Integration Review. May 11, 2013

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1 Integration Review May 11, 2013 Goals: Review the funamental theorem of calculus. Review u-substitution. Review integration by parts. Do lots of integration eamples. 1 Funamental Theorem of Calculus In Calculus 1, much focus was place on the iea of the erivative, or instantaneous rate of change of a function. The instantaneous rate of change of a function f at point, enote f (), is the slope of the tangent line of f at point, as picture: At the en, a tangent (paron the pun) was taken, an another concept was iscusse. We aske the question: how o we calculate the area uner the curve of a function f from a to b? We enote this f(). Let s briefly eplore this problem. b a Observation 1: We only nee to be able to calculate the area from 0 to Why is this? Well, let g() be the function which calculates the area uner f from 0 to, as picture. 1

2 Then, notice that the area uner f from a to b is just g(b) g(a). a b There s really nothing special about 0 either. We coul have mae g calculate the area uner f from k to as well, where k is any fie number. Observation 2: g () = f() Let s preten we have the function g above, even though we have not yet foun it. We can ask: how oes g change as changes? We epect that a little change in the function shoul equal a little change in the area equal to how high the function is. This is han wavy, but of course we have not aresse what the area uner a curve is formally yet anyway. We, however, conclue that we epect g () = f(). Observation 3: If we fin a g such that g () = f() then we are one. The above equation is calle a ifferential equation. We are trying to fin a y such that y = f(). Such a solution y is calle an anti-erivative of f, for obvious reasons. We say an anti-erivative rather than the anti-erivative because it is easy to see there are many. For instance, if y satisfies the above an is an anti-erivative of f(), then y + 1 is as well. This is because ifferentiation is linear so we can take the erivative of each part of the sum separately. Also, constants are constant, an therefore o not change. Therefore, (y + 1) = (y) + (1) = y The beginning part of this course will be to investigate how to fin solutions of y = f(); that is, we will fin the anti-erivative of functions. The main application of this is obvious: anti-erivatives give us the tool to calculate areas by the above 3 observations. We call the funamental link between the fining the area uner the curve an the erivative the funamental theorem of calculus. States formally, it says k f(z)z = f() 2

3 Sai alou, it says the rate of change with respect to of fining the area uner f from some constant to is eactly the value that f has at. We call the area uner the curve functional escribe above the integral. We often say integral instea of anti-erivative. For the most general anti-erivative of f, we write f() This is a function of which, when you take it s erivative, you get f. Note, there is a uniqueness issue as there are many anti-erivatives. Theorem 1. If F an G are both anti-erivatives of f then F G = k where k is a constant. Proof. (F G) = f f = 0. Therefore, the erivative of the ifference is 0. It isn t har to see (or, believe) that constant functions are the only functions with erivatives 0. Therefore, their ifference must be constant. Therefore, if F is an anti-erivative of f then we write: f() = F () + C Clearly anything in the above form is an anti-erivative, an by the theorem any anti-erivative has the above form for some constant C. 2 Easy Integrals We begin with a few easy integrals which we solve essentially by inspection. Theorem 2 (The Power Rule for Integrals). If n 1 then n = 1 n + 1 n+1 + C Proof. By the efinition of anti-erivative, it suffices to show that the erivative of the right han sie is n. This is trivia by the power rule for erivatives. By a similar metho, we can etermine the following integrals: cf() = c f() n = 1 n + 1 n+1 + C [n 1] sin() = cos() + C e = e + C (f() + g()) = 1 = ln + C cos() = sin() + C f() + g() 3 u-substitution Recall that the chain rule for erivatives: This tells us that: f(g()) = f (g()) g () f (g())g () = f(g()) + C 3

4 This is the principle of u-substitution. If we can ientify a g() as above, an set u = g(), then we can o a substitution of u for g() an o a trae of u = g (). So, the above looks like f (g())g () = f (u)u = f(u) + C = f(g()) + C This trick, in these cases, is to ientify a g() that works. Given any integral, one shoul also first begin by trying to ientify easy substitutions. Eample 1. We calculate the following integral: ln() Set u = ln(). Then u = 1. Then ln() = = u ln() u u u 1 = 1 2 u2 + C = 1 2 (ln())2 + C Warning: Remember that you have to go back to worl after oing a u-substitution. If the problem was given to you in terms of, an it s a little rue to give the answer in terms of u, right? 4 Integration by Parts Recall the prouct rule for erivatives: This tells us: [f() g()] = f () g() + g () f() f() g() g () f() = f () g() Or, in the more relevant irection: f () g() = f() g() f() g () The is the principle of integration by parts. If we ientify two parts of an integral, we can essentially integrate one part an take the erivative of the other to calculate the integral. In practice, we usually write this in terms of the letters For eample, to calculate: f () g() We woul set: u = g() u = g () v = f () v = f() An get: f () g() = uv = uv vu = g() f() g () f() Parts is particularly useful when there is one part with a much simpler function as it s erivative (like ln() or the inverse trig functions) or when there is a polynomial time some other function which is not too ifficult to integrate (like e or 3 sin()) as polynomials vanish after successive erivatives. 4

5 Eample 2. We calculate the following integral: e Observe that becomes simpler when you take the erivative, an e oes not become too complicate when you integrate. Therefore, we o parts. u = u = v = e v = e Therefore, we have: u v e = e = e e + C v e u Hint: It is much easier to take erivatives than to take integrals. Therefore, it is in your interest to check the answer if it only takes a few secons to o so. You see that the above when you take the erivative becomes: (e ) (e ) C e + e e + 0 Which is, of course, e which is what we wante. 5 Eamples Now, let s o lots of eamples of fining integrals of functions. For all the following integrals, the only knowlege that you nee is how to calculate erivative of all the usual functions, the chain rule an prouct rule for erivatives, an u-substitution an integration by parts escribe above. 1. sin() 2. ( + 3) ln( 2 ) 4. e sin() e e tan() 10. sin 2 () cos 3 () sec 2 ()