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Internet Videos & Course Commentary
HW Help
Fri Aug 26
(56) Note c^2x=(csqrt(x))(c+sqrt(x)) by difference of squares factorization.
(9) No product rule for integrals. Must fully foil first.
(12) No quotient rule for integrals. Each term needs divided by x^2 from downstairs.
(1518) A value of the antiderivative function is known, so you can now find C precisely (and must for credit).
(28) In parts a and b, note the expression inside the Sigma does not change as i changes. You'll be repeatedly adding that constant.
(29) Can use properties of sums to write Sigma(c*a_k  b_k) as c*Sigma(a_k)  Sigma(b_k).
(30, 33) No product rule for sums. Needs foiled first to be able to use the properties of sums.
(34) Recall the sum from i=1 to N of i is N(N+1)/2.
(3536) Will ultimately need a formula for the sum from i=1 to N of i^2 from notes/book.
Wed Aug 31
(5) M_n uses midpoints of each subinterval as the points that determine function/rectangle height.
(910) Recall that if d is (ba)/N, then R_N is d times the sum from i=1 to N of f(a+dj). You'll need to work backwards algebraically.
(1314) No simple formula for the antiderivate of an absolute value function exists. You'll need to draw a picture and find the area geometrically.
(17) x^2+y^2=r^2 is the equation of a cirle of radius r centered at the origin. So, y=sqrt(r^2x^2) is the top half of that circle. (y=sqrt(r^2x^2) is the bottom half.)
(26) No quotient or product rule for integrals. Needs foiled upstairs and each resulting term divided by 2x from downstairs.
(28) No quotient rule for integrals. Each term needs divided by t^2 from downstairs.
Wed Sep 7
(11) If you perform polynomial long division, you'll get a sum of two terms. One is easy. The other uses inverse tan.
(13, 19) No quotient or product rules for integrals. Need to expand out product(s) and divide each resulting term by what's downstairs to get individual terms to integrate.
(15) Your answer will involve the letter b within it. When working with b, just pretend it's a good oldfashioned number.
(17) The fact sin(2x)=2sin(x)cos(x) is really useful here.
(20) r values in the table give function values that will be used to determine rectangle height.
Fri Sep 16
(7) u is often set to what's inside something else in your function. In this case, sine is inside cosine.
(8) When there are two expressions in a function that have degree differing by 1, u is almost always set to the higher power expression. Here, u is what's inside the parenthesis downstairs.
(9) See 8.
(16) Tricky. Letting u=x^2, this will become an inverse tangent integral.
(18) Two area regions demand two integrals. In general, as many integrals are required as there are area regions.
(19) First determine the number of area regions by finding the intersection points. In this case, you should get 2 area regions.
(2225) When dealing with functions of y, your interval of integration [a,b] needs to come from the yaxis.
(27) With 3 curves, the situation almost always is something like this.
Fri Sep 23
(5) Note that when you solve for y to be of the form sqrt(ax^2), this only accounts for the top half of the unit circle above the xaxis. You'll have to use symmetry of the circle and double this to find the distance from the top to the bottom of the circle.
(15) Integrating the acceleration function yields a sum of rectangles with units of (m/s^2)*(s)=(m/s). Hence, integrating acceleration yields velocity. Initial velocity equal to zero means the C from integrating acceleration will be zero.
(16,17) Do these after having done 1826. These are harder than 1826, and doing the less difficult problems first can warm you up to these types of problems (which are very important and will surely show up on your test.).
(28) Find the intersection point of the two curves. This will give you the x value you need to break the integral into two pieces.
Fri Sep 30
(68, 11, 13) Shifting the axis of revolution from the yaxis to an axis parallel to the yaxis does not change function heights...it only changes the radius of each cylindrical shell.
(9) A function of y revolved about the xaxis naturally lends itself to the shell method.
(12) In order to use shells when revolving about an axis parallel to the xaxis, you will need a function of y...which will demand representing the function in that form in this case.
(14) A function of x revolved about an axis parallel to the xaxis naturally lends itself to disks/washers.
(1525) See videos and commentary above on Work Problems.
(2425) Consider a small slice of the chain at some height x. The mass of this slice can be computed by multiplying the linear density by the thickness of the slice...namely dx. The distance the slice must be lifted is not x, but it is a function of x you can determine by thinking about how far that slice must travel vertically from start to finish.
Fri Oct 14
(4, 9, 11) The integral involves ln(x) inside another function. So, we need to set u equal to that function which has the ln(x) inside it, including the ln(x).
(7, 12) The choice of u is flexible and could be either of the two terms (other than dx) in the product expression of the function in the integral, but ultimately this integral will rely on algebraic manipulation of equations that result when integrating by parts.
(1315, 2335) With an odd power of sin/cos, you can use sin^2(x)+cos^2(x)=1 to reduce the least odd power to a power of 1. Then, you set u equal to the other trig function.
(16, 17, 22) With only even powers of sin/cos, you'll need to apply the double angle identities and simpliffy the functions expression in that new form to be able to integrate.
(19, 21) The integral of sec^3(x) can be evaluated using parts, letting u=sec(x) and dv=sec^2(x)*dx.
(20) If you can leave just a sec^2(x) where you see the sec^4(x), you'll be in a position to use u=tan(x).
Fri Oct 21
(7, 11) Similar to #4.
(10) Similar to #1, #2.
(12, 13) After completing the square, set u equal to the term in the completed square of the form x+c. Substitute so the original integral is written in the form of u/du, and then use trig substitution on that integral.
(14) Note there are two bits of your expression that are polynomials differing by a degree of 1...use substitution.
(15, 16) Powers of sinh/cosh can be handled analogously to power of sin/cos. Just, need to recall cosh^2(x)sinh^2(x)=1.
(1719) 19 is probably easiest of this bunch. Dot it before 17 and 18. If x is expressed as some constant multiple of cosh^2(x), you'll be able to use the fact cosh^2(x)1=sinh^2(x).
(20) When you have only even powers of cosh, use the fact cosh^2(x)=(1/2)+(1/2)*cosh(2x) just like you do with the analogous form for cosine in such a scenario.
(24, 25) Uses method of partial fraction decomposition, just like in #21, to get the function given into a form that can be integrated easily.
Fri Oct 28
(3) Improper fraction demands long division first.
(5) Careful...x^2+cx term is not fully factored. Need to fully factor first to be able to fully decompose.
(13) Need to break into a pair of integrals, say on (inf, 0] and [0, inf)...then use improper integration on both.
(17) Complete square on x^2+4x+10 to get (x+2)^2+6. Then use u=x+2 to get that factor into a more familiar form.
(18, 21, 23) Letting u equal what's inside the sqrt may help you see what's going on.
(22) Use substitution first.
(20, 24, 25) Be certain to account for the chain rule on the expressions inside the e, cos, and ln functions.
Sun Nov 20
(10) Could multiply by [sqrt(n+5)+sqrt(5)]/[sqrt(n+5)+sqrt(5)] to see limit value more clearly.
(11) The exponent of 1/n is awkward to deal with in the exponent. Take ln of both sides to bring the power down. You'll find the limit of ln(b_n). Then, use the exponential function to drop the ln from ln(b_n).
(12) Consider expanding out the expressions on top and on bottom when n is large, say like 100. Writing out enough terms may help you see the trend.
(13) Recall ln(A)ln(B) is equal to ln(A/B).
(14) Since limit evaluates to inf/inf, you can use L'Hopital.
(1922) Use partial fraction decomposition. Them write enough terms from the sum to begin to recognize the cancelling pattern.
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