Lesson Explainer: Area between a Curve and a Line Mathematics • Higher Education

In this explainer, we will learn how to apply integration to find the area between the curve of a function and a horizontal or vertical straight line.

Finding the area underneath the curve of a function is a very useful technique that has many applications. For example, finding the area underneath a velocity–time graph of an object between two points in time gives the total distance traveled over that time.

The area underneath a straight-line graph is fairly straightforward to calculate. For example, consider the area enclosed by the curve of the function, 𝑓(π‘₯)=4π‘₯+7, and the π‘₯-axis between two points, π‘₯=π‘Ž and π‘₯=𝑏.

We can find the enclosed area using the standard formula for the area of a trapezoid: areaoftrapezoid=𝑝+π‘ž2β‹…β„Ž, where 𝑝 and π‘ž are the lengths of the two parallel sides and β„Ž is the distance between them. Applying this formula to the enclosed region on the graph, our two parallel sides have lengths 𝑓(π‘Ž) and 𝑓(𝑏), and the height β„Ž=π‘βˆ’π‘Ž. Therefore, we have areaoftrapezoid=𝑓(π‘Ž)+𝑓(𝑏)2β‹…(π‘βˆ’π‘Ž).

With 𝑓(π‘₯)=4π‘₯+7, we then have areaoftrapezoid=4π‘Ž+7+4𝑏+72β‹…(π‘βˆ’π‘Ž)=(2π‘Ž+2𝑏+7)(π‘βˆ’π‘Ž)=2π‘Žπ‘βˆ’2π‘Ž+2π‘βˆ’2π‘Žπ‘+7π‘βˆ’7π‘Ž=2𝑏+7π‘βˆ’ο€Ή2π‘Ž+7π‘Žο….

This approach works fine when we have a linear function, 𝑓(π‘₯), and we can use simple geometry, but it will not work for more complicated functions. For instance, if the velocity of an object is not changing at a constant rate (i.e., its acceleration is not constant), then the velocity–time graph will not be a straight line, and we cannot easily find the area this way.

For a general curve of the function 𝑦=𝑓(π‘₯), the area between the curve and the π‘₯-axis, and between the vertical lines π‘₯=π‘Ž and π‘₯=𝑏, is given by the fundamental theorem of calculus.

We define a new function 𝐴(π‘₯) to be the area between the curve, 𝑓(π‘₯), the π‘₯-axis, and the vertical lines π‘₯=0 and π‘₯=π‘₯, a general value of π‘₯.

Increasing the upper bound of the area by a small amount, dπ‘₯, the increment of the area function d𝐴 is given by dd𝐴=𝑓(π‘₯)π‘₯.

Rearranging gives the change in 𝐴 divided by the change in π‘₯: dd𝐴π‘₯=𝑓(π‘₯).

This is the derivative of 𝐴 with respect to π‘₯. The original area function 𝐴(π‘₯), the area under the curve between 0 and π‘₯, is given by inverting the differentiation operation on both sides of the equation. On the right-hand side, this gives the antiderivative, 𝐹, of 𝑓: 𝐴(π‘₯)=𝐹(π‘₯), where dd𝐹π‘₯=𝑓(π‘₯).

Now, consider the exact same scenario, but evaluating the area under the curve between two specific values of π‘₯, π‘₯=π‘Ž and π‘₯=𝑏.

The area 𝛽 under the curve and between π‘₯=π‘Ž and π‘₯=𝑏 is given by 𝛽=𝐹(𝑏)βˆ’πΉ(π‘Ž).

Evaluating the antiderivative 𝐹 at two points π‘₯=π‘Ž and π‘₯=𝑏 and taking the difference is called the definite integral between π‘Ž and 𝑏 and is denoted 𝐹(𝑏)βˆ’πΉ(π‘Ž)=𝑓(π‘₯)π‘₯.d

Therefore, the area 𝛽=𝐹(𝑏)βˆ’πΉ(π‘Ž) under the curve and between π‘₯=π‘Ž and π‘₯=𝑏 is given by 𝛽=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

The right-hand side is often denoted 𝐹(𝑏)βˆ’πΉ(π‘Ž)=[𝐹(π‘₯)], where the square brackets with π‘Ž on the bottom right and 𝑏 on the top right indicate that 𝐹(π‘₯) is being evaluated at π‘₯=π‘Ž and π‘₯=𝑏, and then the difference, 𝐹(𝑏)βˆ’πΉ(π‘Ž), is taken. The arbitrary constant of integration, +𝐢, is omitted for a definite integral, since it is the same for both 𝐹(π‘Ž) and 𝐹(𝑏) and is therefore cancelled out.

