Problem 16 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x, y=3-x, x=0$$

Step-by-Step Solution

Answer

The area bounded by the curves is \(\frac{9}{4}\).

Step 1: Identify the bounded area

The area we are interested in is bounded by the curves: \(y = x\), \(y = 3 - x\), and the vertical line \(x = 0\). To find the bounded area, first determine the points of intersection between the curves.

Step 2: Find the points of intersection

To find the intersection of \(y = x\) and \(y = 3 - x\), set the equations equal: \(x = 3 - x\). Solve for \(x\):\[x + x = 3 \implies 2x = 3 \implies x = \frac{3}{2}\]The intersection points of interest for the area are: \((0,0)\) from \(x = 0\), and \(\left(\frac{3}{2}, \frac{3}{2}\right)\).

Step 3: Set up the integral for the area between curves

The area between the two curves from \(x = 0\) to \(x = \frac{3}{2}\) can be found by integrating the difference between the functions, \( (3-x) - x \), over this interval:\[ A = \int_0^{\frac{3}{2}} ((3-x) - x) \, dx = \int_0^{\frac{3}{2}} (3 - 2x) \, dx \]

Step 4: Evaluate the integral

Calculate the integral:\[ A = \int_0^{\frac{3}{2}} (3 - 2x) \, dx = \left[ 3x - x^2 \right]_0^{\frac{3}{2}} \]First, evaluate at the upper bound:\[ \left(3 \times \frac{3}{2} - \left(\frac{3}{2}\right)^2\right) = \left(\frac{9}{2} - \frac{9}{4}\right) = \frac{18}{4} - \frac{9}{4} = \frac{9}{4} \]Next, evaluate at the lower bound:\[ 3 \times 0 - 0^2 = 0 \]Thus, the area is:\[ \frac{9}{4} - 0 = \frac{9}{4} \]

Step 5: Finalize the result

The area bounded by the curves \(y = x\), \(y = 3 - x\), and \(x = 0\) is the result obtained from the integration:\[ A = \frac{9}{4} \]

Key Concepts

Area Between Curves

When you are trying to find the area between two curves, you're essentially looking for the space that lies between two curves on a given interval.
The curves in question 'sandwich' this area. Think of it as the gap between these two lines across the x-axis.
In our example, the curves are given by the equations \(y = x\) and \(y = 3 - x\), and the area we want is to the right of \(x = 0\) until these curves meet.

To find this area, follow these steps:

• First, identify the points where the curves intersect. This helps define the limits over which you will integrate.
• Once you have the intersection points, express the area as a definite integral of the top curve minus the bottom curve.
• This difference helps "slice" the area into tiny vertical strips that when summed up, give the total area.

These strips help you visualize what you're calculating: the gap between two curves at each point along the x-axis within the interval given by the intersection points.

Definite Integral

A definite integral is a type of integral that calculates the accumulation of quantities, like area, between an interval on the x-axis.
This is why they are so handy in finding areas between curves like in our exercise.

Here's how you use a definite integral for finding area:

• Set up the integral to reflect the difference between the curves; in our case, it's \((3-x) - x\).
• Specify the integration limits, which are your intersection points (e.g., from \(x=0\) to \(x=\frac{3}{2}\)).
• Once set up, solve this integral using fundamental calculus techniques. It provides you the net "area" by considering rates of change continuously along the x-axis.

After evaluating, you have the area bounded by the curves within those specific intersection points. Using integration ensures precise calculation of this area, even if the shape is curvy or irregular.

Intersection Points

Intersection points are where the curves cross each other. These points are crucial as they define the bounds for the area you are looking to calculate.
They signify where the curves swap their respective positions (which one is on "top" changes).

In our exercise, to find the intersection, we equate the two functions:

• Take \(y = x\) and \(y = 3 - x\) and solve \(x = 3 - x\).
• Solve the equation to find \(x = \frac{3}{2}\), which gives us one intersection point.
• The process also identifies another limit, \(x=0\), where both curves also meet the y-axis.

These intersection points \((0, 0)\) and \((\frac{3}{2}, \frac{3}{2})\) are essential because they allow us to precisely know where to start and stop the integration to compute the desired area.