Let be a continuous function defined on . The definite integral is the “area under ” on . We can turn this concept into a function by letting the upper (or lower) bound vary.
Let . It computes the area under on as illustrated in Figure 5.4.2. We can study this function using our knowledge of the definite integral. For instance, since .
The and the axes are uncalibrated. There are three distinct positions on the axis, , and in the order from left to right. The function is a curve facing downward, the areas under the curve between and are shaded. lies to the right of outside of the shaded portion.
Consider pictured in Figure Figure 5.4.4 and its associated “area so far” function, . Using the graph of and geometry, find an explicit formula for .
The axis is drawn from to while the axis is drawn without calibration but it has two distinct points and . The graph is a straight line passing through the origin and has a positive slope. The shaded portion is drawn under the graph between and .
Figure5.4.4.The area of the shaded region is
Solution1.
We can see from Figure 5.4.5 that for , the area under the curve can be found by subtracting the area of two triangles. The larger triangle will have a base of and a height of , while the smaller triangle will have a base of and a height of . Therefore, the area under the curve for is given by .
Graph of function same as the one used for the previous example.
The axis is drawn from to while the axis is drawn without calibration but it has two distinct positions and . The graph is a straight line passing through the origin and has a positive slope. The shaded portion is drawn under the graph between and .
The distance from the origin to point on the axis is marked . The distance from origin to is labeled . The left boundary of the shaded region the value is labeled and on the right boundaries the value is labeled .
Figure5.4.5.The area of the shaded region is
Note that this same formula holds for . If , then . The areas to the left of will have opposite signs (since they areas are accumulated before). For example, when ,. This is the same value we get from evaluating for . Also notice that . This integral is clearly since the areas over and will sum to zero. Again, this is the same answer obtained by evaluating for .
Therefore, we can reasonably say that . A plot of both and are given in Figure Figure 5.4.6. You should notice a familiar relationship between these two functions. This relationship is formally stated in Theorem 5.4.7.
The axis is drawn from to and the axis is drawn from to . The graphs of two functions are shown. The first is a line through the origin with slope 2, and the second is a parabola that opens upward, with its vertex at . Both functions intersect at approximately and , it is also between these points that the line is above the curve beyond which the curve is above the line.
The straight line that passes through the origin and has a positive slope. It lies in the first and the third quadrant.
The curve appears to start from intercept . It has a intercept at and the second intercept at .
As Example 5.4.3 hinted, we can apply calculus ideas to ; in particular, we can compute its derivative. In Example 5.4.3, , so . While this may seem like an innocuous thing to do, it has far-reaching implications, as demonstrated by the fact that the result is given as an important theorem.
Using the Fundamental Theorem of Calculus, we have . That is, the derivative of the “area so far” function, is simply the integrand replacing with .
This simple example reveals something incredible: is an antiderivative of ! Therefore, for some value of . (We can find , but generally we do not care. We know that , which allows us to compute . In this case, .)
What we have done in Example 5.4.9 was more than finding a complicated way of computing an antiderivative. Consider a function defined on an open interval containing , and . Suppose we want to compute . First, let
We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives! In fact, this is exactly what we noticed in Example 5.4.3. The “area so far” function was indeed an anti-derivative of the integrand. This is the second part of the Fundamental Theorem of Calculus.
As its name suggests, the Fundamental Theorem of Calculus is an important result. In fact, it’s sufficiently important that it’s worth taking a moment to understand why it’s true. A proof is given in Figure 5.4.12.
Notation: A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writing , the notation is used. Thus the solution to Example 5.4.13 would be written as:
The Constant :Any antiderivative can be chosen when using the Fundamental Theorem of Calculus to evaluate a definite integral, meaning any value of can be picked. The constant always cancels out of the expression when evaluating , so it does not matter what value is picked. This being the case, we might as well let .
(This is interesting; it says that the area under one “hump” of a sine curve is 2.)
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This integral is interesting; the integrand is a constant function, hence we are finding the area of a rectangle with width and height 2. Notice how the evaluation of the definite integral led to . In general, if is a constant, then .
