016A Homework 14 SolutionJae-young Park∗December 6, 2008• 6.2 #4 Determine ∆x and the midpoints of the subintervals for 3 ≤x ≤ 5, n = 5Solution ∆x =5−35=25= .4. The first midpoint is 3 +∆x2= 3.2The other four midpoints are found by adding ∆x = .4 successively.So we get3.2 3.6 4 4.4 4.8• 6.2 #6 Estimate the area under the graph of f(x) = x2for −2 ≤ x ≤ 2using the Riemann sum for n = 4. with midpoints of subintervals.Solution ∆x =2−(−2)4= 1. So the midpoints are given by−1.5 − .5 .5 1.5So the Riemann sum is∆x[(−1.5)2+ (−.5)2+ (.5)2+ (1.5)2] = 5• 6.2 #12 3 ≤ x ≤ 7, n = 4, left endpoints.Solution The length of the subintervals are ∆x = 1. Readingfour values at the left end p oints, we get f(3) = 8, f(4) = 7, f(5) =6, f(6) = 4. So the Riemann sum is∆x[f(3) + f(4) + f(5) + f(6)] = 8 + 7 + 6 + 4 = 25To draw approximating rectangles, sketch 4 rectangles with width 1;The first rests on the interval [3,4] with height 8, the second rests onthe interval [4,5] with height 7, the 3rd is on the interval [5,6] withheight 6 and the last rectangle is on the interval [6,7] with height 4.∗jaypark at m a t h . b e r k e l e y . e d u. GSI for 16A 101,104,1051• 6.3 #10 CalculateR41x2√xdx.Solution Let F (x) be an antiderivative of x2√x. Then F (x) =Rx2√x =Rx5/2=27x7/2+ C where C is an integration constant. Bythe fundamental theorem of calculus, we haveZ41x2√x = F (4) −F (1) = [2747/2+ C] −[2717/2+ C] =28− 27=2547• 6.3 #12 CalculateR10(4x3− 1)dx.Solution Let F (x) be an antiderivative of 4x3− 1. Then F (x) =R4x3− 1 = x4− x + C where C is an integration constant. By thefundamental theorem of calculus, we haveZ104x3− 1 = F (1) − F (0) = [14− 1 + C] − [0 − 0 + C] = 0• 6.3 #22 FindR21(x2+2x+12x2)dx.Solution The antiderivative of (x2+2x+12x2) isx24+ 2 ln x +x−1−2.So we haveZ21(x2+2x+12x2) = [224+2 ln 2−1/4+C]−[14+2 ln 1−1/2+C] = 2 ln 2+3/2• 6.3 #24 CalculateR1−11−x−e−x2dx.Solution Let F (x) be an antiderivative of1−x−e−x2. Then F (x) =R1−x−e−x2=x−(1/2)x2+e−x2+C where C is an integration constant. Bythe fundamental theorem of calculus, we haveZ1−11 − x − e−x2= [1 − (1/2)12+ e−12+C]−[−1 − (1/2)(−1)2+ e−(−1)2+C]So,R1−11−x−e−x2= 1 +e−12−e2• 6.3 #26 CalculateR10(ex/3−2x5)dx.Solution Let F (x) be an antiderivative of (ex/3−2x5). Then F (x) =R(ex/3−2x5) = (3ex/3−x25) + C where C is an integration constant.By the fundamental theorem of calculus, we haveZ10(ex/3−2x5) = F (1)−F (0) = [(3e1/3−125)+C]−[(3e0/3−025)+C] = 3e1/3−1652• 6.3 #28R943t−2dtSolutionZ943t − 2dt = [3 ln(t − 2)]94= 3 ln 7 − 3 ln 2• 6.3 #34 Find the area under the curve y = e3x; x = −1/3 to x = 0Solution The area under the curve is given by the definite integralR0−1/3e3xdx So,Z0−1/3e3xdx = [e3x3]0−1/3=13−13e• 6.3 #38 InterpretR75p(t)dtSolution It is the pollutants discharged into a lake from 1995 to1997.• 6.3 #42 (Profit) Suppose that the marginal profit function for a com-pany is P0(x) = 100 + 50x − 3x2at production level x.