Problem 46 - Calculus Single Variable

Question

Calculate the area of the region between the pair of curves. $$ y=(x-3) / 2 \quad x=y^{2} $$

Step-by-Step Solution

Answer

The area between the curves is \( \frac{32}{3} \).

Step 1: Find Points of Intersection

To find the points of intersection between the curves, set \( y = \frac{x-3}{2} \) equal to \( x = y^2 \). Substitute the expression for \( y \) into the equation \( x = y^2 \): \( x = \left( \frac{x-3}{2} \right)^2 \). Solve this equation for \( x \).

Step 2: Simplify Intersection Equation

Substitute \( y = \frac{x-3}{2} \) to get: \[ x = \left( \frac{x-3}{2} \right)^2 = \frac{x^2 - 6x + 9}{4} \]. Then, multiply everything by 4: \[ 4x = x^2 - 6x + 9 \]. Rearrange to get a quadratic equation: \[ x^2 - 10x + 9 = 0 \].

Step 3: Solve Quadratic for x

Solve \( x^2 - 10x + 9 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=-10, c=9 \). Thus, \( x = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} \), which gives solutions \( x = 9 \) and \( x = 1 \).

Step 4: Determine Corresponding y-Values

For \( x = 9 \), substitute back into \( y = \frac{x-3}{2} \) to find \( y = 3 \). For \( x = 1 \), similarly substitute to find \( y = -1 \). Thus, the points of intersection are \((9,3)\) and \((1,-1)\).

Step 5: Integral Setup for Area of Region

We need to find the area between the curves \( y = \frac{x-3}{2} \) and \( x = y^2 \) from \( x = 1 \) to \( x = 9 \). Here, switch roles: use \( x = y^2 \) and \( y = \frac{x-3}{2} \) to express in terms of \( y: \) \( x = y^2 \) and \( x = 2y + 3. \)

Step 6: Integrate with Respect to y

The integral is: \[ \int_{-1}^{3} ((2y + 3) - y^2) \, dy \]. Simplify within the integral: \[ \int_{-1}^{3} (2y + 3 - y^2) \, dy \].

Step 7: Solve the Definite Integral

Calculate \( \int_{-1}^{3} (2y + 3 - y^2) \, dy = [y^2 + 3y - \frac{y^3}{3}]_{-1}^{3}. \) Evaluate from \(-1\) to \(3\): \([3^2 + 3(3) - \frac{3^3}{3}] - [(-1)^2 + 3(-1) - \frac{(-1)^3}{3}] = [9 + 9 - 9] - [1 - 3 + \frac{1}{3}] = 9 - (-\frac{5}{3}) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3} \).

Step 8: Conclusion

The area of the region between the curves \( y = \frac{x-3}{2} \) and \( x = y^2 \) is \( \frac{32}{3} \).

Key Concepts

Integration

Integration is a mathematical process used to find quantities, like areas under curves, by summing infinitesimally small data points. By integrating, we can determine the total accumulation of a quantity, whether that be distance, volume, or in our case, area. The concept plays a crucial role when calculating the area between two curves, as it allows for the accumulation of the space between them.
To find the area between curves, we typically subtract one function from another across a certain interval. In this exercise, the area is found by integrating the function

• \((2y + 3) - y^2\) from
• \(y = -1\) to
• \(y = 3\).

These specific integration limits correspond to the y-values we found at the intersection points of the curves. This gives us the shaded region, and by solving the integral, we determine the total area.

Quadratic equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They arise frequently in numerous mathematical problems involving curves and surfaces. Solving quadratic equations helps us to find critical points, such as solutions for intersections between curves.
In this particular problem, we encounter a quadratic equation when finding the points of intersection: \(x^2 - 10x + 9 = 0\). We used the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), to find solutions for \(x\) which are 9 and 1. These solutions are crucial because they define the endpoints where the curves intersect. In terms of coordinates, it reveals the x-values essential to fully define the area we are calculating between the curves.

Definite integral

A definite integral is a method to calculate the accumulated sum of values across a specified interval. It is symbolized as \(\int_a^b f(x)\,dx\), where \(a\) and \(b\) are limits of integration.
This problem involved setting up a definite integral to find the area of a region bound by two intersecting curves. The integral used, \(\int_{-1}^{3} (2y + 3 - y^2) \, dy\), helps us determine the total area between curves \(y = \frac{x - 3}{2}\) and \(x = y^2\) over the

• range \(y = -1\) to
• \(y = 3\).

By evaluating this definite integral, we calculated the result as \(\frac{32}{3}\). The definite integral offers a concrete value that represents the complete area, an essential step in solving area between curves problems.

Points of intersection

Points of intersection refer to where two curves meet or cross each other. Finding these points is a fundamental step when solving many mathematical problems involving multiple curves, such as determining the area between them.
In this exercise, the curves \(y = \frac{x-3}{2}\) and \(x = y^2\) intersect each other. By setting \(y = \frac{x-3}{2}\) equal to \(x = y^2\), we derived a quadratic equation that allowed us to find the respective x-values of the intersection points, \(x = 9\) and \(x = 1\). The corresponding y-values were found by plugging these x-values back into the equation \(y = \frac{x - 3}{2}\), resulting in points \((9, 3)\) and \((1, -1)\).
These points outline the boundaries of integration. Points of intersection give us the necessary limits to calculate the area effectively, helping translate the curves into understandable data for our calculations.