Linear Approximation and the Fundamental Theorem of Calculus Recall the Following Deﬁnitions from Lecture

Math 31A Discussion Notes Week 8 November 17 and 19, 2015

Linear Approximation and the Fundamental Theorem of Calculus Recall the following deﬁnitions from lecture:

Deﬁnition. Let f :[a, b] → R be continuous. Then the deﬁnite integral of f over [a, b] is given by n−1 Z b X f(x)dx = lim f (a + i∆x) ∆x, n→∞ a i=0 where ∆x = (b − a)/n.

Deﬁnition. If f and F are continuous functions on [a, b], then we say that F is an antiderivative of f if F 0 = f. Note that unlike derivatives, antiderivatives are not unique.

We want to explore the interaction between these two important ideas, and we begin with a motivating example. Recall the idea of local linear approximation: given a diﬀerentiable function f and a ﬁxed point a, we approximate the value of f(x) by

f(x) ≈ f(a) + f 0(a)(x − a) when x is suﬃciently close to a. For example, suppose you drive for 4 hours in one direction and I am hoping to estimate the distance you have traveled. Say that at time t = 0 you were traveling at a rate of 5 mph; then my estimate for the distance you’ve covered in 4 hours would be

location(4 hours) − location(0 hours) ≈ speed(0 hours)(4 − 0) = 20 miles.

Of course, this (hopefully) isn’t a great estimate. In all likelihood, the 5 mph measurement came while you were pulling out of the garage, and you didn’t travel at this speed for the full 4 hours. I can improve my estimate if I know your speed at some other point along the journey. Suppose I now know that after 2 hours, you were driving 70 mph. Then I can reﬁne my estimate by assuming that you drove 5 mph for the ﬁrst 2 hours, then drove 70 mph for the last 2 hours. For notational purposes, let’s write f(t) for your location at time t and f 0(t) for your speed at time t. Then

f(4) − f(0) = [f(4) − f(2)] + [f(2) − f(0)] ≈ f 0(2)(4 − 2) + f 0(0)(2 − 0) = 140 miles + 10 miles = 150 miles.

1 So my new estimate says that you’ve traveled 150 miles over 4 hours, which is hopefully more realistic. Let’s reﬁne the estimate once more, assuming I now know your speeds at hours 1 and 3 to have been 60 and 45 mph, respectively. Then f(4) − f(0) = [f(4) − f(3)] + [f(3) − f(2)] + [f(2) − f(1)] + [f(1) − f(0)] ≈ f 0(3)(4 − 3) + f 0(2)(3 − 2) + f 0(1)(2 − 1) + f 0(0)(1 − 0) = 45 miles + 70 miles + 60 miles + 5 miles = 180 miles. Notice that we can use our summation notation to write this estimate as n−1 n−1 X X f(4) − f(0) = f((i + 1)∆t) − f(i∆t) ≈ f 0(i∆t)∆t, i=0 i=0 where n = 4 and ∆t = (4 − 0)/4 = 1. It should be the case that as I learn your speed at more and more instances during your journey, I can get a more accurate estimate of the distance you’ve traveled. So we should have

n−1 X Z 4 f(4) − f(0) = lim f 0(i∆t)∆t = f 0(t)dt, n→∞ i=0 0 where ∆t = (4 − 0)/n. This example is actually quite typical, and motivates the following result.

Theorem 1 (The Fundamental Theorem of Calculus). If f :[a, b] → R is continuous and F is an antiderivative of F on [a, b], then Z b f(x)dx = F (b) − F (a). a Proof. As with our example, we begin with the usual local linear approximation F (b) ≈ F (a) + F 0(a)(b − a) and refine it. First, suppose we slice [a, b] into n intervals, each of width ∆x = (b−a)/n. Then the difference between F (a) and F (b) is the sum of the differences between the endpoints of these intervals: F (b) − F (a) = [F (b) − F (a + (n − 1)∆x)] + [F (a + (n − 1)∆x) − F (a + (n − 2)∆x)] + ··· + F (a + ∆x) − F (a) n−1 X = F (a + (i + 1)∆x) − F (a + i∆x), i=0 where ∆x = (b − a)/n. Also, on each of our intervals we can make the linear approximation F (a + (i + 1)∆x) ≈ F (a + i∆x) + F 0(a + i∆x)[(a + (i + 1)∆x) − (a + i∆x)] = F (a + i∆x) + F 0(a + i∆x)∆x,

2 so we have F (a + (i + 1)∆x) − F (a + i∆x) ≈ F 0(a + i∆x)∆x. Substituting this into our sum, we obtain the following estimate for the diﬀerence between F (a) and F (b): n−1 X F (b) − F (a) ≈ F 0(a + i∆x)∆x, i=0 where ∆x = (b − a)/n. As usual, increasing n decreases ∆x. This means we’re making our linear approximations on smaller and smaller intervals, so these approximations are becoming more accurate. So n−1 X F (b) − F (a) = lim F 0(a + i∆x)∆x. n→∞ i=0 Since F 0 = f, we can rewrite this as

n−1 X Z b F (b) − F (a) = lim f(a + i∆x)∆x = f(x)dx, n→∞ i=0 a which is our desired result.

Note. Notice that it doesn’t matter which antiderivative of f we have, it just matters that F is some antiderivative of F . This means that if F and G are both antiderivatives of f on [a, b], then G(b) − G(a) = F (b) − F (a). In fact, it’s not difficult to show that F and G must differ by a constant. Also notice that this theorem justifies our use of the notation R f(x)dx for an antiderivative of F . Of course, the real usefulness of this result is its computational power. The quantities F (b) and F (a) are R b usually quite easy to determine, but a f(x)dx, the signed area between the curve y = f(x) and the x-axis, can be quite difficult to compute. This theorem gives us an easy formula, assuming we can find an antiderivative of f.

