Lesson Explainer: Area between Curves Mathematics • Higher Education

In this explainer, we will learn how to apply integration to find the area bounded by the curves of two or more functions.

Recall that the area enclosed by a curve 𝑦=𝑓(π‘₯), the π‘₯-axis, and the lines π‘₯=π‘Ž and π‘₯=𝑏 is given by the definite integral of 𝑓(π‘₯) with respect to π‘₯ evaluated between π‘Ž and 𝑏 as follows: 𝐴=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž),d where 𝐹 is the antiderivative of 𝑓 such that dd𝐹π‘₯=𝑓(π‘₯).

Space enclosed above the π‘₯-axis evaluates to a positive value, and space enclosed below the π‘₯-axis evaluates to a negative value. To find the area of a region, which is a strictly positive quantity, the absolute value is taken.

Recall also that the integral operator is a linear operator; hence, it is closed under both addition and scalar multiplication. The difference between the integrals of two functions, 𝑓(π‘₯) and 𝑔(π‘₯), is therefore given by the integral of the difference between the two functions as follows: 𝑓(π‘₯)π‘₯βˆ’ο„Έπ‘”(π‘₯)π‘₯=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.ddd

Recall that this property can be used to find the area enclosed by a curve, 𝑦=𝑓(π‘₯), and a horizontal line, where 𝑔(π‘₯) is replaced by the value 𝑐, so that the integrand is 𝑓(π‘₯)βˆ’π‘. Sometimes, however, we need to find the area of more complicated regions that are not necessarily enclosed by the π‘₯-axis or a horizontal straight line.

The above property holds for a general continuous function 𝑦=𝑔(π‘₯). Consider the region enclosed by two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏.

The area 𝐴 will be given by 𝐴=𝑓(π‘₯)βˆ’π‘”(π‘₯)=𝑓(π‘₯)π‘₯βˆ’ο„Έπ‘”(π‘₯)π‘₯.areaunderareaunderdd

Using the previous properties of the linearity of integration, we then have the following theorem.

Theorem: Area between Two Curves

For two functions, 𝑓(π‘₯) and 𝑔(π‘₯), where 𝑓(π‘₯)β‰₯𝑔(π‘₯) within the interval [π‘Ž,𝑏], the area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

Unlike finding the area between a single continuous function, 𝑓(π‘₯), and the π‘₯-axis, it does not matter whether either curve is above or below the π‘₯-axis. However, it does matter which curve is above the other. If instead 𝑔(π‘₯)β‰₯𝑓(π‘₯) within the interval [π‘Ž,𝑏], we can instead take the absolute value of the integral, or reverse the order of the curves and integrate 𝑔(π‘₯)βˆ’π‘“(π‘₯).

For intervals where the curves cross each other, we need to be more careful, as we will see in a later example.

Let’s look at an example of applying this formula to find the area enclosed by two curves and two vertical lines.

Example 1: Finding the Area of a Region Bounded by Trigonometric and Linear Functions

Find the area of the region enclosed by the curves 𝑦=π‘₯, 𝑦=π‘₯sin, π‘₯=πœ‹2, and π‘₯=πœ‹.

Answer

It can be helpful to draw a graph of the functions for the area we are evaluating, especially the points where the line and curve intersect. To find these, recall that the slope of sinπ‘₯ is equal to 1 at π‘₯=0 and then decreases and oscillates, never exceeding 1. The slope for the line 𝑦=π‘₯ is also 1, for all π‘₯. Therefore, the line, 𝑦=π‘₯, will stay strictly above the curve, 𝑦=π‘₯sin, and the two curves will intersect only once at π‘₯=0.

The other two lines are vertical lines at π‘₯=πœ‹2 and π‘₯=πœ‹.

