Problem 14 - 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. $$ y=x^{2}-4 x, \quad y=-x+4 $$

Step-by-Step Solution

Answer

The region bounded by the graphs of the equations \(y = x^2 - 4x\) and \(y = -x + 4\) is enclosed by the points of intersection \((4, 0)\) and \((-1, 5)\). The area of this region can be found by integrating the difference between the two functions over the interval \([-1, 4]\), leading to \(\int_{-1}^{4}(-x^2 + 5x -4)\,dx\). The result of the evaluation is a negative value, so we take the absolute value to get the area of the enclosed region as \(\dfrac{7}{2}\).

Step 1: Find the Points of Intersection

Set the two equations equal to each other to find the points of intersection: \[x^{2} - 4x = -x + 4\] Now, solve the equation to find the x-coordinates of the intersection points: \[x^{2} - 4x + x - 4 = 0\] \[x^{2} - 3x - 4 = 0\] Factor the quadratic equation: \[(x - 4)(x + 1) = 0\] The solutions to the equation are \(x = 4\) and \(x = -1\). Now, find the corresponding y-coordinates: 1. For \(x = 4\): \(y = -(4) + 4 = 0\) 2. For \(x = -1\): \(y = -(-1) + 4 = 5\) Thus, the points of intersection are \((4, 0)\) and \((-1, 5)\).

Step 2: Sketch the Graphs

Now use the information gathered in Step 1 to sketch the graphs of both functions, highlighting the region enclosed by them. The region is defined by \(-1 \leq x \leq 4\).

Step 3: Calculate the Area of the Bounded Region

To calculate the area of the region, we will determine the difference between the two functions and integrate with respect to x over the interval defined by the x-coordinates of the intersection points, which is \([-1, 4]\). The difference between the two functions is: `(-x + 4) - (x^2 - 4x)`, which simplifies to `-x^2 + 5x - 4`. Now, we will integrate this function with respect to x from -1 to 4: \[ \int_{-1}^{4} (-x^{2} + 5x - 4) \, dx \] Integrate the function: \[ \left[-\dfrac{1}{3}x^3 + \dfrac{5}{2}x^{2} - 4x\right]_{-1}^4 \] Evaluate the integral at the bounds: \[ \left[-\dfrac{1}{3}(4)^3 + \dfrac{5}{2}(4)^{2} - 4(4)\right] - \left[-\dfrac{1}{3}(-1)^3 + \dfrac{5}{2}(-1)^{2} - 4(-1)\right] \] Now, simplify the obtained expression: \[ \left[-\dfrac{64}{3} + 40 - 16\right] - \left[\dfrac{1}{3} + \dfrac{5}{2} + 4\right] \] Combine the terms: \[ -\dfrac{64}{3} + 24 + \dfrac{1}{3} - \dfrac{5}{2} - 4 \] \[ \left(-\dfrac{63}{3}\right) + 20 - \dfrac{5}{2} \] \[ -21 + 20 - \dfrac{5}{2} \] \[ -1 - \dfrac{5}{2} \] \[ -\dfrac{7}{2} \] Since the area cannot be negative, the absolute value of the result is the area of the region bounded by the two functions: \[ Area = \dfrac{7}{2} \]

Key Concepts

Area Under Curves

The area under a curve in calculus gives us a way to understand how quantities accumulate over intervals. This is particularly useful when dealing with functions that describe changing rates, like speed or growth. In this exercise, you'll determine the area under the curves described by two functions. The concept involves calculating how much space is enclosed between the curves across a particular interval. By finding where the curves intersect, you can determine this region's bounds. The definite integral, as explained later, is key to calculating this area accurately. Knowing the area gives insights into the cumulative behavior of certain functions or phenomena determined by these functions.

Intersection Points

Finding the points where two curves intersect is crucial to solving area problems like the one we have here. These points tell us where the bounds of our area are located.To find these intersection points, set the two equations equal to each other and solve for the variable(s). For example, in the given problem, equate the functions: - Solve: \(x^2 - 4x = -x + 4\).- Simplify and factor: \((x - 4)(x + 1) = 0\).- Find solutions: \(x = 4\) and \(x = -1\).Once these x-values are found, plug them back into the original equations to find the y-values. This gives us intersection points \((4,0)\) and \((-1,5)\), marking the points where the curves meet.These points are essential as they will define the boundaries of your integral when calculating the area.

Definite Integral

The definite integral is the tool you'll use to calculate the area between curves. It essentially sums up an infinite number of small products of width and height to find the total area. In solving the given problem:- Determine the function that represents the difference between the upper and lower bounds: \((-x^2 + 5x - 4)\).- Set up the integral: \(\int_{-1}^{4} (-x^2 + 5x - 4) \, dx\).- Calculate it to get the accumulated area between these curves.Go through integration step-by-step to ensure accuracy. For those new to calculus, this process might seem complex at first.However, it's a powerful method to quantify spaces under curves precisely.

Function Sketching

Function sketching in calculus allows you to visualize mathematical equations and their interactions. This step helps in understanding not only where intersection points lie but also how the area between curves is shaped.While sketching the functions \(y = x^2 - 4x\) and \(y = -x + 4\):

• Identify critical points, like the vertex or linear slope.
• Mark intersection points found earlier: \((4,0)\) and \((-1,5)\).

Use these points and basic curve characteristics to draw the plot accurately.Sketching functions by hand is a useful practice.It provides a deeper grasp of their behavior beyond just numerical solutions.This understanding aids in visualizing the enclosed area under curves, offering a clearer pathway to solving calculus problems.