Ganesh A Hegde

An Introduction To Integral Calculus Ganesh A Hegde 13-May-2020 Abstract This paper introduces the concepts underlying Integral Calculus for high school students by the way of developing an intuitive understanding of the concepts involved. The reader is assumed to be familiar with the concept of Differential Calculus, but isn’t expected to be having a strong grasp of the same. In this text, we will be using the relation between the velocity of an object and the distance traveled by it to illustrate the concepts of Integral Calculus. Velocity is defined as the distance traveled by an object divided by the time taken for it to cover that distance. Thus, v≡ x t (1) where x is the distance, v is the velocity and t is the time duration. If we rewrite formula 1, we can express distance using velocity and time: x=v×t (2) However, there’s one caveat. Calculating distance in this way is appropriate only if the velocity v is constant during the time duration t. Let’s take the case where the velocity is having a constant value v1 during a duration of time t1 . If we draw the graph of velocity with respect to time, it will look like in figure 1. The area of the hatched region of the graph is the product of velocity v1 , and time t1 . area ≡ v1 × t1 If we plugin the values for velocity and time namely v1 and t1 into equation 2, we can see that the distance traveled by the object during the time interval [0, t1 ] is the area of the hatched region of the graph. This area is also called the definite integral of velocity with respect to time in the interval [0, t1 ]. In this case, since the velocity is constant with respect to time, the value of the definite integral, in other words, the distance traveled is v1 × t1 . When the velocity of the object is not constant, the calculation of the area, the definite integral, is more involved. For example, the graph in figure 2 shows the velocity of an object varying with time. Again, the distance traveled by the 1 Figure 1: Constant Velocity Figure 2: Varying Velocity 2 Figure 3: Instantaneous Velocity Rectangles object in time interval [0, t1 ] is the area of the hatched region of the graph. A simple way to calculate this area is to divide the hatched region into rectangles having minute widths and adding together the areas of these rectangles. Let’s assume that all the rectangles have the same width namely, ∆t. Also, assume that ∆t divides the interval [0, t1 ] evenly. Three such rectangles are shown in figure 3. The rectangles are of varying heights namely, v2 , v3 and v4 . These heights represent the velocities of the object at times t2 , t3 and t4 respectively. The velocity of the object at any point of time, such as t2 in the figure, is called the instantaneous velocity of the object at that point of time. The area of the rectangles are v2 × ∆t, v3 × ∆t and v4 × ∆t respectively1 . Thus, if we add the area of all such rectangles that make up the hatched region, we get2 : area ≡ t1X −4t v∆t (3) t=0 where v is the instantaneous velocity of the object at different points of time namely, 0, 4t, 24t, 34t and so on till t1 −4t. Observe that the instants of time start at 0 and change by an amount of 4t as shown in figure 4 . But, the area that we calculate in formula 3 is only a close approximation to the actual area of the hatched region of the graph shown in figure 2. The reason why it’s only an approximation and not an exact value is that the rectangles don’t exactly fit the hatched region of the graph as shown in figure 5 . The smaller the widths of these rectangles, the more closely will they fit that region and the more closely will the sum of their areas match the actual area, in other words, the definite integral of the velocity function. In other words, the smaller the value of ∆t, 1 The velocity v4 is negative and hence the area is also considered negative the notation Σ means summation of the products of velocity and time duration at different instants of time. 2 Here, 3 Figure 4: Tiling Of Rectangles Figure 5: Blow-up of Graph 4 the better will be the estimate of the definite integral. In math, an infinitesimal value is a very small value that approaches zero. Thus, in formula 3 the area will be equal to the definite integral of the velocity function when 4t is an infinitesimal value. This is expressed as: area ≡ lim t1X −4t 4t→0 v∆t (4) t=0 where v is the instantaneous velocity of the object at different points of time in the interval [0, t1 ]. In other words, the distance covered by the object in time interval [0, t1 ] is t1X −4t x = lim v∆t (5) 4t→0 t=0 The notation of limit of a function allows us to express the fact that the summation function is being carried out under a specific constraint namely, ∆t approaching3 zero. There’s an intuitive way to understand equation 5: the distance covered by an object is the sum of distances covered by it over minute intervals of time. Recall that equation 2 holds only when the velocity of the object is constant during time interval t. Hence, equation 5 can be thought of as an improvisation of equation 2; the former holds for even those cases where the velocity is not constant over a specific duration of time. This improvisation was achieved by choosing infinitesimal intervals of time, ∆t, over which the velocity of the object would essentially be constant. Equation 4 accurately expresses the definite integral of the velocity function with respect to time in the interval [0, t1 ]. The verbosity of the right hand side of equation 4 can be reduced by using the mathematical notation for the definite integral: ˆ t1 x= vdt (6) 0 The values t1 and 0 are called the upper and lower limits of the definite integral respectively. Up until now we have not yet answered the central question: If the velocity v can be expressed as a function of time t, how can we find its definite integral? For