Theorem: The Fundamental Theorem of Calculus

The area enclosed between a curve, 𝑓(π‘₯), the π‘₯-axis, and the two vertical lines π‘₯=π‘Ž and π‘₯=𝑏 is given by aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓. The right-hand side is often denoted 𝐹(𝑏)βˆ’πΉ(π‘Ž)=[𝐹(π‘₯)], where the square brackets with π‘Ž on the bottom right and 𝑏 on the top right indicate that 𝐹(π‘₯) is being evaluated at π‘₯=π‘Ž and π‘₯=𝑏, then the difference, 𝐹(𝑏)βˆ’πΉ(π‘Ž), is taken.

Let’s return to the previous example function, 𝑓(π‘₯)=4π‘₯+7, and this time we will use integration to find the area underneath the curve between two points π‘₯=π‘Ž and π‘₯=𝑏, as shown in the diagram below.

Using the formula for the area under a curve, 𝑓(π‘₯), between π‘₯=π‘Ž and π‘₯=𝑏, aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž)=[𝐹(π‘₯)].

In this case, we have 𝑓(π‘₯)=4π‘₯+7, which gives aread=ο„Έ4π‘₯+7π‘₯.

Integrating with respect to π‘₯ gives area=2π‘₯+7π‘₯.

Evaluating at π‘₯=𝑏 and π‘₯=π‘Ž and taking the difference, area=2𝑏+7π‘βˆ’ο€Ή2π‘Ž+7π‘Žο….

This result agrees precisely with the result found previously from the area of the trapezoid. However, the huge advantage of using integration to find the area under a curve is that it extends to far more complicated functions for which it would otherwise be impossible to calculate the area.

Let’s look at an example of using integration to find the area under a more complicated curve.

Example 1: Finding the Area under the Curve of a Quadratic Function

Let 𝑓(π‘₯)=2π‘₯+3. Determine the area bounded by the curve 𝑦=𝑓(π‘₯), the π‘₯-axis, and the two lines π‘₯=βˆ’1 and π‘₯=5.

Answer

It can be helpful to sketch the area we are asked to calculate. In this case, this is the area bounded by the lines π‘₯=βˆ’1 and π‘₯=5 and the curve 𝑦=2π‘₯+3.

Recall that the area between a curve 𝑓(π‘₯), the π‘₯-axis, and two lines π‘₯=π‘Ž and π‘₯=𝑏 is given by the definite integral aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓 such that dd𝐹π‘₯=𝑓(π‘₯). Substituting in the given function, 𝑓(π‘₯)=2π‘₯+3, and the limits, π‘₯=βˆ’1 and π‘₯=5: aread=ο„Έ2π‘₯+3π‘₯.

Integrating with respect to π‘₯ gives area=23π‘₯+3π‘₯, and evaluating this between the limits π‘₯=βˆ’1 and π‘₯=5 gives areasquareunits=ο€Ό23β‹…5+3β‹…5οˆβˆ’ο€Ό23β‹…(βˆ’1)+3(βˆ’1)=ο€Ό2503+15οˆβˆ’ο€Όβˆ’23βˆ’3=102.

For some functions, there are regions for which the curve lies below the π‘₯-axis. By the nature of how we evaluate the area between a curve and the π‘₯-axis, since the increment dπ‘₯ is always positive, the area element dd𝐴=𝑓(π‘₯)π‘₯ will of course be negative if the function, 𝑓(π‘₯), is negative at this point.

Therefore, any area enclosed below the π‘₯-axis by the curve will evaluate as negative. Since area is a strictly positive quantity, this is often dealt with by taking the absolute value of the definite integral for each region.

Let’s look at an example of how we can use integration to find the area enclosed by a curve below the π‘₯-axis.

Example 2: Using Definite Integration to Find the Area between Two Lines and below the Horizontal Axis

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

Answer

Recall that the area between a curve 𝑓(π‘₯), the π‘₯-axis, and two lines π‘₯=π‘Ž and π‘₯=𝑏 is given by the definite integral aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓 such that dd𝐹π‘₯=𝑓(π‘₯). Substituting in the given function, 𝑓(π‘₯)=1π‘₯, and the limits, π‘₯=βˆ’1 and π‘₯=βˆ’13: aread=ο„Έ1π‘₯π‘₯.οŽͺ

Integrating with respect to π‘₯ gives arealn=[|π‘₯|];οŽͺ we recall that the natural logarithm ln(π‘₯) is undefined for negative values of π‘₯, so we take the absolute value.