Recall that the area enclosed by two curves, 𝑦=𝑓(π‘₯), and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

In this case, we have one line, 𝑓(π‘₯)=π‘₯, that is strictly above another curve, 𝑔(π‘₯)=π‘₯sin, and two vertical lines, π‘₯=πœ‹2 and π‘₯=πœ‹, enclosing a region whose area we will call 𝐴. The area 𝐴 is therefore given by 𝐴=ο„Έπ‘₯βˆ’π‘₯π‘₯.οŽ„ο‘½οŽ‘sind

Integrating with respect to π‘₯ gives us 𝐴=12π‘₯βˆ’(βˆ’π‘₯)=12π‘₯+π‘₯.οŠ¨οŽ„οŠ¨οŽ„coscosο‘½οŽ‘ο‘½οŽ‘

Evaluating the limits gives us 𝐴=ο€Ό12πœ‹+(πœ‹)οˆβˆ’ο€½12ο€»πœ‹2+ο€»πœ‹2=ο€Ό12πœ‹βˆ’1οˆβˆ’ο€Ύπœ‹8+0𝐴=3πœ‹8βˆ’1.coscossquareunits

Sometimes, the upper and lower limits for the integration will not be defined by the vertical lines π‘₯=π‘Ž and π‘₯=𝑏 but instead by intersection points between curves that need to be found.

Let’s look at an example of how to find the area enclosed by curves whose upper and lower limits are not given.

Example 2: Finding an Area Bounded by Cubic and Linear Functions

Find the area of the region bounded by 𝑦=π‘₯ and 𝑦=π‘₯.

Answer

It can be helpful to draw a graph of the functions to see which area we are evaluating. We will need to know the points where the curves intersect, which can be quickly determined by finding the solutions to the equation 𝑓(π‘₯)=𝑔(π‘₯).

In this case, we have π‘₯=π‘₯.

Rearranging and factorizing gives us π‘₯ο€Ήπ‘₯βˆ’1=0.

Solving for π‘₯, we find π‘₯=0π‘₯=Β±1.or

Hence, the curves intersect at π‘₯=βˆ’1, π‘₯=0, and π‘₯=1. The cubic function, 𝑦=π‘₯, will grow faster than the linear function, 𝑦=π‘₯, for large positive or negative values of π‘₯. Using this information, we can sketch the curves as follows.

Recall that the area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

In our case, for negative values of π‘₯, the cubic is above the linear function, and for positive values of π‘₯, the cubic is below the linear function. This means that if we evaluate each of the two regions separately, the difference between the two functions, and therefore the sign of the integral, will be opposite for the two regions.

The functions are both odd; that is, 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯) and 𝑔(βˆ’π‘₯)=βˆ’π‘”(π‘₯), so they are both antisymmetric about the 𝑦-axis. We therefore know that the two areas will be equal and opposite and that they will cancel each other out to evaluate to zero.

We could rectify this by evaluating the integrals for the two regions separately, taking the absolute values, then summing them together. A faster way is to use our knowledge that the functions are both odd and therefore the area in each of the two regions is the same. Hence, we can simply evaluate the area of one region, then double it to find the total area.

Substituting the given functions and the limits π‘₯=βˆ’1 and π‘₯=0 (the region below the negative π‘₯-axis) into the formula for the area, we have aread=ο„Έπ‘₯βˆ’π‘₯π‘₯.

Integrating with respect to π‘₯ gives us area1=14π‘₯βˆ’12π‘₯=0βˆ’ο€Ό14βˆ’12=14.οŠͺ

This is the area for the region to the left of the 𝑦-axis. Doubling this gives the total area as follows: areasquareunits=12.

In the previous example, we used a general result in integration that applies to all odd and even functions.

Theorem: Area between the Curve of an Odd or Even Function and the π‘₯-Axis

For an odd function, 𝑓(π‘₯), such that 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯), the area enclosed by the curve of 𝑓(π‘₯) and the π‘₯-axis within an interval [βˆ’π‘Ž,π‘Ž] is equal to twice the area enclosed by the curve of 𝑓(π‘₯) and the π‘₯-axis within an interval [0,π‘Ž]. That is, 𝑓(π‘₯)π‘₯=2𝑓(π‘₯)π‘₯.dd

The same applies for an even function, 𝑔(π‘₯), such that 𝑔(βˆ’π‘₯)=𝑔(π‘₯). That is, 𝑔(π‘₯)π‘₯=2𝑔(π‘₯)π‘₯.dd

As we saw in the previous example, just as we need to take care when evaluating the area of multiple enclosed regions above and below the π‘₯-axis, we also need to take care when evaluating areas between curves when the curves cross each other. Sometimes, the points of intersection might be harder to find, and the separate regions may not be symmetrical.

Let’s look at an example demonstrating how to find the area enclosed by two curves that cross, where the point of intersection is less obvious and the separate regions are asymmetric.

