Definite Integral

The definite integral helps you find the exact value of the area under a curve between two points, or more specifically, over an interval [a, b]. In this exercise, the definite integral is used to find the area between two curves, which is recalculated as the area under the first curve minus the area under the second curve over the same interval.

When setting up a definite integral, points 'a' and 'b' represent the lower and upper bounds of integration. In our scenario, these bounds are 0 and 4, as we are tasked with finding the area between the curves over this range.

For \[\int_{a}^{b} f(x) \, dx,\]the process involves finding the antiderivative of \( f(x) \), then evaluating it at the upper bound \( b \), and subtracting the value at the lower bound \( a \). This step simplifies the calculation of the area exactly between the curves \( y = \sqrt{x+1} \) and \( y = \sqrt{x} \).

Area Between Curves

Finding the area between curves means calculating the space confined by those curves. More precisely, it involves determining the integral of the top function minus the bottom function over a specified interval. In this exercise, these are the functions \( y = \sqrt{x+1} \) and \( y = \sqrt{x} \), and the interval is given as \( 0 \leq x \leq 4 \).

The goal is to compute:\[\int_{0}^{4} (\sqrt{x+1} - \sqrt{x}) \, dx.\]Here, \( \sqrt{x+1} \) is the function positioned above \( \sqrt{x} \) for each \( x \) in [0,4].

This setup ensures that we correctly account for the height difference between the two curves across the range. After setting up the integral, we integrate each function separately, plugging in the limits, and then subtract the results to find the net area. The solution, as calculated, gives us an accurate depiction of the space between the curves, equivalent to about 1.45 units, as shown in the provided exercise solution.