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WEEK #10: Integrals Integration by Parts, Definition and Application of Goals: • Integration by Parts • Riemann Sums • Definite Integrals • Fundamental Theorem of Calculus Textbook reading for Week #10: Read Sections 4.3, 4.4, 4.5 2 Integration by Parts From Section 4.3 So far in studying integrals we have used • direct anti-differentiation, for relatively simple functions, and • integration by substitution, for some more complex integrals. However, there are Z many integrals that can’t be handled with these techniques. Try to evaluate xe4x dx. Week 10 – Integration by Parts, Definition and Application of Integrals 3 This particular integral can be evaluated with a different integration technique, integration by parts. This rule is related to the product rule for derivatives. Expand d (uv) = dx Integrate both sides with respect to x and simplify. Express Z dv dx relative to the other terms. u dx 4 Integration by Parts For short, we can remember this formula as Z Z udv = uv − vdu Integration by parts: • Choose a part of the integral to be u, and the remaining part to be dv • Differentiate u to get du • Integrate dv to get v Z Z • Replace u dv with uv − vdu • Hope the new integral is easier to evaluate Week 10 – Integration by Parts, Definition and Application of Integrals Use integration by parts to evaluate Z 5 xe4x dx. 6 Here is a problem from the homework that will test both the techniques and understanding. (Variant of Section 4.3 #35) Example: Suppose the mass M of a toad grows according to the differential equation dM = t2e−3t with M (0) = 0 dt Question: When is this toad growing fastest? A. When M (t) is largest. dM B. When is largest. dt C. At t = 0. dM D. At some point where = 0. dt d 2 −3t te = 0. E. At some point where dt Week 10 – Integration by Parts, Definition and Application of Integrals dM = t2e−3t with M (0) = 0 dt Find the value of t when M is growing most quickly. 7 8 dM = t2e−3t with M (0) = 0 dt Find the mass of the toad at t = 1. Week 10 – Integration by Parts, Definition and Application of Integrals 9 Integrals as Sums From Section 4.4 So far we’ve needed integration to undo the effect of a derivative. For example, dy = t2 + 4e2t dt means Z y = t2 + 4e2t dt = However, there is another interpretation of the integral that can be equally important and useful: the integral as a sum or area. 10 Consider an object whose velocity is given by v = 1 + t2 Use a ∆x and ∆t argument to estimate the distance traveled between t = 0 and t = 0.2 seconds. Week 10 – Integration by Parts, Definition and Application of Integrals Estimate the distance traveled between t = 0.2 and t = 0.4. 11 12 Here is a graph of v(t). How would the distances you just calculated appear on this graph? Velocity 2 1 0 0.2 0.4 0.6 Time 0.8 1 If you were to continue in the same manner, show on the graph how you would estimate the total distance covered between t = 0 and t = 1.0. Week 10 – Integration by Parts, Definition and Application of Integrals 13 Sketch on the following graphs two other ways we could use other rectangles to approximate the total distance traveled between t = 0 and t = 1.0.1 2 Velocity Velocity 2 1 0 0.2 0.4 0.6 Time 0.8 1 1 0 Give a name to each of these approximations. 1 More sophisticated approximations are introduced in the assignment. 0.2 0.4 0.6 Time 0.8 1 14 On the graph below, what area represents the exact distance traveled between t = 0 and t = 1.0? Velocity 2 1 0 0.2 0.4 0.6 Time 0.8 1 Week 10 – Integration by Parts, Definition and Application of Integrals 15 Riemann Sum and Summation Notation The original rectangles we used have the advantage that they are easy to use in area calculations. Using the function f (t) = 1 + t2 to simplify calculations, write out long-hand the estimated distance using the left-hand estimate. Go from t = 0 to t = 1, using ∆t = 0.2. 16 Construct the same for approximation using the right-hand estimate. Week 10 – Integration by Parts, Definition and Application of Integrals 17 Define ∆t = 0.2, and t0 = 0, t1 = 0.2, . . . , ti = 0 + i∆t, . . . , t5 = 1.0. Write out both distance estimates using summation notation. These sums, approximating an area using terms of the form (∆t)f (t), are called Riemann Sums. 18 3t between t = 0 Example: Express the area under the graph of a(t) = 3+t and t = 10 as a Riemann Sum, using 5 intervals. Use a left-hand estimate. Week 10 – Integration by Parts, Definition and Application of Integrals 3t a(t) = 3+t Estimate the total area. 19 20 Sketch the interpretation of this total area on the graph below. 3 a(t) 2 1 0 2 4 t 6 8 10 Week 10 – Integration by Parts, Definition and Application of Integrals 21 How could we make our rectangular approximation a better approximation of the exact area under the graph? Let n be the number of intervals used between t = 0 and t = 10. Express ∆t, ti, and the total sum using n. 22 How could we express the exact area under the graph using this sum? We define this exact area calculation as the definite integral: Week 10 – Integration by Parts, Definition and Application of Integrals 23 Write down the integral that represents the amount of oxygen absorbed by a 5t lung from t = 0 to t = 5 seconds, if it is being absorbed at a rate of r(t) = 2+t mol/second. 24 Write down the integral that represents the change in the amount of salt in a lake from t = 30 to t = 60, if • the rate of salt flowing in is rin = 4 kg/day • the rate of salt flowing out is rout = 3 − e−0.02t kg /day Week 10 – Integration by Parts, Definition and Application of Integrals 25 The Definite Integral and Anti-Derivatives We will now tie together our two recent ideas: anti-derivatives of functions, and the specific areas under curves. • The indefinite integral, a function relating to anti-derivatives: dy = f (t) dt is equivalent to Z y = f (t) dt • The definite integral, a number representing the change in a function y between two time values, given the rate, r(t): Z b r(t) dt Change in y between t = a and t = b = a where r(t) is the rate of change or accumulation 26 Since we’re using the same symbol for both, they must clearly be related. The relationship is defined by the Fundamental Theorem of Calculus Fundamental Theorem of Calculus Suppose f is continuous. If you can find an antiderivative F (x) of f (x); that is, a function F (x) such that F 0(x) = f (x), then Z b f (x)dx = F (b) − F (a) a In other words, if we can find an antiderivative F (x), then calculating the value of the integral requires a simple evaluation of the anti-derivative F (x) at two points. This is much easier than computing an area using finite Riemann sums, and also provides an exact value of the integral instead of an estimate. Week 10 – Integration by Parts, Definition and Application of Integrals 27 Consider an object which whose velocity is given by v = 1 + t2 m/s Set up the definite integral that defines the distance traveled over 1 second. Approximate the value of this integral using a Riemann sum with 5 intervals, using the left-hand approximation. 28 Compute the exact value of the integral using the Fundamental Theorem of Calculus. Week 10 – Integration by Parts, Definition and Application of Integrals 29 Use the Fundamental Theorem to evaluate the following three definite integrals: Z 3 e−0.1t dt • I1 = 0 • I2 = • I3 = Z Z 5 e−0.1t dt 3 5 e−0.1t dt 0 30 Using a sketch of the graph, explain why the sum of the first two integrals equals the third.