
Math Tutorial Center
Ayres G012
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HW Help
[Q1]
(40, 41) Like the advice given in 9, it might help to work backwards. Rather that looking at the graph of f and deciding which is a graph of its antiderviative, you might look at the possible antiderivative graphs and ask which of them has f for the graph of its derivative. By process of elimination, you can find the solution. [Q1]
(2328) You want to look at the form for each sum given and notice what values that make up each term are changing. The values that are changing need to use j to represent them in summation notation. The values that are not changing will not require the use of j. [Q1]
(14) The function f(x) will take every region of f(x) that is below the xaxis and reflect it to be above the xaxis, hence making every function value that was negative into a positive of equal magnitude.
(46) If you distribute the integral across the sum, you can then compute the integral of the function x on [0,5] by drawing a graph and computing the area geometrically. [Q2]
(4344) Where the expression inside the absolute value is negative, the absolute value will flip that negative to a positive. For example, if we are looking at f(x)=3x, this function is just 3x when x<3. But, when x>3, the function 3x is the opposite of the negative value inside...so the function acts like x3 for those x>3 values.
(5556) The area between the function and the xaxis on the interval [2,3] is the same as the area on the interval [2,2] plus the area between [2,3]. This is simply a matter of a whole being the sum of its parts.
(5960) The F(1)=3 bit can help you find the C associated with the general form of F when you integrate. Once you find what C needs to be, finding F(4) is purely a matter of computation.
[Q2]
(28) Note the top bound is not just x but x^2. The chain rule will need to be applied to x^2 when we take the derivative to find G'(x).
(2930) Similar to 28. The top bound is not just x. We need to apply the chain rule to that top bound. (Note: Even in the case where the top bound is just x, we can still apply the chain rule if we wish...that will just multiply the expression we get by 1...which is exactly what you usually get in those cases.)
(42) The integral on [a,b] calculates an error free estimate for the area between f(x) and the xaxis. Note that when the function is below the xaxis, the function values will be negative, which will cause the regions of area below the xaxis to be accounted negatively by the integral calculation. This function f(x)=x^29 only has so much negative area. Isolate that negative area with a careful choice of a and b and you will account for all the negative area and none the positive area by this integral, hence achieving a minimum for the value of this integral.
(8086) Note the functions given are all in terms of x. That is to say, all of these functions live in the xyplane...the bounds given for the area regions given along the xaxis. Once you change variable to u in order to integrate, your function will now be living in the uyplane. You will need to change the bounds on your area regions accordingly (using the relation between x and u you used to substitute originally) so that the bounds given along the xaxis now are bounds along the uaxis.
(5.9) 5, 6
[Q3]
(5.8 #2021) Using u=x^2 will get these functions starting to look like the standard forms of tan^1(x) and sec^1(x).
(5.8 #22) You know facts like (2+3)/7 = (2/7)+(3/7). If you apply this kind of idea here, you'll be able to break this single integral into a pair of integrals, each of which will be just like problems you saw previously in 79.
(5.9 #56) k=ln(2)/(doubling time) will give you the constant you need to put into N(t)=(N_0)*e^(kt).
(6.2) 4, 8, 9
[Q3]
(6.1 #19, 35) When encountering three functions, you can observe that moving left to right within your region, the region is always only dependent upon two functions. Breaking into proper subregions accordingly, you can compute the area of the desired region. See Image.
(6.2 #8) Drawing a line from the top vertex of an equilateral triangle to its base, you can use 306090 triangles to show that the area of an equilateral triangle is (sqrt(3)/4)*s^2, where s is the length of a side.
(6.3) 12, 56
[Q4]
(6.2 #19) This is a tough problem. Drawing pictures and keeping the letters straight will be a major concern. To find (a), you can use similar traingles. Modifying the picture used to analyze the similar triangles in (a), you can find (b). If you've found (a) and (b), then (c) is merely a matter of integration. Recall all letters except x will be treating as if they were constants (ie. some number).
(6.3 #12, 56) When revolving a function of x around the xaxis, chop up the xaxis interval and draw a slice. The slice will be a thin round disk.
(6.4) 710, 2122, 3536
[Q4]
(6.3 #3940) The same hint as for 1516, 2728 applies. Additionally here, we must consider not just how far the top and bottom functions are from the xaxis but how far they are from the axis of rotation.
(6.4 #710) Find intersection points to have bounds for your region of interest.
(6.4 #2122) Even thoough there are three functions given, your expression for the area will not need to depend on all three. Even if done roughly, a sketch can be really helpful when trying to visualize the region being revolved.
(6.4 #3536) Be as explicit as possible in your description of the functions that define the boundaries of this region. This will allow you to practically plug and chug when expressing the volume of a slice.
