Problem 10 - Essential Calculus

Question

Find the area of the region by integrating (a) with respect to \(x\) and (b) with respect to \(y\). $$ \begin{array}{l} y=x^{2} \\ y=6-x \end{array} $$

Step-by-Step Solution

Answer

The area of the region can be found by integrating the difference of the functions along the range from the points of intersection. The integral for respect to \(x\) gives the area by integrating from the left x-intercept to the right x-intercept, while the integral for respect to \(y\) gives the area by integrating the transformed functions from y=min to y=max. The process includes solving quadratic equations, transforming functions, setting up and calculating definite integrals.

Step 1: Find the Points of Intersection

Set the two equations equal to each other. \(x^{2}=6-x\). Solve this quadratic equation to find the x-intercepts; these are the points where the two graphs intersect along the x-axis.

Step 2: Transform Equations for y-integration

Rewrite the equations in terms of \(x\). This will be necessary to integrate with respect to \(y\) in part (b) of the exercise.

Step 3: Set Up the Integral

The area between curves is given by the integral of the difference of the functions. For the respect to \(x\), set up the integral from the left x-intercept to the right x-intercept of \((6-x) - x^{2}\). Similarly, for y, set up the integral from y=min to y=max. As the equations are transformed to \(x\), it'll be of \(\sqrt{y} - (6-y)\), also note that when \(y\leq4\), \(-sqrt{y}\) is the lower function

Step 4: Calculate the Integral

Perform the integral calculation and add/subtract appropriately for each case to find the final area.

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept used to find the area under curves and between curves. It helps answer questions like: "How much space is between these two functions?" To find this area, you take the integral of the top curve minus the bottom curve over a specified interval. In this exercise, for integration with respect to \(x\), you calculate the integral from one intersection point to the other. You found these points earlier by solving the equation \(x^2 = 6-x\).

Once you have the intersection points, which are the limits of integration, you set up the integral: \(\int (6-x - x^2) \, dx\). Similarly, for integration with respect to \(y\), you determine the area by setting up the integral of the right function minus the left and integrating from \(y_{\text{min}}\) to \(y_{\text{max}}\).

Overall, integral calculus allows for the computation of the exact area, ensuring an accurate and reliable solution. This is especially useful whenever you need to find the area that isn't a simple shape, like squares or triangles, and require more advanced mathematics.

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\). In this problem, the equation \(x^2 = 6-x\) represents a quadratic equation. Solving this involves rearranging it into standard form: \(x^2 + x - 6 = 0\).

To solve a quadratic equation like this, you can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Alternatively, you can try factoring, if possible. For this particular equation, it factors into \((x-2)(x+3) = 0\).

Key points about quadratic equations:

• They may have zero, one, or two real solutions.
• The solutions are where the quadratic graph crosses the x-axis, called roots or x-intercepts.
• Quadratic equations often appear when dealing with parabolic functions.

Equation Transformation

Equation transformation refers to changing an equation's form to make solving a problem easier. In this exercise, transformation is crucial when switching integration from \(x\) to \(y\).

For integration with respect to \(y\), you transform the original functions defined as \(y\) in terms of \(x\), into \(x\) expressed as functions of \(y\). For the equations \(y = x^2\) and \(y = 6-x\), this means rearranging them to express \(x\) in terms of \(y\).

• The first function becomes \(x = \pm \sqrt{y}\), with \(\sqrt{y}\) being the principal square root.
• The second function solves to \(x = 6 - y\).

These transformed equations allow us to redefine the area calculation in terms of \(y\) limits. They make it possible to integrate with respect to \(y\) accurately, based on the respective transformed curves between which the area is enclosed. This ensures that the integral setup will correctly account for the changing bounds and roles of the functions as the integration variable changes.