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Math 113, Calculus IITest 1: Review Solutions1. Use the limit definition of the definite integral to evaluate52(4x-x2)dx.Solution:By definition,52(4x-x2)dx= limn→∞∑ni=1f(x*i)Δx, where Δx=b-an=5-2n=3n,f(x) = 4x-x2and we may takex*i=xito be the right-hand endpoints of thensubintervals, which are given asxi=a+iΔx= 2 +i3n= 2 +3in. Thus,52(4x-x2)dx= limn→∞ni=1f(2 +3in)·3n= limn→∞ni=1[4(2 +3in)-(2 +3in)2]·3n= limn→∞ni=1[8 +12in-(4 +12in+9i2n2)]3n= limn→∞ni=1[4-9i2n2]·3n= limn→∞ni=1[12n-27i2n3]= limn→∞[(12nni=11)-(27n3ni=1i2)]= limn→∞[12n(n)-27n3(n(n+ 1)(2n+ 1)6)]= limn→∞[12-276(n+ 1)(2n+ 1)n2]= 12-276limn→∞(n+ 1n)(2n+ 1n)= 12-92[(1)(2)] = 12-9 = 3.Hence,52(4x-x2)dx= 3. [Here’s a quick check using FTC(II):52(4x-x2)dx= 4x22-x33|52=(50-1253)-(8-83) = 42-1173= 42-39 = 3.]2. There is a rule of thumb that the area of any parabolic arch is two-thirds the product ofits heighthand the length of its baseb.Confirm this formula by using the FundamentalTheorem of Calculus on a suitable function to calculate the area in question.Solution:The functionf(x) =4hb2(bx-x2) has graph which is a parabola passing throughthe points (0,0),(b,0) (so that its base isb) and heighth=f(b/2). Thus we want to showthatb0f(x)dx=23bh. So considerb04hb2(bx-x2)dx=4hb2b0(bx-x2)dx=4hb2[bx22-x33]b0=4hb2[(bb22-b33)-(0)]=4hb2[b3(12-13)] = 4hb =23bhas desired.
Math 113, Calculus IITest 1: Review Solutions3. Suppose you know that a certain functionfis twice differentiable and that its graph over[0,2] is given in the figure on your review sheet. As you see, the printer was sloppy and spilleda lot of ink on the graph. Decide, if possible, whether each of the following definite integralsis positive, equal to zero, or negative.(a)20f(x)dx(b)20f(x)dx(c)20f(x)dxSolution:By theFundamental Theorem of Calculus(Part I), we know that20f(x)dx=f(2)-f(0) becausef(x) is an antiderivative off(x). Nowfmeasures the slope of thecurve, and we see that the slope offis negative whenx= 2 (because the curve is decreas-ing there!)so thatf(2)<0.Similarly, sincefis increasing whenx= 0,f(0)>0, sof(2)-f(0), whatever the actual value of the numbers may be, must be negative as it is anegative number (f(2)) MINUS a positive number (f(0)). Thus,20f(x)dx <0.Using theFundamental Theorem of Calculus(Part I) again,20f(x) =f(2)-f(0) sincef(x) is an antiderivative of its derivative,f(x). Looking at the graph ofy=f(x), we seethatf(2) is asmallpositive number whilef(0) is alargepositive number, so thatf(2)-f(0)must then be negative. Therefore,20f(x)dx <0.Lastly, we can see a fair bit of the graph, and the portion of it to the left of the ink spot is allwell above thex-axis and by itself contributes a positive area larger than the total area of theink blot BELOW thex-axis. Thus, as20f(x)dxmeasures thesigned areaoff(x) over theinterval [0,2] and the part of this area to the left of the ink blot is larger than any negativeamount the ink blot could be covering and the part to the right of the blot is again positive,