# Lecture 3.pdf - LECTURE 3 FUNDAMENTAL THEOREM OF CALCULUS...

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LECTURE 3, FUNDAMENTAL THEOREM OF CALCULUSPO-NING CHENIn this lecture, we first look at a few more examples of the Riemann sum and use its limitto define the definite integral. Then, we will state the fundamental theorem of calculus.1.More Examples of Riemann SumLet us look at the example before more closely. We want to approximateZ10x2dxwith 4 intervals. When we divide [0,1] into four intervals, each interval has lengthΔx=1-04=14.The right end points of the intervals arex1=14,x2=24,x3=34andx4= 1.The height of the four rectangles aref(x1),f(x2),f(x3) andf(x4), respectively. Hence,the sums of the area is12×116+12×14+12×916+12×1or simplyΔx[f(x1) +f(x2) +f(x3) +f(x4)]If we approximate the same interval withnintervals, then each rectangle has width2n.Hence, we haveΔx=1nLetxkbe the right end point of the k-th interval. We havex1=1n,x2=2nand so on. Ingeneral,xk=knThe total area of thenrectangle is simplyΔx[f(x1) +f(x2) +f(x3) +·f(xn)] =1n[(1n)2+ (2n)2+· · ·+ 12]Let us look at more examples:Exercise 1ApproximateZ42cosxdx
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2PO-NING CHENusing the Riemann sum with 4 equally spaced intervals.Answer to Exercise 1Letf(x) =x2. When we divide [2,4] into 4 intervals, each interval has lengthΔx=4-24=12.We have a total of 4 vertices along [0,2]. Namelyx1= 2,x