# Calculus II - notes 9/11 # # Ostobee Zorn # # 11. page 47 # Evaluate limit 2/n *sum((2*j/n)^3,j=1..n) as n goes to infinity. > s := n -> 2/n *sum((2*j/n)^3,j=1..n); > n ----- / 3 \ \ | j | ) |8 ----| / | 3 | ----- \ n / j = 1 s := n -> 2 -------------- n > seq(s(3*10^i),i=1..7); 961 90601 9006001 900060001 90000600001 9000006000001 ---, -----, -------, ---------, -----------, -------------, 225 22500 2250000 225000000 22500000000 2250000000000 900000060000001 --------------- 225000000000000 # By inspection it appears that the limit of this sequence will be # 900/225=4. # We can check that by noticing that the nth term is the right sum Rn # approximating the integral of x^3 over the interval 0 to 2. So the # limit is the # area under x^3 from 0 to 2. By FTC, this is x^4/4 evaluated at 2 # or 16/4=4. # # # speed versus velocity. It is important to distinguish between # speed and velocity. # The velocity of an object moving along a coordinatized line is a # number, whose sign # (positive or negative) tells which direction the object is moving on # the line and whose # absolute value tells the speed of the object. If we know the # velocity of an object # at all times and we know its position at one time, then we can get its # position at all # times by integration. We can also calculate the distance it # traveled in any given time # interval and find its average speed and average velocity in the time # interval. # # Problem Given the table of velocities and positions of a particle # moving on the x-axis # below, estimate the position at each time in the table (Use a left # Riemann sum). Also # estimate the distance travelled in the time interval 0..6 and the # average speed and average # velocity of the particle in that time interval. Draw approximate # graphs of the velocity, # position and speed functions over the time interval 0..6. # # t 0 1 3 4 6 # dx/dt 10 5 -5 -1 # x 7 # # Area between curves. # # Ostobee-Zorn # 15, page 56 Find the area bounded by y = 2^x, y = 5^x, x = -1 , x=1 > plots[implicitplot]({y=2^x, y=5^x, x=-1,x=1},x=-10..10,y=-10..10); > simplify(int(abs(2^x-5^x),x=-1..1)); 5 ln(5) - 32 ln(2) - 1/10 ------------------ ln(5) ln(2) > evalf("); 1.266924271 # Methods of finding antiderivatives: # # The fastest method of finding antiderivatives is inspection, when it # works. Just look at the # function f and write down a function F whose derivative is f. You # get better at this as you # practice differentiating functions. # # # When inspection fails, try a substitution to simplify the function we # are trying to antidifferentiate. # So to find an antiderivative for f(x), and it is seen that f(x) = # g(u(x))*u'(x), we can let # u = u(x), and the problem Int(f(x),x , reduces to the problem # Int(g(u(x))*`u'`(x),x=Int(g(u),u) # As you can see, substitution is based on the chain rule. If G' = g # and f(x) = g(u(x))*u'(x), # then F(x) = G(u(x)) is an antiderivative of f, since (by the chain # rule) F'(x) = G'(u(x))*u'(x) # and G'(u(x))*u'(x)= g(u(x)) *u'(x) = f(x). # # You can use the word changevar in the student package in Maple # perform the substitutions, # but I recommend using this sparingly at first. You need to have a # facility for doing this # with pencil and paper. # # Problem: Integrate f(x) = x^2 * sin(x^3). > f := x^2*sin(x^3); 2 3 f := x sin(x ) > Int(f ,x); / | 2 3 | x sin(x ) dx | / > > with(student); [D, Diff, Doubleint, Int, Limit, Lineint, Product, Sum, Tripleint, changevar, combine, completesquare, distance, equate, extrema, integrand, intercept, intparts, isolate, leftbox, leftsum, makeproc, maximize, middlebox, middlesum, midpoint, minimize, powsubs, rightbox, rightsum, showtangent, simpson, slope, trapezoid, value] # We could try letting u be x^3 > changevar(x^3=u,Int(f,x)); / | | 1/3 sin(u) du | / # Now I can do this by inspection, using my knowledge that the # derivative of -cos is sin, # to get -1/3 cos(u) = -1/3 (cos(x^3)) + constant. # # You can use Maple to check your pencil and paper work. I recommend # using this # unsparingly. > int(f,x); 3 - 1/3 cos(x ) >