(6.5) 4, 6, 12, 13
[Q4]
(6.5 #1213) Work problems involve determining a general expression for the work on a given slice of the figure of interest. In particular, if the column or cone has height of 4, you could think of these objects existing between y=0 and y=4. The slice you take would have to be at a generic y between 0 and 4. You can use known methods from sections 6.3 or 6.4 to find the volume of tye slice. Multiply by density to find mass. Multiply by g to find weight. Multiply by the height of the lift (namely y) to find work. Integrate along the region of interest (y=0 to y=4), and you'll have it.
[Q5]
(7.1 #8, 12, 14) For integrals of the form x^n times cosine, sine, or e, you need to set u equal to x^n and dv equal to everything else. The power of x will indicate how many times you will need to use parts before you will have fully evaluated this kind of integral.
(7.1 #20, 24, 50) For integrals that involve ln(x), you can set u equal to ln(x), or (ln(x))^2, and let dv be everything else.
(7.1 #52) This is a ticky problem. You may want to use substitution and let u=x^3 since the denominator has an expression of degree 2 within it. However, if you try this out, you will notice it does not work. To exploit substitution, you can make dv equal (x/sqrt(x^2_9))*dx...you will be able to utilize substitution on that when you find v. Hence, u will have to be everything else...namely x^2. If you now run through parts, you will be able to make the integration work. Tough problem!
[Q5]
(7.1 #82) See hint aboce from 7.1 #20, 24, 50
(7.1 #85) Treat b^x almost like e^x. Just known that the derivative of b^x is not simply b^x, but b^x*ln(b). With this and the hint from 7.1 #8, 12, 14 in mind, you can solve this problem without too much trouble.
(7.3) 15, 7, 8
(7.2 #9, 10) Integrals involving powers and sine and cosine that do not have an odd power demand the half angle formulas of cos^2x=(1/2)+(1/2)*cos2x and sin^2x=(1/2)(1/2)*cos2x. Repeated use of these formulas will allow integrals of sine and cosine without odd powers to be solved.
(7.3) 15, 17, 19
(7.2 #57) Taking cos^2x+sin^2x=1 and dividing both sides by cos^2x yields 1+tan^2x=sec^2x. Applying this once to the expression sec^4x should give you a pair of integrals to work with, each of which can be dealt with with (at worst) substitution.
(7.3 #15, 17, 19) Integrals involving sqrt(a+x^2) need the substituion x=sqrt(a)*tan(theta). Integrals involving sqrt(x^2a) need the substitution x=sqrt(a)*sec(theta). Integrals involving sqrt(ax^2) need the substitution x=sqrt(a)*sin(theta). These three facts should be memorized, and it will probably be worth your time to become familiar with integrals of these types.
[Q6]
(7.5 #16) Note that x^2+x is not fully factored yet. You'll need to factor it fully before you attempt to decompose into partial fractions.
(7.5 #1718) The repeated factors in the denominator of these expressions need to be dealt with by including a fraction in the partial fraction decomposition for each power of the repeated factor (up to the power present).
[Q6]
(7.5 #2829) Similar to the hint immediately above, but with two fractions in your decomposition for each of the repeated factors of degree 2.
(7.5 #30) Note the numerator degree is not less than the denominator degree, so we will need to perform long division before using partial fraction decomposition.
[Q7]
(7.7 #2223) Similar to above. The bounds on these integrals appear on the surface to present no issue. Looking further reveals an undefined nature at x=1 for #22 and at x=3 for #23. The bounds corresponding to these asymptotes will need to be accounted for with a limit.
(7.7 #24) Here, we also have bounds that appear harmless on the surface. Moreover, neither x=4 nor x=0 are locations of an asymptote for this function. Be careful though. There is an asymptote between x=4 and x=0...in particular, at x=2. To properly account for this, we'll need to split the integral into two regions: one on [4,2], and another on [2,0]. For each integral, we will need to take a limit for the bound at x=2.
(7.7 #38) ln(1)=0.
(7.7 #4547) To deal with the doubly improper integral region, we will have to break this into a pair of singly improper integrals. Pick any value along the number line to split the integral. For simplicity, let's say x=0. This will give us two regions: one on (inf,0], and another on [0,inf). Dealing with each infinity via a limit, we can properly account for the area between this curve and the xaxis.
[Q7]
(7.7 #6162) In 61, since we have an x^5 expression square rooted, the effective power of x is 5/2. You can use this for comparison. #62 is similar, but with an effective power of 3/2.
(7.7 #63) The effective power here is 1/2. Knowing what we know about the ptest for integrals, we should expect divergence and work by comparison to show this.