Example. Suppose a ball is thrown straight up with initial velocity v0 > 0. Then its velocity at time t is given by v(t) = v0 + gt, where g ≈ −9.8 m/s is the acceleration due to gravity. Assuming that the ball reaches its maximum height at time t = −v0/g, use a Riemann sum to compute the diﬀerence between the initial height of the ball and the maximum height of the ball.

(Solution) From our discussion, it should be the case that

n−1 X height(−v0/g) − height(0) = lim v(i∆t)∆t, n→∞ i=0

3 where ∆t = (−v0/g − 0)/n = −v0/(ng). Then the difference in heights is given by n−1 n−1 X X i −v0 −v0 lim v(i∆t)∆t = lim v n→∞ n→∞ n g ng i=0 i=0 n−1 2 n−1 ! X i −v0 −v0 X i 1 = lim v0 − v0 = lim 1 − n→∞ n ng n→∞ g n n i=0 i=0 n−1 n−1 −v2 1 X n − i −v2 1 X = 0 lim = 0 lim n − i g n→∞ n n g n→∞ n2 i=0 i=0 −v2 1 n(n + 1) −v2 n(n + 1) = 0 lim = 0 lim g n→∞ n2 2 g n→∞ 2n2 −v2 = 0 . 2g Example. Confirm the above result by finding a trajectory whose velocity function is v(t) = v0 + gt (that is, find an antiderivative of v(t)) and evaluating this trajectory at t = 0 and t = −v0/g. (Solution) Consider the trajectory 1 r(t) = x + v t + gt2, 0 0 2

where x0 is a constant. Then 0 r (t) = v0 + gt = v(t),

so r is an antiderivative of v. We have r(0) = x0 and 1 1 v2 r(−v /g) = x + v (−v /g) + g(−v /g)2 = x − 0 , 0 0 0 0 2 0 0 2 g so −v2 r(−v /g) − r(0) = 0 , 0 2g

as desired. Notice that the constant x0 doesn’t matter — we can choose any antiderivative of v. This example serves to illustrate the power of the fundamental theorem; ﬁnding an antiderivative of v(t) and evaluating at the endpoints was much simpler than computing the Riemann sum. Finally, there’s an application of Theorem1 that is very important, and is often referred to as the second fundamental theorem of calculus. Theorem 2. For nice enough functions f :[a, b] → R, d Z x f(t)dt = f(x), dx a for every x ∈ (a, b).

4 Note. There are two typical objections to this theorem:

1. If you’re a little too familiar with the FTC, this might look like we’re saying “the derivative of the antiderivative of a function is the function you started with,” which seems trivial. This is, in eﬀect, how we justify the statement, but it’s a little deeper than that.

2. Even more likely, you might look at this equation and think it’s completely useless.

Both of these concerns can (hopefully) be cleared up by thinking of an “area function”

Z x A(x) = f(t)dt. a This function represents the (signed) area under f between a and x. The theorem says that the derivative of this function at a point x is precisely f(x). So if f(x) > 0, the area under f is increasing at x, if f(x) < 0, the area is decreasing at x, and if f(x) = 0, we have a critical point for the area function. Hopefully this helps us to see that the theorem is saying something, and that it has some use.

Proof. By a “nice enough” function, we mean that f has an antiderivative, say F . Then, using Theorem1,

d Z x d f(t)dt = (F (x) − F (a)) dx a dx d d = (F (x)) − (F (a)) dx dx = F 0(x) = f(x).

d 0 We have dx (F (a)) = 0 because F (a) is a constant. We have F (x) = f(x) because F is an antiderivative of f. We can actually prove the second FTC using the same local approximation ideas we used to prove the ﬁrst FTC, as the following, alternative proof demonstrates. So the second FTC isn’t just a corollary to the ﬁrst FTC; it comes out of the same ideas. R x Proof. Let A(x) = a f(t)dt. We want to compute A(x + h) − A(x) A0(x) = lim . h→0 h Notice that Z x+h Z x A(x + h) = f(t)dt and A(x) = f(t)dt, a a

5 so, using properties of the integral,

Z x+h Z x Z x+h A(x + h) − A(x) = f(t)dt − f(t)dt = f(t)dt. a a x For small h we can approximate the value of this last integral by h · f(x). This is because the integral is being taken over an interval of width h, and the height of the function will stay close to f(x) when h is small. In particular, we should have

Z x+h 1 Z x+h f(t)dt ≈ hf(x), so lim f(t)dt = f(x). x h→0 h x But this just means

A(x + h) − A(x) 1 Z x+h lim = lim f(t)dt = f(x), h→0 h h→0 h x and this is our desired result.

Example. Use the second fundamental theorem of calculus and integration by substitution to ﬁnd " 2 # d Z x f(t)dt . dx x

(Solution) First, let’s split our integral into two integrals, each of which has a constant as a limit of integration:

" 2 # " 2 # " 2 # d Z x d Z 0 Z x d Z x Z x f(t)dt = f(t)dt + f(t)dt = f(t)dt − f(t)dt . dx x dx x 0 dx 0 0

Now the second FTC expects a linear term as a limit of integration, not a quadratic term. To ﬁx this, we introduce a variable u for which u2 = t. Then dt = 2udu and our limits of integration become 0 and x, because u = 0 when t = 0 and u = x when t = x2. Then

" 2 # d Z x d Z x Z x f(t)dt = 2uf(u)du − f(t)dt dx x dx 0 0 = 2xf(x2) − f(x),

where the second step comes from the second FTC.