Evaluating this between the limits π‘₯=βˆ’1 and π‘₯=βˆ’13: =|||βˆ’13|||βˆ’|βˆ’1|=ο€Ή3ο…βˆ’0=βˆ’(3)β‰ˆβˆ’1.099().lnlnlnlnto3d.p.

Notice that the result is negative. This is because the area between the curve and the π‘₯-axis is beneath the π‘₯-axis. Taking the absolute value, we find that the true value for the area of the shaded region is ln(3)β‰ˆ1.099 (to 3 d.p.) square units.

Sometimes, deducing the true value of the area enclosed by a curve is not as straightforward as taking the absolute value of the definite integral. Let’s look at an example where we need to be more careful with this method.

Example 3: Using Definite Integration to Find the Area of Regions above and below the Horizontal Axis

The figure shows the graph of the function 𝑓(π‘₯)=2π‘₯βˆ’8π‘₯. Evaluate the area of the shaded region.

Answer

Recall that the area between a curve 𝑓(π‘₯), the π‘₯-axis, and two lines π‘₯=π‘Ž and π‘₯=𝑏 is given by the definite integral aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓 such that dd𝐹π‘₯=𝑓(π‘₯).

In this example, the function, 𝑓(π‘₯)=2π‘₯βˆ’8π‘₯, is an odd function. That is, it is perfectly antisymmetric about the 𝑦-axis, so 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯). This in turn means that the area between π‘₯=0 and any value π‘₯=π‘Ž will be exactly the same as the area between π‘₯=βˆ’π‘Ž and π‘₯=0. When using definite integration to find the area, however, the integral on one side of the 𝑦-axis will have the opposite sign to the integral on the other.

This means that when integrating between two equal and opposite values as we are here, between βˆ’2 and +2, we will find that the integral on one side of the 𝑦-axis will cancel out the integral on the other side, giving a total of zero: aread=ο„Έ2π‘₯βˆ’8π‘₯π‘₯.

Integrating with respect to π‘₯ gives area=12π‘₯βˆ’4π‘₯.οŠͺ

We are now evaluating an even function (i.e., 𝑓(βˆ’π‘₯)=𝑓(π‘₯)) at two equal and opposite values before taking the difference, so the result will be zero: area=12β‹…2βˆ’4β‹…2βˆ’ο€Ό12β‹…(βˆ’2)βˆ’4β‹…(βˆ’2)=8βˆ’16βˆ’(8βˆ’16)=0.οŠͺοŠͺ

Clearly, however, the area of the complete shaded region is not zero. To correct this problem, we need to evaluate the area of each part of the shaded region separately.

In this example, we have two separate regions enclosed by the curve, 𝑓(π‘₯), and the π‘₯- and 𝑦-axes.

So, we need to evaluate the integral between π‘₯=βˆ’2 and π‘₯=0, and the integral between π‘₯=0 and π‘₯=2 separately, take the absolute value of each integral to give the area, and then add them together to give the area of the complete shaded region.

For the first region, we have area1=12π‘₯βˆ’4π‘₯=0βˆ’ο€Ό12β‹…(βˆ’2)βˆ’4β‹…(βˆ’2)=0βˆ’(8βˆ’16)=8.οŠͺοŠͺ

And for the second region: area2=12π‘₯βˆ’4π‘₯=ο€Ό12β‹…2βˆ’4β‹…2οˆβˆ’0=8βˆ’16βˆ’0=βˆ’8.οŠͺοŠͺ

The negative value is expected, since the region between π‘₯=0 and π‘₯=2 is below the π‘₯-axis. Taking the absolute value to give the area: area2=|βˆ’8|=8.

Therefore, the total area of the complete shaded region is the sum of these two areas: areaareaareasquareunits=1+2=8+8=16.

The fundamental theorem of calculus is of course not restricted to functions of the independent variable (usually π‘₯). It can also be used to find the area enclosed by a function of 𝑦.

