Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the area bounded by \(y=2-x^{2}\) and \(y=x\).

Step-by-Step Solution

Answer

So, the area bounded by \(y=2-x^{2}\) and \(y=x\) is \(7/3\) square units.

Step 1 - Finding Points of Intersection

The points of intersection of the two curves can be found by setting \(y=2-x^{2}\) and \(y=x\) equal to each other:\(2-x^{2} = x \)This gives us a quadratic equation that can be solved for \(x\). Rearrange the terms:\(x^{2} + x - 2 = 0 \)This factors to:\( (x-1)(x+2) = 0 \)Setting each factor equal to zero gives the solutions \(x=1\) and \(x=-2\). Plugging these back into \(y=x\) gives the corresponding y-values, \(y=1\) and \(y=-2\). So the points of intersection are \((1, 1)\) and \((-2, -2)\).

Step 2 - Setting Up the Integral

The integral for the area bounded by the two curves from \(x=-2\) to \(x=1\) is given by:\[ \int_{-2}^{1} (2-x^{2}) - x dx \]This expression is obtained by subtracting the equation of the lower curve (\(y=x\)) from the equation of the upper curve (\(y=2-x^{2}\)).

Step 3 - Evaluating the Integral

Now, compute the integral:\[ = -(x^{3}/3) + x^{2} + x -2 | _{-2}^{1} \]Substituting the lower and upper limits of the integral, we compute:\[ = -(1/3) + 1 + 1 - 2 - [(-8/3) + 4 +2 -2 ] = 11/3 -4/3 = 7/3 \]

Key Concepts

Integration

Integration is a fundamental tool in calculus, especially useful when determining areas bounded by curves. When you have two functions that intersect and you want to find the area between them, integration helps you do just that. To find the area, you must integrate the difference between the two functions over a particular interval where they intersect.

Here’s a simpler breakdown:

• First, identify the two functions or curves involved.
• Next, determine the interval where these curves intersect.
• Set up the integral by subtracting the function that lies below from the one that lies above, across the intersection interval.

The definite integral will give us the net area between the curves over the specified interval. In our exercise, we subtracted the curve represented by the linear function from the quadratic function and integrated from -2 to 1.

Remember: If you get a negative area, this likely means that the upper and lower curves were switched in your integral set-up. Double-check your functions to ensure they are in the correct order.

Quadratic Equations

Quadratic equations take the form of x^2 + bx + c = 0, and are crucial in finding where curves intersect, especially when one of them is a parabola. The standard approach to solve for the variable, x, in these equations is to factorize (if possible), use the quadratic formula, or complete the square.

In our example, we set the two equations equal to each other to represent their intersection, leading to a quadratic equation: x^2 + x - 2 = 0. This specific equation is relatively simple and can be factored easily since it decomposes into two binomials: (x - 1)(x + 2) = 0.

With factoring, we set each factor equal to zero. This gives solutions x = 1 and x = -2, which are the x-coordinates of the intersection points.

Quadratic equations are powerful in analyzing and predicting the behavior of parabolas and other curves, especially in understanding their points of intersection with other lines or curves.

Points of Intersection

Finding the points of intersection between two curves is a commonly encountered problem in calculus. This step is indispensable when determining areas between curves. These points are essential as they define the interval over which you will integrate.

The process involves:

• Equating the two expressions or equations of the curves.
• Solving this resultant equation, often quadratic, to find the x-values for the points of intersection.

In our example, setting y = 2 - x^2 equal to y = x allowed us to solve for x-values by forming a quadratic equation. The solutions, x = 1 and x = -2, were then used to find corresponding y-values, yielding the full intersection points (1, 1) and (-2, -2).

These intersection points not only define the limits of integration but also provide necessary insight into the geometry of the interacting curves.