We established, starting with Key Idea 2.2.3, that the derivative of a position function is a velocity function, and the derivative of a velocity function is an acceleration function. Now consider definite integrals of velocity and acceleration functions. Specifically, if is a velocity function, what does mean?
A ball is thrown straight up with velocity given by ft/s, where is measured in seconds. Find, and interpret, and .
Solution.
Using the Fundamental Theorem of Calculus, we have
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Thus if a ball is thrown straight up into the air with velocity , the height of the ball, 1 second later, will be 4 feet above the initial height.
Note that the ball has traveled much farther. It has gone up to its peak and is falling down, but the difference between its height at and is 4ft.
If we wish to find the total distance traveled, we must evaluate (noting that negative velocities will reduce the diplacement, but we want distance, not displacement). In this case, we know that the velocity changes sign once when ,so seconds. The velocity is positive over and negative over . Therefore
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So the total distance traveled over is feet .
As we can see in Figure 5.4.16, the positive area between and the -axis, , while the negative area, . When we add these two areas, we get the displacement of ft. But when we add the absolute value of both of these areas (as in Figure 5.4.17), we get the total distance of ft.
The axis is drawn from to and the axis is drawn from to . The function is a straight line with a negative slope. The function has an intercept of and an intercept of .
The line forms two shaded regions with the axis. Between to on the axis the region under the graph is shaded and labeled . The second area is between and that lies in the fourth quadrant and the area is labeled .
Figure5.4.16.The area between and the -axis can be used to represent displacement
The axis is drawn from to and the axis is drawn from to . The function is a straight line with a negative slope. The function has an intercept of and an intercept of .
The line forms two shaded regions with the axis. Between to on the axis the region under the graph is shaded and labeled . Between and lies the second shaded region labeled .
Figure5.4.17.The area between and the -axis can be used to represent distance
Integrating a rate of change function gives total change. Velocity is the rate of position change; integrating velocity gives the total change of position, i.e., displacement.
Integrating a speed function gives a similar, though different, result. Speed is also the rate of position change, but does not account for direction. That is, the speed an object is the absolute value of its velocity. This is what we saw in Example 5.4.15 when we evaluated . So integrating a speed function gives total change of position, without the possibility of “negative position change.” Hence the integral of a speed function gives distance traveled.
As acceleration is the rate of velocity change, integrating an acceleration function gives total change in velocity. We do not have a simple term for this analogous to displacement. If miles/h and is measured in hours, then
While we have just practiced evaluating definite integrals, sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. Functions written as are useful in such situations.
Consider continuous functions and defined on , where for all in , as demonstrated in Figure 5.4.22. What is the area of the shaded region bounded by the two curves over ?
The and the axes are uncalibrated but there are two positions and marked on the axis. There are two functions and graphed. The graph shows the area that forms on between positions and . The two functions intersect at one point.
The function is above , it first curves up and reaches a maxima after which it dips down for a minima and then continues to increase. The function has a dip between and then it increases.
(a)
Graph same as previous with additional details. The and the axes are uncalibrated but there are two positions and marked on the axis. There are two functions and graphed. The graph shows the area that forms on between positions and . The two functions intersect at one point.
The function is above , it first curves up and reaches a maxima after which it dips down for a minima and then continues to increase.The function has a dip between and then it increases. The graph shows areas under both curves as different shaded regions.
It will help to sketch these two functions, as done in Figure 5.4.25.
The axis is drawn from to and the axis is drawn from to . The first function is a curve that has a intercept at and the intercept near . The second function is a straight line. The two lines intersect at points and . The area bounded by the two functions is shaded.
Figure5.4.25.Sketching the region enclosed by and in Example 5.4.24
The region whose area we seek is completely bounded by these two functions; they seem to intersect at and . To check, set and solve for :
One of the things we have to be careful about when finding the area between curves is that the curves might cross, so that the distinction between “upper curve” and “lower curve” can change. The video example in Figure 5.4.26 illustrates this phenomenon.
Consider the graph of a function in Figure 5.4.27 and the area defined by . Three rectangles are drawn in Figure 5.4.28; in Figure 5.4.28.(a), the height of the rectangle is greater than on , hence the area of this rectangle is is greater than .
