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Lecture Commentary
HW Help
Tue Jan 19
(5.3 #25) Since there is no division rule for integrating, you might want to divide each numerator term by the x^2 first.
(5.3 #41) Here, f is the derivative of the other functions graphed. So, if f takes a value of zero, the other functions would need to have a critical pt.
(5.3 #53) Similar to #23 above.
(5.3 #61) No simple way for your answer not to involve a power of e.
(5.3 #63) A twostepper. You might try finding each missing constant at each step rather than involving two contants at once.
(5.3 #71) The first sentence essentially says that s(1)=0. Getting familiar with this kind of phrasing will help for these types of problems.
(5.1 #32b) This is just the sum a_1+...+a_10 minus the a_1 term. You can think about what the difference will be.
(5.1 #39) Similar to 5.3 #23.
(wksht #1) Similar to 5.3 #25, but with x^(1/2).
(wksht #3) Similar to 5.3 #63.
Tue Jan 26
(5.1 #43) 1/N^2 can be factored out of the summation since it does not involve the variable i.
(5.1 #58) The summation will distribute across the sum and factor out the constant to yield formulafriendly Simga(x^2) and Sigma(x) sums.
(5.1 #63) The constant multiple is the width of each rectangle. The terms of the sum are the rectangle heights/function values.
(5.1 #73) Squaring y=sqrt(1x^2) and simplifying, you get x^2+y^2=1^2, the equation of a circle centered at the origin of radius 1. Since you are only given the positive sqrt, you are only conisdering the semicircle portion where y>=0.
(5.2 #7) Similar to 5.1 #73.
(5.2 #61) You have the area under f(x) on [2,9] and are subtracting the area under f on [2,5]. What amount of the xaxis remains considered?
(AP #2) Similar to 5.1 #63.
(AP #4) Similar to 5.1 #73.
(AP #6) Consider the min and max xvalues given as the domain (ie. [1,8]). Then, break the area region into pieces as given.
Tue Feb 3
(5.4 #39) In general, the integral of 1/(mx+b) equals (1/m)*lnmx+b+C. The 1/m is to account for the chain rule.
(5.5 #29) d/dx of F(x^2) requires the chain rule. It computes to f'(x^2)*2x.
(5.5 #33) Similar to 5.5 #29, but you need to compute d/dx of F(x^2)F(sqrt(x)). Since this expression is a sum, just take it term by term.
(5.6 #5) The integral from a to b of v(t) is V(b)V(a), which is just s(b)s(a)...or, change in position.
(5.7 #33) In general, when in the function you see a pair of polynomial expressions differing in degree by 1, set u equal to the higher power expression.
(5.7 #59) Hint: u=x^(1/2). When you compute du, you'll see why this works.
(5.7 #72) Hint: u equals either x, 1/x, lnx, or 1/lnx.
(5.7 #91) Similar to 5.7 #72.
(5.8 #11) In other words, x=3u.
(5.8 #25) Comparing the expressions in the numerator and denominator, one is the derivative of the other. Substitution can come to you aid.
(5.8 #44) Since you don't have an integration rule for tan(x), you might rewrite as sin(4x+1)/cos(4x+1) to get in terms of more familiar trig functions.
(5.8 #50) This one is not easy to see. Using u=sqrt(25x^21) and multiplying the top and bottom of the fraction by 25x, it is doable as an invtan integral (not invsec...which doesn't apply without absolute value bars).
(AP #13) You need x to be some multiple of u that when squared cancels the 9 and allows a sqrt(4) to be factored out of the denominator.
5.5 #29
5.7 #54
5.8 #15
Tue Feb 9
(6.1 #38) You can drop the absolute value and find the point of intersection where x is positive. Then, using the fact thatthe parabola and aboslute value functions are symmetric about the yaxis, you have the other intersection point
(6.1 #41) Similar to 6.1 #29
(6.1 #47) Need to break the region between the three curves into a pair of regions each between two curves. See image.
(6.2 #9) For any given y, what is the width of the unit circle from lefttoright? This expression holds the key.
(AP #2) Similar to 6.1 #29
Tue Feb 16
(6.2 #27) Multiply out x(6x) to be able to integrate.
(6.3 #1519) On the region of interest, you need to determine which function is above and which is below.
(6.3 #2731) Note region A is between y=x^2+2 and y=6, or (as a function of y) between x=2 and x=sqrt(y2)
(6.3 #3335) Note region B is between y=x^2+2 and y=0.
(AP #5) Even though this region is technically bounded by three curves, you should be able to complete the computation using only two of those functions in your integral.
Tue Feb 23
(6.4 #47) Function of y about yaxis. This is not shells, is it?!
(6.4 #51) Similar to 6.4 #47 above.
(6.5 #37) Note that k is given in N/m, but stretching/compressing measurements are given in cm.
(6.3 #1729) Videos posted for Fri 2/19 may be of assistance. In #29, note the kg/m chain density can be computed directly from the information given.
(AP #2) Similar to 6.4 #43 above.
(AP #6) Similar to 6.3 #17 above.
Wed Mar 2
(7.1 #22, #23) When you see ln(x) inside an integral, it is extraordinarily frequent that you set u=ln(x) and use integration by parts.
(7.1 #35) With du=(1/2)*x^(1/2)*dx, you can rearrange to write dx in terms of just u and du to be able to integrate (by parts).
(7.1 #77) With u=x^4, you can write x^7*dx in terms of u and du to be able to integrate (by parts).
(7.2 #53) Recall that 1/cos(x) is just sec(x). To integrate sec(x), use the (sec(x)+tan(x))/(sec(x)+tan(x)) trick from class.
(7.2 #55) When you see an integral involving only tan(x), a path to consider always is rewriting as sin(x)/cos(x).