Example 3: Finding the Area Enclosed between Two Curves That Cross Each Other

The curves shown are 𝑦=1π‘₯ and 1π‘₯. What is the area of the shaded region? Give an exact answer.

Answer

Let’s first establish which curve is which function on the graph. For large π‘₯, π‘₯ will grow larger faster than π‘₯, and therefore 1π‘₯ will grow smaller faster than 1π‘₯. Therefore, the curve that is lower to the right-hand side of the graph is the curve of 𝑦=1π‘₯.

Recall that the area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

In this example, the two curves cross each other at a point partway through the region whose area we need to find. Recall that since we are taking a difference of two functions in the integral, 𝑓(π‘₯)βˆ’π‘”(π‘₯), the sign of the evaluated integral will depend on which curve is above the other. In the region where 𝑓(π‘₯) is above 𝑔(π‘₯), the integral will be positive, and vice versa. This means that if we evaluate the integral between the limits, the evaluation from one region will be canceled out with the evaluation of the other and will not give us the total area.

This means that we will need to evaluate the area of the two regions separately; that is, one to the left of the point where the curves intersect and one to the right. To do this, we need to find the point where the curves intersect, which we can find by solving the equation 𝑓(π‘₯)=𝑔(π‘₯).

In our case, we have 1π‘₯=1π‘₯.

Solving for π‘₯, we find π‘₯=1.

Hence, the curves intersect at π‘₯=1. This can be clearly seen on the graph, but it is always a good idea to find the exact value.

We now need to evaluate two separate definite integrals. From the graph, we see that the regions are also bounded by the vertical lines π‘₯=0.5 and π‘₯=2.

The lower and upper limits of integration for the region on the left are therefore π‘₯=0.5 and π‘₯=1, and likewise the lower and upper limits for the region on the left are π‘₯=1 and π‘₯=2.

If we were to proceed without splitting the regions, we would evaluate the area. Thus, areadlnlnlnln=ο„Έ1π‘₯βˆ’1π‘₯π‘₯=ο”βˆ’1π‘₯βˆ’|π‘₯|=ο€Όβˆ’12βˆ’|2|οˆβˆ’ο€Όβˆ’10.5βˆ’|0.5|=32βˆ’22β‰ˆ0.11.οŠ¨οŠ¦οŽ–οŠ«οŠ¨οŠ¨οŠ¦οŽ–οŠ«

Looking at the graph, we can see that this cannot be the correct value for the area of the shaded region, which looks much larger than this. To find the true area, we instead need to split the integration into the two separate regions on either side of the intersection point, taking the absolute value of each and then summing them together.

We have already found the indefinite integral above, which is given by ο„Έ1π‘₯βˆ’1π‘₯π‘₯=βˆ’1π‘₯βˆ’|π‘₯|+.dlnC

So, for the first region between π‘₯=0.5 and π‘₯=1, we have arealnlnlnln1=ο”βˆ’1π‘₯βˆ’|π‘₯|=ο€Όβˆ’11βˆ’|1|οˆβˆ’ο€Όβˆ’10.5βˆ’|0.5|=1βˆ’2β‰ˆ0.31.οŠ§οŠ¦οŽ–οŠ«

For the second region between π‘₯=1 and π‘₯=2, we have arealnlnlnln2=ο”βˆ’1π‘₯βˆ’|π‘₯|=ο€Όβˆ’12βˆ’|2|οˆβˆ’ο€Όβˆ’11βˆ’|1|=12βˆ’2β‰ˆβˆ’0.19.

The integral has evaluated to a negative value because the curve of 1π‘₯ is above the curve of 1π‘₯, and we have subtracted the former function from the latter function.

Since area is a strictly positive quantity, we take the absolute value. So, we have arealnln2=|||12βˆ’2|||=2βˆ’12β‰ˆ0.19.

Now, summing these two areas together, we find the exact value of the area of the shaded region as follows: areaareaarealnlnareasquareunits=1+2=1βˆ’2+2βˆ’12=12.

For some problems, we may need to find the area enclosed by more than two curves. Let’s look at an example of how to find the area enclosed by three curves.

Example 4: Finding the Area Bounded by Two Linear Functions and a Reciprocal Function Involving Dividing the Region of Integration

Consider the region in the first quadrant enclosed by the curves 𝑦=4π‘₯, 𝑦=π‘₯, and 𝑦=π‘₯4. Find the area of this region.