Finally, in Figure 5.4.28.(c) the height of the rectangle is such that the area of the rectangle is exactly that of . Since rectangles that are “too big”, as in (a), and rectangles that are “too little,” as in (b), give areas greater/lesser than , it makes sense that there is a rectangle, whose top intersects somewhere on , whose area is exactly that of the definite integral.
The axis is uncalibrated and the axis is drawn from to . The curve first decreases from to then it increases until . The area under the curve between to is shaded. A rectangular boundary is drawn on the axis with height above the curve.
(a)
The axis is uncalibrated and the axis is drawn from to . The curve first decreases from to then it increases until . The area under the curve between to is shaded. A rectangular boundary is drawn on the axis with height a little below the curve.
(b)
The axis is uncalibrated and the axis is drawn from to . The curve first decreases from to then it increases until . The area under the curve between to is shaded. A rectangular boundary is drawn on the axis such that from to the curve is above the rectangle and from to the curve is below the rectangle.
This is an existential statement; exists, but we do not provide a method of finding it. Theorem 5.4.29 is directly connected to the Mean Value Theorem of Differentiation, given as Theorem 3.2.4; we leave it to the reader to see how.
Consider . Find a value guaranteed by the Mean Value Theorem.
Solution.
We first need to evaluate . (This was previously done in Example 5.4.14.)
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Thus we seek a value in such that .
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The axis has two positions marked and , the axis has ,, and marked in order. The function is an inverted parabola of height and has intercepts at and . The area under the parabola until the axis is shaded. A rectangular boundary is drawn with height and width
Figure5.4.32.A graph of on and the rectangle guaranteed by the Mean Value Theorem
In Figure 5.4.32 is sketched along with a rectangle with height . The area of the rectangle is the same as the area under on .
When is shifted by , the amount of area under above the -axis on is the same as the amount of area below the -axis above ; see Figure 5.4.33 for an illustration of this. In this sense, we can say that is the average value of on .
The and the axes are uncalibrated. The axis is labeled as . The axis has three distinct positions marked , and in the order.
The function represents a parabola and the graph shows the area under the parabola. and are the two ends for the area. is to the left of the vertex of the parabola almost at half the distance from .
The rectangle guaranteed by the Mean Value Theorem lies above the vertex of the parabola.
The and the axes are uncalibrated. The axis is labeled as . The axis has three distinct positions marked , and in the order.
The function represents a parabola and the graph shows the area under the parabola. and are the two ends for the area. is to the left of the vertex of the parabola almost at half the distance from .
Sine the function is shifted by , there are three parts to the area, two are positive and are above the axis, from to and from a little before to . From a little before to the area is drawn in the fourth quadrant from above the parabola on the axis.
Figure5.4.33.On the left, a graph of and the rectangle guaranteed by the Mean Value Theorem. On the right, is shifted down by ; the resulting “area under the curve” is 0
Examining this last line closely, the expression represents adding up sample values of and then dividing by . This is exactly what we do when we calculate the average of a set of numbers. Now when we consider taking the limit as goes to ,, we are adding up all of the function’s output values over and dividing by the “number of numbers”. In a sense, we are adding up an infinite number of output values and then dividing by the number of terms we summed (which is again infinite).
We can understand the above example through a simpler situation. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/time = 100 miles/2 hours = 50 mph.
What was the displacement of the object in Example 5.4.36? We calculate this by integrating its velocity function: ft. Its final position was 3 feet from its initial position after 3 seconds: its average velocity was 1 ft/s.
This section has laid the groundwork for a lot of great mathematics to follow. The most important lesson is this: definite integrals can be evaluated using antiderivatives. Since Section 5.3 established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than “area under the curve,” convert the sums to definite integrals, then evaluate these using the Theorem 5.4.10. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more.
The downside is this: generally speaking, computing antiderivatives is much more difficult than computing derivatives. Chapter 6 is devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated. Before that, Section 5.5 explores techniques of approximating the value of definite integrals beyond using the Left Hand, Right Hand and Midpoint Rules. These techniques are invaluable when antiderivatives cannot be computed, or when the actual function is unknown and all we know is the value of at certain -values.