Problem 24 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=2 x, \quad y=x \sqrt{x+1} $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the equations \(y=2x\) and \(y=x\sqrt{x+1}\) is \(\frac{137}{15}\) square units. This is determined by finding the points of intersection, sketching the region, and evaluating the integral of the difference between the two functions over the interval [0, 3].

Step 1: Find the points of intersection

To find the points of intersection between the two curves, we can set the two equations equal to each other: \(2x = x\sqrt{x+1}\) Solve for x: First, observe that x=0 is a solution since both expressions are equal to 0 when x=0. For non-zero x, we can divide both sides by x: \( 2 = \sqrt{x+1} \) Now, square both sides: \( 4 = x+1 \) And solve for x: \( x = 3 \) So we have the two points of intersection \(x=0\) and \(x=3\).

Step 2: Sketch the region bounded by the graphs

To sketch the region bounded by the graphs, we first draw the graph of \(y=2x\) and the graph of \(y=x\sqrt{x+1}\). Then, locate the points of intersection at x=0 and x=3, and shade the region that is bounded by the graphs.

Step 3: Set up the integral

We'll find the area of the region by integrating the difference between the two functions over the interval [0,3]. Since \(y=2x\) is above \(y=x\sqrt{x+1}\) in the shaded region, we'll subtract the lower function from the higher one: Area = \(\int_{0}^{3} (2x - x\sqrt{x+1}) dx\)

Step 4: Evaluate the integral

Next, we'll evaluate the integral: \( \int_{0}^{3} (2x - x\sqrt{x+1}) dx = \int_{0}^{3} 2x dx - \int_{0}^{3} x\sqrt{x+1} dx \) Evaluating the first integral: \(\int_{0}^{3} 2x dx = x^2\Big|_{0}^{3} = 9\) Now, we'll evaluate the second integral using substitution. Let u = x+1, then x = u - 1, so dx=du: \(\int_{0}^{3} x\sqrt{x+1} dx = \int_{1}^{4} (u - 1)\sqrt{u} du\) Solve the integral using substitution: \(\int_{1}^{4} (u - 1)\sqrt{u} du =\int_{1}^{4}u\sqrt{u}du- \int_{1}^{4}\sqrt{u}du \) \(=\frac{2}{5}u^\frac{5}{2}\Big|_{1}^{4} - \frac{2}{3}u^\frac{3}{2}\Big|_{1}^{4}\) \(= \frac{2}{5}(32 - 1) - \frac{2}{3}(8 - \sqrt{1})\) \(= \frac{62}{5} - \frac{14}{3}\) Area = \(9 - (\frac{62}{5} - \frac{14}{3})\)

Step 5: Find the area

Now we'll find the area: Area = \(9 - (\frac{62}{5} - \frac{14}{3}) = \frac{27}{5} + \frac{28}{15}\) Area = \(\frac{137}{15}\) The area of the region bounded by the graphs of the given equations is \(\frac{137}{15}\) square units.

Key Concepts

Intersection Points

Finding the intersection points of given functions is the first step in calculating the area between curves. Intersection points show where the graphs of two functions meet, helping to define the limits for any calculations. For our example, we are working with the equations \( y = 2x \) and \( y = x\sqrt{x+1} \). To find these intersection points, we first set the equations equal to each other: \( 2x = x \sqrt{x+1} \). By solving this equation, we find two values for \( x \): 0 and 3. These represent the points where the curves intersect. It’s important to double-check that these \( x \) values satisfy both original equations. Intersection points help locate the section of the graph we are interested in. In our context, it marks the boundaries of the region whose area we need to find.

Definite Integral

A definite integral calculates the accumulation of quantities, such as areas under or between curves. In terms of areas, a definite integral takes two points on a graph, usually intersection points, and measures the area between these points under the curve.This involves an understanding of limits, as the definite integral is effectively finding the difference of sums from a starting point to an endpoint. For the problem at hand, the definite integral \( \int_{0}^{3} (2x - x\sqrt{x+1}) \, dx \) is used. The limits 0 and 3 are derived from our previously calculated intersection points.When working with definite integrals, breaking down complex expressions into simpler integrals can be beneficial. Each part usually needs solving separately, ensuring an accurate overall computation. In this task, the calculation split into two parts for simplicity: the integral of \( 2x \) and the more complex \( x\sqrt{x+1} \) term.

Graph Sketching

Sketching graphs is an essential skill in visualizing how functions interact and determining the area between them. For this, we plot both functions: \( y = 2x \) and \( y = x\sqrt{x+1} \).- First, draw the linear function \( y = 2x \), which is a straight line with a slope of 2, passing through the origin.- Then, plot \( y = x\sqrt{x+1} \). This function curve can be trickier as it involves a square root, indicating a radical curve with slower growth compared to the linear function.Identifying intersection points, now at \( x = 0 \) and \( x = 3 \), helps outline the region we need. By sketching the area between these two lines from x = 0 to x = 3, we can clearly see which function lies above the other and should be subtracted in our integration setup.

Function Subtraction

To find the area between two curves, we calculate the difference above and below the x-axis or each other. Function subtraction involves taking the top curve and subtracting the bottom curve over a specific interval. For this example:- Our top function is \( y = 2x \).- Our bottom function is \( y = x\sqrt{x+1} \).We set up the integral \( \int (2x - x\sqrt{x+1}) \, dx \) over the intersection points from 0 to 3. Subtraction of functions ensures that we capture only the area between the curves, disregarding excess below or outside our focus area.This simple subtraction transforms into several integral problems which include finding the net area. Thus, function subtraction leads directly to the proper setup in defining our integral, key in tasks involving calculation of area between complicated curves.