Answer

It can be helpful to draw a graph of the functions to see which area we are evaluating. We will need to know the points where the curves intersect. In this particular case, we need to be sure to draw the curves accurately. Determining the points where one curve is above another (and which one) and the points where the curves intersect is crucial to ensuring we calculate the area correctly.

To begin with, consider the lines of the linear functions. 𝑦=π‘₯ is a straight line that passes through the origin and has a slope of 1. 𝑦=π‘₯4 is a straight line that passes through the origin and has a slope of 14. Therefore, it stays strictly below 𝑦=π‘₯ in the first quadrant.

𝑦=4π‘₯ is a reciprocal function, so its curve is a hyperbola that never touches the π‘₯- or 𝑦-axes but has asymptotes at both. When π‘₯=1, 𝑦=4, and when π‘₯=4, 𝑦=1, so the curve passes through the points (1,4) and (4,1).

We now have enough information to sketch the curves in the first quadrant as follows.

Recall that the area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

For this to evaluate to the true area, we require 𝑓(π‘₯) to be above 𝑔(π‘₯) between the lines π‘₯=π‘Ž and π‘₯=𝑏. In our case, however, which of our functions is above which is dependent on the value of π‘₯. Looking at the sketch, we can see that we have two different regions within 𝐴 as follows.

In region 1, the area is enclosed by the lines 𝑦=π‘₯ and 𝑦=π‘₯4.

In region 2, the area is enclosed by the lines 𝑦=4π‘₯ and 𝑦=π‘₯4.

The lower limit for region 1 is the origin, π‘₯=0. The upper limit for region 1 is the π‘₯-value of the point of intersection between 𝑦=π‘₯ and 𝑦=4π‘₯.

Likewise, the lower limit for region 2 is this same π‘₯-value of the point of intersection between 𝑦=π‘₯ and 𝑦=4π‘₯, and the upper limit for region 2 is the π‘₯-value of the point of intersection between 𝑦=4π‘₯ and 𝑦=π‘₯4.

We therefore need to find the π‘₯-values of two points of intersection.

Point 1: Intersection between 𝑦=π‘₯ and 𝑦=4π‘₯

To find the π‘₯-value of the point of intersection between these two curves, we need to equate the two functions and solve the following equation for π‘₯: π‘₯=4π‘₯.

Solving for π‘₯ gives us π‘₯=Β±2.

Since we are considering the curves only in the first quadrant, the value of π‘₯ that is relevant to this problem is the positive value. Therefore, the two curves intersect at π‘₯=2. Note that we do not need to find the 𝑦-value, as it is not necessary for the integration.

Point 2: Intersection between 𝑦=4π‘₯ and 𝑦=π‘₯4

Likewise, to find this point of intersection, we need to solve the following equation for π‘₯: 4π‘₯=π‘₯4.

Solving for π‘₯ gives us π‘₯=Β±4.

Again, the value of π‘₯ that will be relevant is the positive value, π‘₯=4.

In region 1, since the line 𝑦=π‘₯ is above that of 𝑦=π‘₯4, we let 𝑓(π‘₯)=π‘₯ and 𝑔(π‘₯)=π‘₯4. Hence, the area of region 1 is given by areadd1=ο„Έπ‘₯βˆ’π‘₯4π‘₯=ο„Έ3π‘₯4π‘₯=3π‘₯8=ο€Ό128βˆ’0=32.

In region 2, since the curve 𝑦=4π‘₯ is above that of 𝑦=π‘₯4, we let 𝑓(π‘₯)=4π‘₯ and 𝑔(π‘₯)=π‘₯4. Hence, the area of region 2 is given by areadlnlnlnlnlnlnlnln2=ο„Έ4π‘₯βˆ’π‘₯4π‘₯=4|π‘₯|βˆ’18π‘₯=ο€Ό44βˆ’2βˆ’42βˆ’ο€Όβˆ’12=44βˆ’42βˆ’32=82βˆ’42βˆ’32=42βˆ’32.οŠͺοŠͺ

The total area of the region, 𝐴, is therefore given by areaareaarealnlnsquareunits=1+2=32+42βˆ’32=42.

Sometimes, we may need to find the area enclosed by implicit functions. Let’s look at an example of how to find the area enclosed by two implicit functions.

Example 5: Finding the Area of aRegion Bounded by Two Quadratic Functions Defined with Respect to 𝑦

Find the area of the region bounded by π‘₯=βˆ’5𝑦+1 and π‘₯=2π‘¦βˆ’5.

