Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-30, sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer: $$ y=x^{2}-2 x, y=-x^{2} $$

Step-by-Step Solution

Answer

The area between the curves is \(\frac{1}{3}\).

Step 1: Identify Intersection Points

To find the region bounded by the curves, we first need to determine where the graphs intersect. Set the equations equal to each other to find the intersection points: \(x^2 - 2x = -x^2\). Solving this equation, we get \(2x^2 - 2x = 0\) or \(2x(x-1) = 0\), which gives us the intersection points \(x = 0\) and \(x = 1\).

Step 2: Sketch the Graphs

Sketch the graphs of \(y = x^2 - 2x\) and \(y = -x^2\) to visualize the enclosed area. The function \(y = x^2 - 2x\) is a parabola that opens upwards and \(y = -x^2\) is a parabola that opens downwards. The region bounded by these parabolas is between \(x = 0\) and \(x = 1\).

Step 3: Set Up the Integral

To find the area between the curves, we evaluate the integral of the top function minus the bottom function from \(x = 0\) to \(x = 1\). Here, \(y = -x^2\) is the top function and \(y = x^2 - 2x\) is the bottom function within the interval. Set up the integral: \[\int_{0}^{1} (-x^2 - (x^2 - 2x)) \, dx = \int_{0}^{1} (-x^2 - x^2 + 2x) \, dx \] or \[\int_{0}^{1} (-2x^2 + 2x) \, dx.\]

Step 4: Calculate the Integral

Compute the integral: \[\int_{0}^{1} (-2x^2 + 2x) \, dx = \left[ -\frac{2}{3}x^3 + x^2 \right]_0^1.\]Evaluating this at the boundaries gives: \[\left( -\frac{2}{3}(1)^3 + (1)^2 \right) - \left( -\frac{2}{3}(0)^3 + (0)^2 \right) = -\frac{2}{3} + 1 = \frac{1}{3}.\]

Step 5: Estimate and Confirm Area

An estimate of the area, considering the geometric shape and nature of curves involved, can confirm the calculation. This geometric calculation should yield approximately \(\frac{1}{3}\) as well, verifying our integral calculation.

Key Concepts

Integral Calculus

Integral Calculus is a branch of calculus focusing on the concept of integration, which is essentially the process of finding the whole from a sum of parts. It deals with the accumulation of quantities, such as areas under curves, volumes, and other physical quantities. Integrals can be seen as the reverse operations of derivatives, and they allow us to compute total areas through a method called definite integration.
In this exercise, Integral Calculus is used to calculate the area of a region bound by two curves. We set up an integral by determining the functions that represent the area we are interested in. In particular, we integrate the difference between two functions over a specified interval. This calculation provides the precise area of the region in question.

Area Between Curves

The calculation of the area between curves is a common application of integral calculus. It involves identifying two distinct functions and calculating the area that lies between them over a certain range.

• To find this area, it's crucial to establish which function is on top and which is on the bottom within the given interval, as the top function minus the bottom function represents the vertical distance between them.
• Integration then helps to sum up these distances over the defined range, producing a total area measurement.

In our exercise, we have two parabolic curves. By evaluating the integrals of their differences, allowing us to compute the entire area enclosed, effectively capturing the space betwixt the curves, without manually counting ridges or measurements.

Parabolas

Parabolas are U-shaped curves on a graph and are the graphical representations of quadratic equations. They can open either upwards or downwards, depending on the sign of their leading coefficient.Here we have two parabolas represented by the functions:

• For the curve \(y = x^2 - 2x\), the parabola opens upwards because the coefficient of \(x^2\) is positive.
• Conversely, for the curve \(y = -x^2\), the parabola opens downwards due to the negative coefficient of \(x^2\).

The intersection of these parabolas creates the region of interest. Their behavior at the intersection points determines the boundaries of integration, leading to a clearer picture of the area that must be calculated using integral calculus.

Intersection Points

Intersection points occur where two equations or lines on a graph meet. These points are crucial when calculating areas bounded by multiple functions, as they define limits of integration.To find these points, set the equations equal to each other and solve for \(x\). For instance:

• The equation \(x^2 - 2x = -x^2\) simplifies to \(2x^2 - 2x = 0\), providing solutions at \(x = 0 \) and \(x = 1\).
• These solutions indicate where the curves intersect, pointing out crucial starting and stopping points for our integration.

Recognizing and accurately calculating these points builds the foundation for determining the integrals' bounds, enabling us to capture the phenomena or areas encompassed by curves on a graph.