(a) Find the extra profit earned from the sale of 3 additional unitsif 5 units are currently being produced.Solution We want to find P (8) −P (5). By the fundamental theo-rem of calculus, we haveP (8) − P (5) =Z85P0(x)dx =Z85100 + 50x − 3x2dxSo, the extra profits areR85100 + 50x −3x2dx = [100x + 25x2−x3]85=888(b) Describe the answer to part (a) as an area. (Do not make asketch)Solution The area under the marginal profit curve y = P0(x) fromx = 5 to x = 8.• 6.3 #48 (a) ComputeRb11tdt, where b > 1.Solution Note that ln t is an antiderivative of1t. So by the funda-mental theorem of calculus, we getRb11tdt = ln b − ln 1 = ln b3(b) Explain how the logarithm of a number greater than 1 may beinterpreted as the area of a region under a curve.(What is the curve?)Solution We know ,from (a), ln b =Rb11tdt So we may interpret itas a area under the curve y =1tfrom t = 1 to t = b.• 6.3 #50 For each number x > 2, let A(x) be the area of the regionunder the curve y = x3from 2 to x. Find A0(6).Solution We can express A(x) as definite integral, that is, A(x) =Rx2t3dt = F (x) − F (2) where F (x) is an antiderivative of t3. Takingthe derivative, we getA0(x) = F0(x) = x3So, A0(6) = 63= 216.• 6.4 #2 Write down a definite integral that gives the area of the shadedregion of Figure 11.Solution The area of the shaded portion is given byA =Z32[f(x) − g(x)]dx.Note that this rule doesn’t depend on whether f(x) and/or g(x) arepositive or negative. It is valid so long as the graph of f(x) lies abovethe graph of g(x) for all x from x = 2 to x = 3.• 6.4 #8 Find the area of the region.Solution Note that 2 ≥ x(2 − x) for 0 ≤ x ≤ 2. So the area isgiven by a definite integralR202 − x(2 − x)dx =R202 − 2x + x2dx =[2x − x2+ x3/3]20= 8/3• 6.4 #10 Find the area of the region.Solution Note that y = e2xis an increasing function, and its valueat x = 0 is 1. On the other hand, y = 1 − x is a decreasing function,and its value at x = 0 is also 1. So both the functions have the samevalue at x = 0. But after that the first function increases continuouslyto e2at x = 1, while the second function goes to 0 at x = 1. Hencefor all x between x = 0 to x = 1, e2x≥ 1 −x. So the area between the4curves from x = 0 to x = 1 isA =Z10[e2x− (1 − x)]dx=Z10[e2x+ x − 1]dx=he2x2+x22− xi¯¯¯10= (e22+12− 1) − (12)=e22− 1• 6.4 #18 Find the area of the region bounded by two graphs; y = 4/xand y = 5 − xSolution We first find the points of the intersection. By setting4x= 5 − x and solving for x, we get x = 1, 4. Noting that 5 − xis greater than 4/x for x between 1 and 4, the area is given by thisintegral;R41[5 − x] −4x. Therefore the desired area isZ41[5 − x] −4x= [5x − x2/2 − 4 ln x]41= 15/2 − 8 ln 2• 6.4 #20Solution (a) Note that y = x2is an increasing function, and itsvalue at x = 1 is 1. On the other hand, y = 1/x2is a decreasingfunction, and its value at x = 1 is also 1. So both the functions havethe same value at x = 1. But after that the first function increasescontinuously to 16 at x = 4, while the second function goes to116atx = 4. Hence for all x between x = 1 to x = 4, x2≥ 1/x2. So the areabetween the curves from x = 1 to x = 4 isA =Z41[x2− 1/x2]dx=Z41[x2− x−2]dx=hx33+ x−1i¯¯¯41= (643+14) − (13+ 1) =8145(b) The moment we go to the left of 1, things become little tricky.To the