Theorem: Area Enclosed by a Curve of a Function of 𝑦

The area enclosed between a curve 𝑓(𝑦), the 𝑦-axis, and the two horizontal lines 𝑦=π‘Ž and 𝑦=𝑏 is given by aread=𝑓(𝑦)𝑦=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓. The right-hand side is often denoted 𝐹(𝑏)βˆ’πΉ(π‘Ž)=[𝐹(𝑦)], where the square brackets with π‘Ž on the bottom right and 𝑏 on the top right indicate that 𝐹(𝑦) is being evaluated at 𝑦=π‘Ž and 𝑦=𝑏, then the difference, 𝐹(𝑏)βˆ’πΉ(π‘Ž), is taken.

In our next example the area we seek is bounded by a function of 𝑦 and two horizontal lines.

Example 4: Using Definite Integration to Find the Area between an Implicit Function and Two Horizontal Lines

Find the area enclosed by the graph of π‘₯=9βˆ’π‘¦οŠ¨, the 𝑦-axis, and the lines 𝑦=βˆ’3 and 𝑦=3.

Answer

It can be helpful to sketch the area we are asked to calculate. In this case, this is the area bounded by the lines 𝑦=βˆ’3 and 𝑦=3 and the curve π‘₯=9βˆ’π‘¦οŠ¨.

π‘₯ is a negative quadratic function of 𝑦, so the curve is an n-shaped parabola around the π‘₯-axis. To find the π‘₯-intercept, we substitute 𝑦=0 into the equation of the function and solve it for π‘₯, which gives π‘₯=9.

So the curve intercepts the π‘₯-axis at π‘₯=9. Next, to find the 𝑦-intercepts of the curve, we substitute π‘₯=0 into the equation of the function and solve it for 𝑦, which gives 0=9βˆ’π‘¦π‘¦=Β±3.

Hence, the curve intercepts the 𝑦-axis at 𝑦=3 and 𝑦=βˆ’3. We now have enough information to sketch the curve.

Recall that the area between a curve 𝑓(π‘₯), the π‘₯-axis, and two lines π‘₯=π‘Ž and π‘₯=𝑏 is given by the definite integral aread=𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž), where 𝐹 is the antiderivative of 𝑓 such that dd𝐹π‘₯=𝑓(π‘₯).

In this case, however, we cannot simply rearrange the function of the curve for 𝑦 and integrate with respect to π‘₯, since the area above the π‘₯-axis is equal to the area below it, and the result from the integration will be zero.

If we try to proceed this way, we start with 𝑦=√9βˆ’π‘₯,⟹=ο„Έβˆš9βˆ’π‘₯π‘₯.aread

The problem in this specific case is that 𝑦=√9βˆ’π‘₯ is not a one-to-one function, since the square root can be either positive or negative. If a function is not a one-to-one function, then it does not map one unique value to another single value; therefore, it is impossible to know which area we are evaluating.

In this case, 𝑦=√9βˆ’π‘₯ maps a unique value of π‘₯ to both the positive square root +√9βˆ’π‘₯and the negative square root βˆ’βˆš9βˆ’π‘₯.

Taking the positive square root gives the area enclosed by the curve above the π‘₯-axis.

Since we have changed the function to a strictly one-to-one function, we can now evaluate this area in the usual way with a definite integral: aread=ο„Έ+√9βˆ’π‘₯π‘₯=βˆ’ο•23(9βˆ’π‘₯)=18.

Likewise, taking the negative square root gives the area enclosed by the curve below the π‘₯-axis.

Then, we can once again evaluate this area with a definite integral: aread=ο„Έβˆ’βˆš9βˆ’π‘₯π‘₯=23(9βˆ’π‘₯)=βˆ’18.

Adding these areas together gives a result of 0, but we can use our knowledge that the true area is given by the sum of these absolute values: ⟹=|18|+|βˆ’18|=36.areasquareunits

A simpler alternative is to integrate with respect to 𝑦 instead. Essentially, this switches the variables around.

We can visualize this by taking the original graph of the function, but reflected along the line 𝑦=π‘₯.

This can be integrated with respect to 𝑦, since there is nothing special about 𝑦 compared with π‘₯, and it is merely convention that π‘₯ is the independent variable. Furthermore, this is now a one-to-one function, so it can be integrated without needing to split the integral into separate regions. Thus, we have π‘₯=𝑓(𝑦)=9βˆ’π‘¦.

So, the area may be given by instead integrating with respect to 𝑦: aread=ο„Έ9βˆ’π‘¦π‘¦.

Integrating with respect to 𝑦 gives areasquareunits=9π‘¦βˆ’13𝑦=ο€Ό9β‹…3βˆ’13β‹…3οˆβˆ’ο€Ό9β‹…(βˆ’3)βˆ’13β‹…(βˆ’3)=27βˆ’9βˆ’(βˆ’27+9)=36.

