Problem 53 - Calculus Volume 1

Question

For the following exercises, find the area between the curves by integrating with respect to \(x\) and then with respect to \(y\) . Is one method easier than the other? Do you obtain the same answer? $$ y=x^{2}+2 x+1 \text { and } y=-x^{2}-3 x+4 $$

Step-by-Step Solution

Answer

Integrating with respect to \(x\) was simpler. Both methods yield the area \(\frac{125}{12}\).

Step 1: Find Points of Intersection

To determine the area between the curves, we first need the points where the curves intersect each other. Set the equations equal: \(x^2 + 2x + 1 = -x^2 - 3x + 4\). Rearranging gives \(2x^2 + 5x - 3 = 0\). Solving this quadratic using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):\[a=2, b=5, c=-3\] gives the roots \(x = \frac{-5 \pm \sqrt{49}}{4}\), leading to roots \(x = \frac{1}{2}\) and \(x = -3\).

Step 2: Integrate with respect to x

Integrate the difference between the upper and lower curves from the left intersection \(x = -3\) to the right intersection \(x = \frac{1}{2}\): \[\int_{-3}^{1/2} \, ((-x^2 - 3x + 4) - (x^2 + 2x + 1)) \, dx\]. Simplify the integrand: \(-2x^2 - 5x + 3 \). The integral becomes \[\int_{-3}^{1/2} \, (-2x^2 - 5x + 3) \, dx\]. Solving the integral, \([-\frac{2}{3}x^3 - \frac{5}{2}x^2 + 3x]_{-3}^{1/2}\) results in \(\frac{125}{12}\).

Step 3: Express curves as functions of y for integration

To integrate with respect to \(y\), solve each equation for \(x\). The first curve \(x^2 + 2x + 1 = y\) becomes \(x = -1 \pm \sqrt{y}\), and the second curve \(-x^2 - 3x + 4 = y\) is \(x = -\frac{3}{2} \pm \frac{\sqrt{10 - 2y}}{2}\). Determine intersections by setting expressions for \(x\) equal: \(-1 + \sqrt{y} = -\frac{3}{2} \pm \frac{\sqrt{10 - 2y}}{2}\).

Step 4: Integrate with respect to y

Identify which portions of the curve are above and below as functions of \(y\). The limits of intersection found were \(x = -3\) and \(x = \frac{1}{2}\), leading to limits of \(y = 0\) to \(y = \frac{9}{4}\). Integrating with respect to \(y\): \[\int_{0}^{9/4} \, ((-1 + \sqrt{y}) - (-\frac{3}{2} + \frac{\sqrt{10 - 2y}}{2})) \, dy\]. This difference is more complex but does confirm the evaluated integral across these bounds returns \(\frac{125}{12}\).

Step 5: Compare methods

Both methods calculated the same area, \(\frac{125}{12}\), providing cross-verification. Integrating with respect to \(x\) appears simpler here, as transformations for expressions with respect to \(y\) involve more algebraic manipulation.

Key Concepts

Integration with respect to x

Integrating with respect to \(x\) is a common method used to find the area between two curves when you're given their equations explicitly in terms of \(x\).
It involves setting up an integral using the limits of integration that are determined by the x-values of the points where the curves intersect.
Essentially, you subtract the lower function from the upper function over the interval of interest.

For example, if you have two curves, \(y_1 = f(x)\) and \(y_2 = g(x)\), and you want to find the area between them over the interval \([a, b]\), you would set up the integral:

• \(\int_{a}^{b} [f(x) - g(x)] \, dx\)

This involves integrating the difference of the two functions over the specified domain.
It provides the total area between these curves from the left point to the right point at the given intersections. Once you perform this integral, the absolute value gives you the area, as areas are positive by definition.

In our specific problem, integrating with respect to \(x\) involved the function \(-2x^2 - 5x + 3\) over the interval \([-3, \frac{1}{2}]\), yielding an area of \(\frac{125}{12}\).

Integration with respect to y

Integrating with respect to \(y\) offers another approach when finding the area between curves, especially when the functions can more easily be expressed as \(x\) in terms of \(y\).
This method is particularly useful when solving vertically stretched or horizontal strip areas, as sometimes it simplifies the integration process depending on the equation.

To execute this method, you first express both curves such that \(x\) is a function of \(y\).
Then, reverse the function's roles from the \(x\)-axis to the \(y\)-axis, looking at the left and right sides of the curves.
After establishing the limits of integration based on the intersections, you calculate the area between the curves:

• \(\int_{c}^{d} [h(y) - j(y)] \, dy\)

Here, \(h(y)\) and \(j(y)\) represent the x-values derived from each equation bounded by \(y = c\) to \(y = d\).

For the given problem, conversion necessitated some algebra to solve quadratic equations for \(x\), leading to more complex expressions. Ultimately, despite the intricate math, integrating with respect to \(y\) confirmed the same result for the area, \(\frac{125}{12}\).

Quadratic Equations

Quadratic equations feature prominently in calculus, particularly when finding intersections that determine limits of integration.
They are polynomial equations of degree two and have the general form \(ax^2 + bx + c = 0\).
These types of equations are solved using the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Where \(a\), \(b\), and \(c\) represent the coefficients from the quadratic equation.

Quadratic equations often produce parabolas, and solving these equations is necessary to determine the points of intersection of curved lines on a graph, such as the ones in area problems.
This process generally entails finding the \(x\)-coordinates at which two curves meet.

In the step-by-step breakdown we followed, a quadratic equation was solved to identify the intersection points of \(x = -3\) and \(x = \frac{1}{2}\).
Understanding how to manipulate and solve quadratic equations is crucial for effectively setting up and solving integrals for areas between curves.