answering this question, we must understand the concept of the indefinite integral 4 of a function. If we look back at formula 1 and equation 2, we see that the distance function is the inverse function of velocity. Similar to the caveat accompanying equation 2, the velocity formula also has a caveat. Formula 1 gives the average velocity of the object. The actual velocity of the object is the same as the average velocity only if the former is constant during the interval of time t. If the velocity is not constant, we can still find the actual velocity of the 3 Another mathematical term for this is tending to. concept of definite integral is derived from the indefinite integral. In this text, however, the definite integral is introduced before the indefinite integral so that the reader can appreciate the motivation behind integral calculus in terms of its practical application. 4 The 5 object at any instant by choosing an interval of time that is so minute that the velocity of the object is essentially constant during that interval. Thus, when the interval of time becomes infinitesimal, formula 1 gives the actual velocity at different instants of time. This is expressed as: v ≡ lim 4t→0 4x 4t (7) where 4x is the distance traveled by the object during the time interval 4t. Formula 7 accurately expresses the actual velocity of the object at any instant of time. As we have already seen, this velocity is also known as the instantenous velocity of the object. We can reduce the notational clutter in formula 7 and succintly express it as: dx (8) v≡ dt This notation introduces the concept of the derivative of a function or differentiation. The reader is already assumed to be familiar with this notation. The effort made here is to explain the intuitive meaning of the derivative of a function and its relation to the indefinite integral. We have seen earlier that the distance function is the inverse of the velocity function. This fact along with Equations 6 and 8 suggests that the definite integral is the inverse of the derivative5 . The only problem in conclusively stating this is that the definite integral specifies an upper and lower limit for the values of its variable t, whereas the derivative doesn’t specify any such limits for its variable. The definite integral is the area of the hatched region of the graph bounded by the axes t = 0 and t = t1 . The specification of the limits for the values of its variable is why the definite integral is characterized as definite. In contrast, the indefinite integral does not specify any such limits. The whole point of introducing the indefinite integral was to be able to express the inverse operation of differentiation. Since there’s no range of values for its variable to specify, the indefinite integral is expressed as: ˆ x = vdt (9) Now that we have defined the indefinite integral as the inverse of the derivative of a function, let’s take an example to see it in action. If the velocity function v is given by: v ≡ 2t (10) then the indefinite integral of this function with respect to time is given by: x ≡ t2 + C1 (11) where C1 is some constant that we are free to choose6 . To arrive at the distance function, formula 11, all we did was to find a function whose derivative with 5 This is formally part of the Fundamental Theorem of Calculus. The other part being the relationship between the indefinite and the definite integral. 6 Such constants are commonly called arbitrary constants 6 respect to time was the velocity function in formula 10. We can verify that the derivative of the distance function is indeed the velocity function. Note especially that the derivative of the constant C1 with respect to time is zero. Hence, it doesn’t really matter which value we choose for C1 . This means, there are many distance functions whose derivatives are the velocity function in formula 10. By choosing a different constant, we still end up having the same derivative: x ≡ t2 + C 2 dx ≡ 2t dt But, this is apparently problematic. Without having a concrete value for the constant in formula 11, it’s not possible to calculate the distance traveled by the object at any specific instant of time. But, wait! The actual problem that we had set out to solve was: If the velocity v can be expressed as a function of time t, how can we find its definite integral? The definite integral, as you may recall, is the same as the indefinite integral, but additionally specifies the upper and lower limits for the time variable t. In other words, the definite integral is the distance traveled by the object during a specific duration of time and not at a specific instant of time. In our example distance function (formula 11), let’s designate the distance traveled by the object at time t = 0 as x0 and at time t = t1 as x1 . Hence, the distance traveled in the duration [0, t1 ] is given by: 4x = x1 − x0 Substituting the values for the variable t in the formula 11, we get: 4x = (t21 + C1 ) − (0 + C1 ) 4x ≡ t21 What we see here is that the constant C1 is of no consequence in calculating the definite integral. In general, if the indefinite integral of a function f (t) is F (t), then the definite integral is given by the formula7 : ˆ b f (t) ≡ F (b) − F (a) a Thus, given the velocity as a function of time, we can always find the definite integral, the distance covered by the object in a specifc interval of time. We have also seen the role that the indefinite integral of a function plays in the calculation of the definite integral and how the indefinite integral is the inverse of the derivative of a function. This introduction has also illustrated the practical application of the concepts of Integral Calculus in solving a problem in physics. 7 This is the second part of the Fundamental Theorem Of Calculus that establishes the relation between the definite and indefinite integrals. 7