(7.7 #64) Note we are dealing with a vertical asymptote at x=0. When x gets small (in absolute value), the smaller powers of x will dominate the expression. Comparison with x^(1/3) will allow you to show convergence.
[Q7]
[Q8]
(11.2) 15
[Q8]
(8.1 #7) You can use usubstitution, letting u equal what is inside the sqrt.
(8.1 #8) Note that ((1/2)x^(1/2) + (1/2)x^(1/2))^2 is equal to 1+(f')^2.
(8.1 #11, #13) Let your function be the arc length formula, sqrt(1+(f')^2). This, use the trapezoidal rule or simpsons rule on that formula (on the interval given).
(11.2 #4) Similar to 8.1 #7.
(11.2 #5) If you factor out the sqrt(t^2)...or just t, you can use usubstitution letting u equal the degree 2 expression remaining inside the sqrt. The t your factored out will work well with du.
[Q8]
[Q8]
[Q8]
(12) Decompose 1/[n(n+1)] into partial fractions. Then, many of the terms will cancel just like in (11) above.
(25) Negative exponents can be expressed as positive exponents by bringing numerator terms to denominator (and vice versa). Hence, (a/b)^(n) is equal to (b/a)^n.
(40) Note that .217217217... is equal to .217 + .000217 + .000000217 + ..., and this is equal to 217/1000 + 217/(1000)^2 + 217/(1000)^3 + ... . Factoring out the first term, you get something of the form c*(r^0 + r^1 + r^2 + ...).
(19) You know 1/(2^n) is the same as (1/2)^n. A series of the form r^n is geometric, and can be easily determined to be convergent or divergent based on the absolute value of r. You can use the (1/2)^n as comparison in this case.
(10.4) 16, 1720
(10.3 #50, #62) Recall that sin(Any Angle) and cos(Any Angle) is between 1 and 1. In #50, you encounter a cos^2 term, which must always be less than 1...so cos^2()/n^2 must be less than 1/n^2. Compare. In #62, the cos(n) being between 1 and 1 will not dictate the limit as n > infinity. The n^2 term in the denominator will.
(10.4 #4) The terms are of the form n^4/n^3 and will not approach zero.
(10.4 #6) Recalling that sin(Any Angle) is between 1 and 1, then this sum takes a value between that of 1/n^2 and 1/n^2. And you know the value of those two sums (convergent, by ptest).
(10.4 #17) Similar to 10.2 #25. In particular, 7^(n) is equal to (1/7)^n.
(10.4 #19) Observe that 5^n3^n is greater than 2^n when n>1. This will allow you to compare.
(10.4 #20) Highest degree terms take the form 1/n. Limit compare.
(10.6) 9, 10
(10.5 #4) This expression is not conducive to the ratio test (it lacks an n! and an exponent of n). Use limit comparison to get a conclusive result.
(10.5 #11) Similar to 10.5 #3, but with a factor of e as well. Just recall that e is a fixed number. The n+1 term though grows boundlessly as a limit when n > infinity.
(10.5 #12) Note that [n^40/(n+1)^40] expands to be of the form [n^40/(n^40+...)], where the dot dot dot is terms of degree lower than 40. Since the numerator and denominator are of equal degree, the limit as n > infinity will be the ratio of the leading coefficients...namely 1.
(10.6 #9, 10) Since power series necessarily have an exponent of n, the ratio test is going to be the test of choice with any power series radius of convergence problem. The a_(n+1)/a_n>1 and <1 cases are conclusive. Since the ratio test does not yield a conclusive result when a_(n+1)/a_n=1, we have to be sure to test a_(n+1)/a_n=1 and a_(n+1)/a_n=1 cases separately.
(10.7) 3, 8, 29, 30, 42, 53, 55
(10.7 #3) Note that 1/(12x) is of the form 1/(1r), where r=2x. This series takes the form r^0 + r^1 + r^2 + ...
(10.7 #8) We need to use Taylor's formula and find a trend in the terms f^(n)(0)/n!. Note that f'(x)=4e^(4x) via the chain rule. Hence f^(n)(x)=4^n*e^(4x).
(10.7 #29, #30) Each coefficient of a_0 + a_1(xc) + a_2(xc)^2 + ... is of the form a_n = f^(n)(x)/n!. Compute the first few terms of this form to see the trend in the expression of the coeffcients.
(10.7 #42) Similar to 10.7 #3. 1/(1x) is of the form 1/(1r), where r=x.
(10.7 #53) Use cos(t) = 1  t^2/2! + t^4/4!  t^6/6! + ..., then subtract 1 and divide out the t.
(10.7 #55) Use ln(1+t) = t  t^2/2! + t^3/3!  t^4/4! + ...