(AP #1, #4) Similar to 7.1 #22 and #23 above.
(AP #6) Will have to use the doubleangle identity cos^2(x)=(1/2)+(1/2)*cos(2x) twice to reduce the cos^4(x) to a cos^2(2x). Then, you'll have to use the identity again to get everything down to single powers of cos.
(AP #9) Similar to 7.2 #55 above.
Tue Mar 8
(7.3 #15) When you boil down to a sin^2 integral, you'll need to use the doubleangle identity (like on last HW).
(7.3 #17) Converting from sec and tan into sin and cos, you can get down to integrals of csc(x). Similarly tricky to integrating sec(x), this time multiply top and bottom by (csc(x)cot(x)).
(7.3 #21) Recall that cot has a derivative of csc^2. This will help you integrate csc^2.
(7.3 #48) You'll need your doubleangle identities to make sense of the sin^2*cos^2 integral.
(7.4 #9) Just like lone powers of tan, you might try dealing with this lone power of tanh by writing in terms of sinh and cosh.
(7.5 #31, #37) Expressions given are not proper rational functions. Need to divide down the numerator degree first.
(AP #1) Rewriting 1/cos^2 as sec^2 can help you integrate more straightforwardly.
(AP #2) Similar to AP #1, but with 1/sin^2 and csc^2.
(AP #4) When you get down to sin^2/cos^2, you might write as tan^2 and then swap for sec^2 using 1+tan^2=sec^2 to get something straightforward to integrate.
(AP #5, #6) The identity cosh^2(x)sinh^2(x)=1 will be helpful on these. Analogous to odd powers of sin & cos problems.
Tue Mar 22
(7.6 #41) If you FOIL, you can deal with the sin(x)cos(2x) term by using the doubleangle identity in reverse...that cos(2x)=2cos^2(x)1.
(7.6 #45) After using integration by parts, the 2x^2/(x^2+9) term can be dealt with via division algorithm and partial fraction decomposition.
(7.6 #55) The notation ln^2(x) means ln(ln(x)). And, letting u equal this gets you half way home via parts.
(7.7 #1) Does the integral have a bound of +/ infinity? Does the integral have a bound at a vertical asymptote?
(7.7 #11) u=4x will get the function in the form of a familiar pintegral.
(7.7 #15) Similar to 7.7 #11.
(7.7 #45) Substitution with u=1+x^2 can evaluate the integral.
(7.7 #59) Since 1sin(x) is never more than 2, our function is sandwiched between 2/x^2 and 2/x^2. Sketching a picture of this might help.
(AP #2) Similar to 7.5 #13.
(AP #3, #4) Similar to 7.7 #11.
Tue Apr 5
(7.9 #27) The data given contains your function values (ie. f(0 min)=550, f(5 min)=575, etc.).
(8.4 #33) To find our K, we need the maximum absolute value of the 6th derivative of f on [0,0.25]. This is the maximum of cos(x) on [0,0.25], which can be determined definitively.
(8.4 #41) We just need to find a single n. So, we could use a K that we know bounds all absolute derivative values for cos(x) and then chase the n by substituting n=1,2,3,... into our error formula.
(8.4 #45) The expression on the right hand side is just x^4/4!. So, the K being used is 1. Why does K=1 work for f(x)=e^x on [0,inf)?
(8.4 #46) Similar to 8.4 #45 above. Note that 5!=120. So, again, we must show that K=1 is a valid choice of K.
(10.1 #23) Multiplying top and bottom by 1/sqrt(x^2) might help.
(10.1 #25) Since ln() is a continuous function, it distributes in and out of a limit. That is to say, "the limit of ln(f)" is just equal to "ln(the limit of f)".
(AP #2) Similar to 7.9 #27.
(AP #3) Recall the derivative of tan^1(x) is 1/(1+x^2). For subsequent derivaives, just use the chain rule.
(AP #4, #5) Similar to 8.4 #33.
(AP #6) Similar to 8.4 #41.
Tue Apr 12
(10.1 #45) With trig function upstairs, this is a squeeze theorem candidate.
(10.1 #55) May be easier to see ratio when separated as a sum of two fractions, each written over the common denominator.
(10.2 #11) Note cancellation from telescoping terms.
(10.2 #12) Use partial fraction decomposition to see the telscoping terms of the sum.
(10.2 #25, 29) Note sum does not start at n=0. Must account for missing terms.
(10.2 #3335) Can factor out common factor so the terms of the sum start with a 1 and can use the goemetric series formula.
(10.3 #5) u substitution?
(10.3 #17) Note that when n>=1, it is true that n>=sqrt(n).
(AP #1, #2) L'Hopital can be of service.
(AP #3) Similar to 10.2 #3335.
(AP #6) Bounding the numerator and using comparison can make the numerator a lot easier on the eyes.
(AP #7) Recall your result from AP #1.
(AP #8) Note the terms of this sum form a decreasing sequence of positive numbers.
Mon Apr 18
(10.3 #51) Difference in powers between top and bottom is 1/2. Limit compare to 1/n^(1/2).
(10.4 #1213) Using the fact that for an alternating series SS_n
(10.4 #25) Might first separate into two fractions, each over the common denominator to get two geometric sums, one with r=3/5 and the other with r=2/5.
(10.5 #7) Note that (n+1)^100 has leading term of n^100, and all other terms are of lower degree (and hence do not affect the limit as n>inf).
(10.5 #50) u=ln(x)?
(AP #3) Rather involved problem. Using ratio test, note the expression (n+1)^n/n^n can be rewritten as [(n+1)/n]^n...which can further be rewritten as [1+(1/n)]^n. Now use the result from AP #2 in HW 11 that says this expression has a limit of e^1.
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