Area Between Two Curves Calculus

The easiest way to think about the area between two curves: the area between the curves is the area below the upper curve minus the area underneath the lower curve. You can figure out the area between two curves by calculating the difference between the definite integrals of two functions. In 2-D geometry, the area is a volume that describes the region occupied by the two-dimensional figure. Two functions are needed to determine the area, say f(x) and g(x), and the integral limits from 'a’ to ‘b’ (b should be >a) of the function, that acts as the bespoke of the curve.

Formula to Find the Area between Two Curves

The basic mathematical expression written to compute the area between two curves is as follows:

If P: y = f(x) and Q : y = g(x) and x1 and x2 are the two limits,

Now the standard formula of- Area Between Two Curves, A=∫x2x1[f(x)−g(x)]

Through this topic, you should be able to:

ü  find the area between two curves

ü  find the area between two curves by integration

Calculating Areas Between Two Curves by Integration

1. Area under a curve – Region encircled by the given function, vertical lines and the x –axis.

2. Area Under a Curve – region encircled by the given function, horizontal lines and the y –axis.

3. Area between curves expressed by given two functions.

In case f(x) is a nonnegative and continuous function of x on the closed interval [a, b], then the area of the region enclosed by the graph of ‘f’, the x-axis and the vertical lines x=a and x=b is given by:

b

a

Area f (x)dx

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When computing the area under a curve f(x), follow the below set of instructions:

1. Shade the area.

2. Identify the boundaries a and b,

3. Establish the definite integral,

4. Integrate.

Calculating Areas Between Curves Using Double Integrals

The common application of the single variable integral is to compute the area under a curve f(x) over some interval [a,b] by integrating f(x) over that interval. That being said, you can sometimes also apply double integrals to compute areas between curves. However, the proposition is not the same. It's fairly simple to understand the tactic to achieve this once you can envision how to use a single integral to find the length of the interval.

Now you must be thinking as to What happens if you integrate the function f(x)=1 over the interval [a,b]? You can compute that

∫baf(x)dx=∫ba1dx=x∣∣ba=b−a.

The integral of the function f(x) =1 is merely the length of the interval [a,b]. Fact is that it also comes about as the area of the rectangle of height 1 and length (b−a), but we can explain it as the length of the interval [a,b].

You can apply the similar trick for finding areas with double integrals. The integral of a function f(x,y) over a region D can be simplified as the quantity beneath the surface z=f(x,y) over the region D. As executed above, we can attempt the tactic of integrating the function f(x,y)=1 over the region D. This would give the volume under the function f(x,y)=1 over D. But the integral of f(x,y)=1 is also the area of the region D. This can be a nifty way of calculating the area of the region D. Hence, if we If we entitle ‘A’ be the area of the region D, we can write it in the form of :-

A=∬DdA.

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Solved Example

As we said above, practice is the key to master over calculating area curves. So let’s begin with some fun exercises.

Problem

Find the area encircled by the following curves: 4,= 0, x = y - x =y

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Solution

Determining the boundaries: y = x² - 4, y=0 which implies x²- 4=0, therefore, (x-2) (x+2) = 0 x = - 2 or x = 2.

With the preview of the graph we can observe that 2=x is the boundary at ‘a’. The assessment of 2- =x is long away from encircling the area of the region. This is why the graph here plays a crucial part in helping identify the appropriate outcome to the problem. The value of the other boundary is provided by the equation of the vertical line 4,=x .

Boundaries are:  2,=a  and  4,=b  

Now, Establish the integral:

A= ò ò A f (x)dx (x 4)dx

Solving,

ò (x² - 4) dx= \[(\frac{1}{3} x^{3} - 4x)\]  ò = (\[\frac{1}{3}.(4)^{3}) - 4.4) - (\frac{1}{3}.{2}^{3} - 4.2)\]

= \[(\frac{64}{3} - 16) - (\frac{8}{3} - 8) = \frac{64}{3} - 16 - \frac{8}{3} + 8 - \frac{56}{3} - 8 - \frac{32}{3}\]

Thus, the area encircled by the curves y - x² -4, y=0, x-4 = 32ö ç3 square units.

Fun Facts

1. Multiple integrals are much easier to use than single integrals’ in finding area with integrals 

2. Drawing the sketch or graph beforehand makes it easy to find areas of the region that should be subtracted.

3.  It may be a requisite to find the areas of curves in several parts and add up the outcomes to achieve the final result.