# Lecture 6 Notes - Chapter 6 Notes The Definite Integral 1...

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Chapter 6 NotesThe Definite Integral1Area1.1What is Area?
Aregionis a set of points in a plane and apolygonalregion isa polygon, together with its interior.We can always find the area of a polygon by cutting it up into triangles andfinding the sum of the areas of the individual triangles.
LetRbe a region in the plane. We say thatRis bounded if thereis a polygonal region that entirely containsR.
LetRbe a region in the plane. IfPis a polygonal region thatcontainsR, we sayPis acircumscribedpolygon forR. IfPis a polygonalregion that lies entirely inside the regionR, we callPaninscribed polygonofR.
We call the greatest lower bound of areas of circumscribed poly-gons ofRtheouter areaofR.1
Definiton 5.We call the least upper bound of the areas of inscribed polygonsofRtheinner areaofR.
If the inner and outer area’s of a regionRare equal, we call thelike value the area ofR.1.2The Region under a GraphThe set of points in the plane given byR={(x, y)|axb,0yf(x)}is theregionunderfover [a, b]Example 1.We will compute the area of the region under the graph off(x) =x2over the interval [0,2]. We first consider the outer area of this object. Con-sider the rectangle whose base is [0,2] and whose height is 4. This rectangle is acircumscribed polygon forR. The area of the region underfmust be less thanthis value.Now letnNand divide [0,2] intonequal subintervals of length 2/n. Thek-thsubinterval is given by [(2k-2)/n,2k/n]. Now formnrectangles whose union isa circumscribed polygon forRwhose height isf(2/n) = 4/n2. The area of thecircumscribed polygon is simply the sum of the areas of each rectangle. Sincethe height of each rectangle is determined by its right end-point, we call thissum then-thright-handedsum, denotedRf(n).We find
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Rf(n) =f2n2n-0+f4n4n-2n+· · ·+f2nn2-2n-2n=4n22n+16n22n+· · ·+4n2n22n=4n2(1 + 4 + 9 +· · ·+n2)2n=8n3(12+ 22+· · ·+n2)=8n3n(n+ 1)(2n+ 1)6=431 +1n2 +1nTaking the limit asn→ ∞we findRf(n)8/3. Thus the outer area can beno greater than 8/3.We now concentrate on finding the inner area ofR. We divide [0,2] intonsubin-tervals as above. Thus time we construct rectangles contained inR. We definetheir sum to be then-thleft handedsum since their values are determined bythe left endpoints. We calculateLf(n) =f(0)2n-0+f2n4n-2n+· · ·+f2n-2n2-2n-2n= 02n+4n22n+· · ·+4(n-1)2n22n=4n2(0 + 1 + 4 +· · ·+ (n-1)2)2n=8n3(12+ 22+· · ·+ (n-1)2)=8n3(n-1)n(2n-1)6=431-1n2-1nAsn→ ∞Lf(n)8/3. Thus the inner area ofRcannot be less than 8/3.Was does this tell us the area underRmust be?
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1.3Area as the Limit of a SumWe can use the above approach for any function that is continuous and non-negative. For a functionf(x) on an interval [a, b], the algorithm is thus:1. Subdivide the interval [a, b] intonpieces of equal length and letΔx=b-anThen definex0=ax1=a+ Δxx2=a+ 2Δxxk=a+kΔxxn=a+nΔx=bThe set{a=x0, x1,· · ·, xn-1, xn=b}forms a
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