Math 231E. Fall 2013. Lecture 6C.Indefinite Integrals, SubstitutionWe first lay out the example of the indefinite integral.Let us consider the integralZ40exdx.From the Fundamental Theorem of Calculus, we know that the antiderivative of exis itself,soZ40exdx = exx=4x=0= e4− e0= e4− 1.But we also haveZ32exdx = exx=3x=2= e3− e2,andZ7−2exdx = exx=7x=−2= e7− e−2.And we see that in all these cases, the formula into which we are plugging in is the same; wejust use different values to plug in. Moreover, as we will see below, it is typical that the vastmajority of the work that we need to do will involve finding this function, and the pluggingin of the boundary values is the easy part.As such, we would like to just refer as this antiderivative function, and when we need to doso we will talk about the indefinite integral, which is just another word for antiderivative.The one thing to point out here is that an antiderivative is not unique. For example, we knowthatddx(x3) =ddx(x3+ 2) =ddx(x3− 12) =ddx(x3+ π2e17cos sin(4)) = 3x2.In particular, this means that we can always add any constant we like to an antiderivative,since all of the functionsx2, x3+ 2, x3− 12, x3+ π2e17cos sin(4)are all antiderivatives of 3x2. In general, we will write this asx3+ C,where C is an arbitrary constant.And, again, we stress that the choie of constant C will not matter when computing a definiteintegral, sinceZ7−2exdx = ex+ Cx=7x=−2= (e7+ C) − (e−2+ C) = e7− e−2+ (C − C) = e7− e−2.So, the hard part is finding antiderivatives. Basically, all of the rules we come up with arejust rules for differentiation run backwards! Examples:1.Zxn=xn+1n + 1+ C, n 6= −12.Zex= ex+ C3.Z1x= ln |x| + C4.Zcf(x) dx = cZf(x),5.Z(f(x) + g(x)) dx =Zf(x), +Zg(x),Unfortunately, there isn’t a rule likeZf(x)g(x) dx 6=Zf(x) dxZg(x) dx.There is a way to use the product rule in reverse, which we will see below — it is called“integration by parts”.However, we can use the Chain Rule is reverse pretty easily. Say that F0(x) = f(x), and thenddxF (g(x)) = F0(g(x)) · g0(x) = f(g(x)) · g0(x).This means thatZf(g(x)) · g0(x) dx = F (g(x)) + C.This gives rise to the technique of u-substitution.Example 27. We want to computeZxp3 + x2dx.Let us try u = 3 + x2, which gives du = 2x dx, so that x dx = du/2. We then haveZxp3 + x2dx =Z12√u du = u3/2+ C = (3 + x2)3/2+ C.A crucial feature that made this work is that there was an extra power of x in the integrand.For example, if we were givenZp3 + x2dx,and tried this trick, we would getZp3 + x2dx =Z√udu2x,and now what do we do?Example 28. Let’s try the integralZex+exdx.Now, we might first think that our sub should be u = x + ex, which gives du = (1 + ex) dx,so we haveZex+exdx =Zeudx1 + ex,which doesn’t seem that helpful. But, let us go back to the original integral and multiplyout, recalling that ea+b= eaeb, so we haveZex+exdx =Zex· eexdx,and now we might see that it makes sense ot use u = ex, du = exdx, and then we haveZex+exdx =Zeudu = eu+ C = eex+ C.Example 29. Sometimes we might have to do a lot more work than we originally thought!Let’s tryZx1 +√1 − x2− x2dx.Let us try u = 1 − x2, so du = −2x dx, and we haveZx1 +√1 − x2− x2dx = −12Zduu +√u.Since we have a√u term, let us choose v =√u, and thus dv = 1/(2√u) du, or du = 2v dv,and we haveZx1 +√1 − x2− x2dx = −Zv dvv2+ v= −Zdvv + 1.Finally, choose w = v + 1 and dw = dv, and we haveZx1 +√1 − x2− x2dx = −Zdww= −ln |w| + C.Unpacking all of the definitions, we haveZx1 +√1 − x2− x2dx = −ln |w| + C= −ln |v + 1| + C= −ln |√u + 1| + C= −ln |p1 − x2+ 1| + C.Also, if we are doing definite integrals, then we can also use integration by parts, but weshould change the limits of integration as we go.Example 30. Let us consider the definite integralZe1ln xx.It seems clear that we should make the substitution u = ln x, which gives du = dx/x. But alsonotice that we should change the limits on the integral: when x = 1, we have u = ln 1 = 0,and when x = e, we have u = ln e = 1. So we haveZe1ln xx=Z10u du =u22u=1u=0=12.This, for example, will allow us to derive the even–odd relations. For example, we can showthat if f(x) is odd, thenZa−af(x) dx = 0for any a. This is becauseZa−af(x) dx =Z0−af(x) dx +Za0f(x) dx,and if we make the substitution u = −x in the first integral, this becomesZ0−af(x) dx =Z0af(−u)(−du) = −Za0(−f(u))(−du) = −Za0f(u) du.Similarly, if f (x) is even, thenZ0−af(x) dx =Z0af(−u)(−du) = −Za0(f(u))(−du) =Za0f(u) du.and soZa−af(x) dx = 2Za0f(x)
View Full Document
Unlocking...