Answer

It can be helpful to draw the graph of the functions for the area we are evaluating, especially the points where the curves intersect.

In this case, both functions are implicit in 𝑦; that is, we have π‘₯ in terms of 𝑦. Of course, there is nothing intrinsic about the convention of π‘₯ being the independent variable and 𝑦 being the dependent variable. We can therefore simply reverse the typical π‘₯- and 𝑦-axes and sketch the curves of the two functions of 𝑦.

π‘₯=2π‘¦βˆ’5 is a U-shaped parabola that is symmetric about the π‘₯-axis and that intersects the π‘₯-axis at π‘₯=βˆ’5.

π‘₯=βˆ’5𝑦+1 is an n-shaped parabola that is symmetric about the π‘₯-axis and that intersects the π‘₯-axis at π‘₯=1.

Orienting the axes in the conventional way, the graph looks as follows.

Recall that the area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by aread=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.

In our case, however, we have two curves defined such that π‘₯ is a function of 𝑦. We therefore need to modify the integrand by switching these variables. Also, note that the limits of the integral will now be 𝑦-values rather than π‘₯-values.

Hence, the area enclosed by the two curves, π‘₯=𝑓(𝑦) and π‘₯=𝑔(𝑦), and the two horizontal lines, 𝑦=π‘Ž and 𝑦=𝑏, is given by aread=𝑓(𝑦)βˆ’π‘”(𝑦)𝑦.

We are now integrating with respect to 𝑦.

Looking at the first sketch with π‘₯ on the vertical axis, we can see that the curve of π‘₯=βˆ’5𝑦+1 is always above the curve of π‘₯=2π‘¦βˆ’5 for the enclosed area, so we will let 𝑓(𝑦)=βˆ’5𝑦+1 and 𝑔(𝑦)=2π‘¦βˆ’5. The upper and lower limits of integration will therefore also be values of 𝑦 at the points of intersection between the two curves.

To find the intersection points, we need to solve the equation 𝑓(𝑦)=𝑔(𝑦) for 𝑦; that is, 𝑓(𝑦)=𝑔(𝑦)2π‘¦βˆ’5=βˆ’5𝑦+1.

Now, solving for 𝑦, we have 7𝑦=6𝑦=Β±ο„ž67.

The 𝑦-values of the intersection points, and therefore the lower and upper limits of integration, are 𝑦=βˆ’ο„ž67 and 𝑦=ο„ž67.

Now we have everything we need to find the area of the enclosed region as follows: areadd=ο„Έβˆ’5𝑦+1βˆ’ο€Ή2π‘¦βˆ’5𝑦=ο„Έβˆ’7𝑦+6𝑦.ο„žοŠ±ο„žοŠ¨οŠ¨ο„žοŠ±ο„žοŠ¨οŽ₯οŽ₯οŽ₯οŽ₯

Integrating with respect to 𝑦 gives us areaareasquareunits=ο”βˆ’73𝑦+6𝑦=οβˆ’73ο€Ώο„ž67+6ο€Ώο„ž67ο‹οβˆ’οβˆ’73ο€Ώβˆ’ο„ž67+6ο€Ώβˆ’ο„ž67=ο€Ώβˆ’2ο„ž67+6ο„ž67ο‹βˆ’ο€Ώ+2ο„ž67βˆ’6ο„ž67=8ο„ž67=8√427.οŠ©ο„žοŠ±ο„žοŠ©οŠ©οŽ₯οŽ₯

Let’s finish by recapping some of the key points from this explainer.

Key Points

  • The area enclosed between two curves, 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯), with 𝑓(π‘₯) strictly above 𝑔(π‘₯) within the interval [π‘Ž,𝑏], and two vertical lines, π‘₯=π‘Ž and π‘₯=𝑏, is given by 𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯d.
  • Regions where the curve of 𝑓(π‘₯) is above the curve of 𝑔(π‘₯) will evaluate to a positive area, and regions where the curve of 𝑓(π‘₯) is below the curve of 𝑔(π‘₯) will evaluate to a negative area. To find the true area, the absolute value is taken.
  • Where the curves of 𝑓(π‘₯) and 𝑔(π‘₯) cross each other within the interval, the integration must be split for each separate region, with the absolute value of each region taken, then summed together to give the true area.

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