Definite integrals can also be used to find the area enclosed between the curve of a function 𝑦=𝑓(π‘₯) and the curve of another function 𝑦=𝑔(π‘₯), by using the fact that the integral operator, like the differential operator, is a linear operator, meaning that it is closed under addition: 𝑓(π‘₯)π‘₯+𝑔(π‘₯)π‘₯=𝑓(π‘₯)+𝑔(π‘₯)π‘₯.ddd

This extends to subtraction too, since integration is also closed under scalar multiplication: 𝛼𝑓(π‘₯)π‘₯=𝛼𝑓(π‘₯)π‘₯,dd where 𝛼 is a scalar. Taking the specific case of 𝛼=βˆ’1, ο„Έβˆ’π‘“(π‘₯)π‘₯=βˆ’ο„Έπ‘“(π‘₯)π‘₯,dd and combining this with the first property, we can express the difference between two integrals as a single integral: 𝑓(π‘₯)π‘₯βˆ’ο„Έπ‘”(π‘₯)π‘₯=𝑓(π‘₯)π‘₯+ο„Έβˆ’π‘”(π‘₯)π‘₯=𝑓(π‘₯)βˆ’π‘”(π‘₯)π‘₯.ddddd

For the simpler case of the second function being a constant, horizontal line 𝑦=𝑔(π‘₯)=𝑐, we then have 𝑓(π‘₯)π‘₯βˆ’ο„Έπ‘π‘₯=𝑓(π‘₯)βˆ’π‘π‘₯.ddd

Therefore, we can find the area enclosed by a curve 𝑦=𝑓(π‘₯) and a horizontal line 𝑦=𝑐 between two points π‘₯=π‘Ž and π‘₯=𝑏 using the formula aread=𝑓(π‘₯)βˆ’π‘π‘₯.

Let’s look at an example of how we use this to find the area enclosed between a curve and a straight line.

Example 5: Using Definite Integration to Find the Area Enclosed by a Curve and a Horizontal Line

Find the area of the shaded region.

Answer

The area of this shaded region could be found by evaluating the definite integral of the curve 𝑦=3π‘₯+4π‘₯βˆ’2 between the limits π‘₯=1 and π‘₯=2, then subtracting the area of the rectangle below it, which we can find easily with the standard formula for the area of a rectangle.

However, recall that we can also find the area enclosed by a curve 𝑦=𝑓(π‘₯) and a horizontal line 𝑦=𝑐 between two points π‘₯=π‘Ž and π‘₯=𝑏 using the formula aread=𝑓(π‘₯)βˆ’π‘π‘₯.

In this case, we have the curve 𝑓(π‘₯)=3π‘₯+4π‘₯βˆ’2 and the horizontal line 𝑦=5. Substituting these into the formula above gives areadd=ο„Έ3π‘₯+4π‘₯βˆ’2βˆ’5π‘₯=ο„Έ3π‘₯+4π‘₯βˆ’7π‘₯.

Integrating with respect to π‘₯, areasquareunits=π‘₯+2π‘₯βˆ’7π‘₯=ο€Ή2+2β‹…2βˆ’7β‹…2ο…βˆ’ο€Ή1+2β‹…1βˆ’7β‹…1=8+8βˆ’14βˆ’(1+2βˆ’7)=6.

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

Key Points

  • The area enclosed by a curve 𝑦=𝑓(π‘₯), the π‘₯-axis, and two horizontal lines π‘₯=π‘Ž and π‘₯=𝑏 is given by 𝐴=𝑓(π‘₯)π‘₯.d
  • The area enclosed by a curve π‘₯=𝑓(𝑦), the 𝑦-axis, and two vertical lines 𝑦=π‘Ž and 𝑦=𝑏 is given by 𝐴=𝑓(𝑦)𝑦.d
  • A region enclosed above the π‘₯-axis/𝑦-axis gives a positive integral, and a region enclosed below the π‘₯-axis/𝑦-axis gives a negative integral. Since area is a strictly positive quantity, to find the area enclosed we take the absolute value of the integral for each individual enclosed region.
  • The area enclosed by a curve 𝑦=𝑓(π‘₯), another horizontal line 𝑦=𝑐, and two lines π‘₯=π‘Ž and π‘₯=𝑏 is given by 𝐴=𝑓(π‘₯)βˆ’π‘π‘₯.d

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