The Fundamental Theorem of Calculus

Transcript

0:00Introduction

Hello everyone! One small puzzle first. One over x squared is never negative. Yet a student integrates it from minus one to one, and gets minus two. Negative area, from a positive function? Something is wrong, right? Today: the fundamental theorem of calculus, the main tool for definite integrals in JEE, okay? We will catch exactly where that student slipped. So, let's begin.

0:29What a definite integral means

So, what does a definite integral mean? Take f, continuous on the closed interval from a to b; the square brackets mean both ends are included. Now cut this interval into n equal strips. Each strip has width b minus a, over n. Strip number r has height f of x sub r; the small r written below the x is a subscript. And x sub r is a plus r times b minus a, over n. Each strip is a thin rectangle, but rectangles never fit a curve exactly. How do we remove that error? Yes, exactly! Make the strips thinner and thinner. The long S is the integral sign, with the lower limit a at the bottom, the upper limit b at the top, and d x for the variable. The Greek capital sigma means add up, from r equals one to n, and lim, with the arrow, means the limit as n tends to infinity. So: the limit of a sum of rectangles.

1:28Signed area

Got it? Now, the picture. Here is sine x, from zero to two pi. At the bottom left, the first line says: the integral from zero to two pi of sine x, d x, is zero. The second line we will read in a moment. But the curve clearly encloses area. So how can the integral be zero? Above the x axis, f of x is positive, so the rectangles add: the upper hump gives plus two. Below the axis, f of x is negative, so they subtract: the lower hump gives minus two. So the integral is a signed area. With me? The integral is two, plus minus two: zero. The humps cancel. So a definite integral is not always the area you see. Okay? For the total area, integrate the modulus, mod sine x, with the two vertical bars. It flips the lower hump up: two plus two, four. Area is never negative.

2:23The fundamental theorem

Clear? Now the shortcut: the fundamental theorem of calculus. Suppose capital F prime, with the little dash, equals f, on the closed interval a to b; so capital F is an antiderivative of f. Then the integral from a to b of f of x, d x, is just capital F of b, minus capital F of a. With me? But wait. Indefinite integrals always get plus capital C, the constant of integration. So where did it go? Add a constant to capital F, and it appears in both F of b and F of a, so the subtraction cancels it. Any antiderivative works. Right? Now the condition many students forget. The antiderivative must be valid, that is, continuous, on the whole interval, not just at the ends. Our puzzle hides right here.

3:15Back to the puzzle: check continuity

So, back to our puzzle, question mark and all. The antiderivative of one over x squared is minus one over x, in square brackets from minus one to one: top value minus bottom value. Put in one: minus one. Put in minus one: plus one. Minus one, take away one: minus two. Every step looks fine. So where is the mistake? First, a sanity check. One over x squared is greater than zero wherever it is defined, so a negative answer is impossible. Okay? Now the real reason. The point x equals zero sits inside the interval, and there one over x squared is not continuous, not even bounded. So minus one over x is not valid on the whole interval. Clear? So the theorem does not apply. In fact, this integral diverges; it has no finite value. The lesson: check continuity on the whole interval before you put in the limits. Got it?

4:15Four rules for the limits

Now, four quick rules for the limits. First: the integral from a to a is zero. Width zero, so no area. Okay? Second: swap the limits, and the sign flips. The integral from a to b is minus the integral from b to a. Right? Third: split at any point c. The integral from a to b equals a to c, plus c to b. This is how we handle an antiderivative that breaks inside the interval. Clear? Fourth: x is only a dummy variable. The integral of f of x, d x, equals the integral of f of t, d t, with the same limits. That one is really simple.

4:56Example 1: x squared + 2x from 0 to 1, ln x from 1 to e

Now let's use the theorem. Example one. Evaluate the integral from zero to one of x squared plus two x, d x, and the integral from one to e of the natural log of x, d x. Here e is the base of the natural log. Pause the video and try it first. Done? Okay, let's check. First, our new habit: x squared plus two x is continuous on zero to one, and log x is continuous on one to e. So the theorem applies. First one: the antiderivative is x cubed over three, plus x squared. Put in one: one third plus one. Put in zero: zero. So, four thirds. Easy, right? Now the second one. Which function has derivative log x? Yes, exactly: x log x minus x, found by parts in the last chapter. We take it from one to e. With me? At the top, log e is one, so e minus e is zero. At the bottom, one times log one is zero, so we get minus one. Zero minus minus one: the answer is one. Watch that double minus!

6:09Example 2: 1/(5 + 4 cos x) from 0 to 2 pi is 2 pi/3, not 0

Now a harder one, with a trap inside. Example two. Evaluate the integral from zero to two pi of, one over five plus four cos x, d x. Remember our puzzle. Pause the video and try it first. Done? Okay, let's check. Using the substitution t equals tan x over two from the last chapter, the antiderivative is: capital F of x equals two thirds, tan inverse of, one third tan x over two. The minus one up top means inverse, okay? Many students just put in the limits. Tan pi and tan zero are both zero, so the answer is zero, right? No, no, careful here! It is our puzzle again. Since cos x is at least minus one, five plus four cos x is at least one, greater than zero. Positive function, so zero is impossible. Okay? So what broke? Tan x over two is undefined at x equals pi, right inside our interval. So capital F jumps there, and the theorem cannot be used across pi. With me? So we split at pi, our third rule. As x comes up to pi from the left, tan x over two grows without limit, and tan inverse climbs to pi over two. So the first half is two thirds times pi over two, minus zero: pi over three. Right? Now, pi to two pi. The arrow means replace x by two pi minus x. Cos x stays the same, and pi to two pi turns into zero to pi. So by symmetry, the second half is also pi over three. Clear? Add them: pi over three plus pi over three is two pi over three. It is not zero at all. The same trap, hidden in a trig function. Always check continuity first, okay?

7:58Quick revision: signed area, F(b) - F(a), check continuity

So, quick revision. The definite integral is the signed area; the modulus gives the total area. The fundamental theorem: the integral from a to b is capital F of b minus capital F of a, where F prime equals f. And first check that capital F is continuous on the whole interval; if not, split. Now try the practice problems on this page. All the best!