Math 231E. Fall 2013. Lecture 6B.Fundamental Theorem of CalculusArea is positive by definition, but how should we interpret the following area?[picture of curve both above and below axis?]Definition 6. The signed area “below” a curve is defined to be +1 times the areas that occurabove the curve and −1 times the areas that occur below the curve.In general, we are interested in computing the signed area (but only because it is more mathe-matically natural!)Example 26. There are some integrals that we know how to do, since we already know the area.For example,Z10p1 − x2dx =π4,and we can see this by drawing a circle.[picture of quarter circle]But how to we computeZ1/2−1/3p1 − x2dx?We can also computeZ3−3sin(x)e−x17542dx = 0,because it is an odd function, and using symmetry. But how do we computeZ2−3sin(x)e−x17542dx?A nice fact about definite integrals is that they are linear. For example, if we have a functionf(x) and we want to computeRbaf(x) dx, then we choose some large n, define ∆x = (b −a)/n, andx∗kin the kth subdomain, and writeZbaf(x) dx = limn→∞nXk=1f(x∗k) ∆x.But what if h(x) = f(x) + g(x)? Then we haveZbah(x) dx = limn→∞nXk=1h(x∗k) ∆x= limn→∞nXk=1(f(x∗k) + g(x∗k)) ∆x= limn→∞lef t[nXk=1f(x∗k) ∆x +nXk=1g(x∗k) ∆x= limn→∞nXk=1f(x∗k) ∆x + limn→∞nXk=1g(x∗k) ∆x=Zbaf(x) dx +Zbag(x) dx.We can do the same argument (Work it out yourself for practice!) to show thatZbaαf(x) dx = αZbaf(x) dx.Unfortunately, there is no nice rule to break upZbaf(x)g(x) dx.Application. Consider that we know the velocity of a particle for all t ∈ [a, b], and let usdenote this velocity by v(t). Denote the position of the particle at time t by s(t). We want tocompute the total displacement s(b) − s(a).If we consider a small time ∆t, then the displacement in that small time interval is roughlyDk:= v(t∗k) ∆t. The total displacement is then approximatelynXk=1Dk=nXk=1v(t∗k) ∆t.Thus we should haves(b) − s(a) = limn→∞nXk=1v(t∗k) ∆t =Zbav(t) dt.But notice that s0(t) = v(t)! So we haveZbas0(t) dt = s(b) − s(a).Similarly, let us write x for b in the previous equation:s(x) − s(a) =Zxav(t) dt,and differentiate with respect to x, to obtains0(x) =ddxZxav(t) dt,orv(x) =ddxZxav(t) dt.These two boxed equations are known as2The Fundamental Theorem of Calculus:Theorem 12 (FTC, Part I). If f(x) is continuous on [a, b], then the function g(x) defined byg(x) =Zxaf(t) dtis continuous on [a, b] and g0(x) = f(x).Theorem 13 (FTC, Part II). If f(x) is continuous on [a, b], thenZbaf(x) dx = F (b) − F (a)where F is any antiderivative of f, i.e. F0= f.2If there was a way for me to typeset and echo, this is where I would do
View Full Document
Unlocking...