Applications of Integrals

Applications of Integrals There are two methods for solving definite integrals using substitution. The first is to use substitution to solve the indefinite integral, and rewrite the solution in terms of the original variable before evaluating the limits of integration. The other method is to actually modify the limits of integration based on the substitution. We have already evaluated Z 5 6.48e−0.09t dt 2 using the first method. Now let us do so by changing the limits of integration. First we write u = −0.09t and 1 du = dt 0.09 When t = 2 then u = −0.09 · 2 = −0.18 and when t = 5 we have u = −0.09 · 5 = −0.45. Substituting in Z 5 Z 6.48 −0.45 u −0.09t 6.48e dt = − e du = −72eu |−0.45 −0.18 ≈ −45.9 + 60.1 = 14.2 0.09 2 −0.18 − This is exactly the result as calculated before. It is often simpler to evaluate definite integrals in this way, because it does not require the original variable be reintroduced after integration. Z 1 Example 1 Evaluate (1 + 2t)2 dt 0 Solution Using substitution we let u = 1 + 2t so 1 du = dt 2 When t = 0 we have u = 1 and when t = 1 we have u = 3. Thus Z 1 Z 3 u3 27 1 26 2 u2 du = |31 = − = (1 + 2t) dt = 3 3 3 3 0 1 We can also use integrals to find the average rate of change, given the instantaneous rate of change over a period of timme. In order to do so, we simply integrate the instantaneous rate of change over some interval, and divide by the length of that interval. If f (t) describes the instantaneous rate of change of F (t) (an antiderivative), then the averate rate of change from a to b is given by Z b F (b) − F (a) 1 ∆F = = f (t)dt b−a b−a a Example 2 Suppose that water is flowing into a vessel at a rate of (1 − e−t ) cm3 /s. What is the average flow from times t = 0 to t = 2? Solution We find the average rate of change (or flow) is Z 2 1 1 1 (1 − e−t )dt = (t − e−t )|20 = (2 − e−2 + e−0 ) ≈ 0.568 2−0 0 2 2 The area underneath the constant 0.568 from 0 to 2 is the same as the area underneath (1 − e−t ). Finally, we can use definite integrals in order to calculate mass in terms of density. Since right now we are only working in one-dimension, we will have to constrain ourselves to working with one-dimensional objects (but there is no reason that we cannot expand this method to consider 3-dimension objects). For a thin rod with density ρ(x) as a function of x, the total mass is given by Z b m= ρ(x)dx a where the endpoints of the bar are at a and b. In general, it is convenient to define our coordinate system such that one of the endpoints of the bar resides at 0. Example 2 Consider a 100-cm vertical bar composed of a substance that has settled over time, so it has higher density near the bottom of the bar. Let z denote height, and ρ(z) = e−0.01z in grams per centimeter. Find the mass of the bar. Solution The mass is given by Z 1 e−0.01z dz = e−0.01z |10 = −100(e−1 − 1) ≈ 63.21 −0.01 Given this mass we can calculate the average density, given by Z 1 e−0.01z dz ≈ 0.632 100 − 0 This average density lies between the extremes ρ(0) = 1 and ρ(100) ≈ 0.368. Thus, this seems to be a reasonable solution for the mass. We can apply the same idea with densities of another type. Example 3 Suppose the density of otters along the coast of California is given by f (x) = 3 · 10−4 x(1000 − x) in otters per mile, where x is measure in miles from the Mexican border, and can take on values between 0 and 1000. Find the average density of otters. Solution To solve this problem we can find the total number of otters using integration, and divide by the length of the region over which they are found, in order to find the average density. First note that there are no otters at x = 0 and x = 1000, so the maximum is somewhere in the middle (in this case exactly in the middle). The total number of otters T is the definite integral of the density, so Z 1000 x2 x3
1000 10002 10003
−4 T = 3·10 x(1000−x) = 3·10−4 1000 − |0 = 3·10−4 1000 − = 50, 000 2 3 2 3 0 Dividing by 1000, we find that the average density is 50 otters/mile, which is clearly not the average of the values at the endpoints (which is 0).