How to Solve the Area Between Curves Using Integration? - Calculus

Finding the area by integration is the most basic type of application when encountering integrals. This post explains the theory behind it. Let's find the area between two generic curves.

Approximation

To understand how integration works, we must know how to approximate its area. Using Elementary Geometry, this is how we do it:

First, divide the area between the two functions into a definite number of geometric figures \(n\) like a series of rectangles with uniform width.
Then, determine each area of each division \(A_1, A_2, A_3, …, A_n\)
Finally, add all computed areas to find the total approximate area.

We can have as many divisions as we'd like. One important thing to note is that these shapes' widths are the difference between two \(x\) positions along the grid line \(w_1=x_1-x_0\), \(w_2=x_2-x_1\), and \(w_3=x_3-x_2\)

If we let the areas of these three divisions be \(A_1\), \(A_2\), and \(A_3\), the approximate area \(A_T\) would be the sum of these three areas.

\(A_T \approx A_1 + A_2 + ... + A_n \approx \sum_{i=1}^n A_i\)

This solution is an excellent approach to the problem. How well the approximation depends on the number of divisions made for the figure. From this example, we can observe the following:

The more rectangles \(n\), the more closely it would lead to the actual value.
As \(n\) increases, these rectangles' width and area become smaller.
We approach the exact value more closely as the width and area get smaller.

Exact Solution

The solution above is good if we want a rough answer; however, if we want a more precise answer, then integration comes in.

From the approximate solution, we know that if we want an exact answer, the width and area of its divisions must be so small. Building from this, we can solve the precise area. We will only need to modify two things in the approximate solution:

We need to set up a rectangular division that is so small that we can express its area as the differential \(dA\)
We need an operation that will sum up all of these tiny divisions

We can combine all \(dA\) through integration; hence, the general equation to solve for the region \(A_T\) between two functions is:

\(A_T=\int_{a}^{b}{dA}\)

\(A_T\) is the total area between the two functions
\(dA\) is the differential area of the division.
\(a\) is the lower limit position (endpoint with the lowest value)
\(b\) is the upper limit position (endpoint with the highest value)

Differential Strip

Let's expand on this small division with differential area \(dA\). We call this rectangle the differential strip, which consists of a base and height that we call the differential width and length, respectively. The two will depend on the orientation of the strip.

This differential width \(dw\) is the difference between two positions \(x_A\) and \(x_B\), and it is so tiny that it's nearly equal to zero but not precisely zero. It's so small that rather than using \(x_A\) and \(x_B\) to represent the two positions, we can describe the width with a single position \(x\)

Vertical Orientation

When the width is along the x-axis, we say that the differential strip is in the vertical position. The differential width and length are oriented as follows:

The differential width \(dx\) takes positions along the x-axis. It would take the endpoints along this axis as its lower and upper limit.
We take the length from the y-axis. In this orientation, it's the height difference \(y\) between the two functions.

From this, we can say the length is equal to \(y=f(x)-g(x)\). The area of the differential strip is then: \(dA=ydx\). We can further expand the integration equation as follows:

\(A_T=\int_{x_1}^{x_2}ydx\)

\(A_T\) is the total area of the region
\(x_1\) is the lower limit along the x-axis
\(x_2\) is the upper limit along the x-axis
\(y\) is the length between the two functions (difference between functions along y)
\(dx\) is differential along x

Horizontal Orientation

When the width is along the y-axis, we say that the differential strip is in the horizontal position. The differential width and length are oriented as follows:

The differential width \(dy\) takes positions along the y-axis. It would take the endpoints along this axis as its lower and upper limit.
We take the length from the x-axis. In this orientation, it's the width difference \(x\) between the two functions.

From this, we can say the width is equal to \(x=f(y)-g(y)\). The area of the differential strip is then: \(dA=xdy\). We can further expand the integration equation as follows:

\(A_T=\int_{y_1}^{y_2}xdy\)

\(A_T\) is the total area of the region
\(y_1\) is the lower limit along the y-axis
\(y_2\) is the upper limit along the y-axis
\(x\) is the width between the two functions (difference between functions along x)
\(dy\) is differential along y

Summary

Solving the area by integration expands on the approximation solution.

The logic is to subdivide the space needed with multiple divisions, solve for the area of each part, and add all of these areas.

To improve accuracy, the area of a part must be small enough. To find the exact answer, we must express the area as differential \(dA\)

We can combine all \(dA\) through integration; hence, the general equation to solve for the region \(A_T\) between two functions is: \(A_T=\int_{a}^{b}{dA}\). \(A_T\) is the total area between the two functions, \(dA\) is the differential area of the division, \(a\) is the lower limit position (endpoint with the lowest value), and \(b\) is the upper limit position (endpoint with the highest value)

The rectangle with area \(dA\) is the differential strip, which consists of a base and height that we call the differential width and length.

For a vertically oriented differential strip, the area is equal to: \(A_T=\int_{x_1}^{x_2}ydx\). The variable \(y\) is the length of the strip (difference between functions), while \(dx\) is the differential width along the x-axis.