Problem 13 - 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. $$ y=-x^{2}+4, \quad y=3 x+4 $$

Step-by-Step Solution

Answer

The region bounded by the graphs of the given equations is enclosed between the downward-opening parabola \(y = -x^2 + 4\) and the straight line \(y = 3x + 4\), with the points of intersection at \((-3, -5)\) and \((0, 4)\). To find the area of this region, we'll integrate the difference of the two functions from the lower limit of \(-3\) to the upper limit of \(0\). That is, \(A = \int_{-3}^0 (-x^2 - 3x) \,dx\). Solving the integral, we find that the enclosed area is 9 square units.

Step 1: Find the points of intersection

To find the points where the two graphs intersect, we'll set the two functions equal to each other: \(-x^2 + 4 = 3x + 4\) Simplify the equation: \[-x^2 - 3x = 0\] Now let's factor the equation to solve for x: \[x(-x-3) = 0\] Setting each factor equal to 0, we get two values for x: \[x = 0 \quad \text{or} \quad -x-3 = 0 \Rightarrow x = -3\] Now, we'll find the corresponding y-values: For \(x = 0\): \(y = 3(0) + 4 = 4\) For \(x = -3\): \(y = 3(-3) + 4 = -5\) So, the points of intersection are \((-3, -5)\) and \((0, 4)\).

Step 2: Sketch the region

Now, let's prepare to sketch the region. Both function graphs are parabolas. The first equation is a parabola that opens downwards with the vertex at (0, 4). The second equation is a parabolic curve with a slope of 3 and intercept 4. Sketching the graph using these equations and the points of intersection, we'll see a region enclosed between the parabolas.

Step 3: Set up the integral to find the area

To find the area of the enclosed region, we'll integrate the difference of the two functions from the lower limit \(-3\) to the upper limit \(0\). Since for the interval from \(-3\) to \(0\), the quadratic function lies above the linear function, we will have the area as: \[A = \int_{-3}^0 (-x^2 + 4 - (3x + 4)) \,dx\] Simplify the integrand: \[A = \int_{-3}^0 (-x^2 - 3x) \,dx\]

Step 4: Calculate the integral

Now we will calculate the integral: \[A = \left[\frac{-1}{3}x^3 - \frac{3}{2}x^2\right]_{-3}^0\] Applying the limits of integration: \[A = \left[\frac{-1}{3}(0)^3 - \frac{3}{2}(0)^2\right] - \left[\frac{-1}{3}(-3)^3 - \frac{3}{2}(-3)^2\right]\] Which will simplify to: \[A = 9\] The area of the region bounded by the given graphs is 9 square units.

Key Concepts

Area under curve

When you're looking to calculate the area under a curve, it's important to understand what the function represents. In this case, we have two functions: a quadratic and a linear function. When you compute the area between two curves, you're essentially finding where one curve differs from another over a certain interval.

In our specific problem, the area between the curves of

• The quadratic function: \( y = -x^2 + 4 \)
• The linear function: \( y = 3x + 4 \)

represents the "region bounded by the graphs". This area is the difference between the functions, integrated over the range determined by the points of intersection.

Calculus lets us find this area by integrating the difference of these functions, evaluated between the points where they intersect.

Points of intersection

Understanding where two curves intersect is central to many calculus problems. Here, finding intersections helps us determine the limits of integration — the points from which and to which we'll calculate the area.

To find these intersection points, we equate the given functions \( y = -x^2 + 4 \) and \( y = 3x + 4 \). Solving for \( x \) gives us two solutions: \( x = 0 \) and \( x = -3 \). These solutions are where the parabolas cross each other on the graph. Using these \( x \)-values in either function gives corresponding \( y \)-values. The complete coordinates are:

• The point \((0, 4)\)
• The point \((-3, -5)\)

These intersection points are crucial as they define the region over which we need to integrate to find the enclosed area.

Graph sketching

Visualizing problems can often make them easier to solve. Sketching graphs is a great way to understand intersections and areas.

In this problem, you're working with two functions:

• \( y = -x^2 + 4 \) is a downward-opening parabola, with its vertex at \((0, 4)\).
• \( y = 3x + 4 \) is a straight line with a slope of 3, crossing the y-axis at \( y = 4 \).

By plotting these two functions, you can see the regions where they overlap. The intersections at \((-3, -5)\) and \((0, 4)\) specify the boundaries of such a region. Looking at these two graphs on the same coordinate plane helps identify areas of interest quickly, especially when calculating regions enclosed by the two equations.

Integration techniques

Calculating the area between curves typically involves integration. Here, we use integration to find the exact size of the area enclosed between the two intersections calculated earlier:

• From \(x = -3\)
• To \(x = 0\)

The integral \[ A = \int_{-3}^0 (-x^2 + 4 - (3x + 4)) \, dx \] represents the area.

Simplifying, we get \[ A = \int_{-3}^0 (-x^2 - 3x) \, dx \]. Next, we perform the integration:

• Calculate \( \int -x^2 \, dx \), which is \( \frac{-1}{3}x^3 \)
• Calculate \( \int -3x \, dx \), which gives \( \frac{-3}{2}x^2 \)

Then, evaluate the result from \(-3\) to \(0\) to obtain the final area, calculated to be 9 square units. Understanding these integration steps is vital for solving calculus problems involving areas between curves, as they provide accurate measurements of "how much space" is contained within defined boundaries on a graph.