Recent content by CaptainK

Finding the derivative of cos(x4+4) I got -4x3sin(x4+4) Do I then follow the formula above with f(x) and t and then solve for the Integral.

Chain Rule states dy/dx = dy/du * dx/du ∫8x3e-cos(x4+4)sin(x4+4) 8∫x3∫e-cos(x4+4)∫sin(x4+4) 8∫x4/4∫e-cos(x4+4)∫cos(x4+4)+4x3

so f(x) = cos(x4+4) then f'(x) = sin(x4+4) + 4x3 which gives ∫cos(x4+4) +4x3 *sin(x4+4) + C

So using substitution du = sin(x4+4) because cos converted to sin doesn't change its sign, I thought it did but its for converting from cos to sin. t = u ∫f'(g(x))g'(x) dx = ∫f'(t) dt/dx = ∫ f'(t)dt = f(t) + C = f(g(x)) + C when substituting back in f(x)...

So I've found du and put it into the form ∫udv = uv - ∫vdu so for v I have 8x3 du = -sin(4x3) dx dv = 24x2 dx Which gives me ∫cos(x4+4)24x2dx = cos(x4+4)8x3 - ∫8x3(-sin4x3) dx But I feel like I'm going down the wrong path, especially since the e isn't present

Homework Statement ∫8x3e-cos(x4+4)sin(x4+4)dx Homework Equations Let u = cos(x4+4) The Attempt at a Solution I know the answer does not have the sin in it and only the e remains, because when the integral is found e stays unchanged. I could find somewhere online to calculate it...

Alright I did some more work and this is what i have so far, but I'm not 100% sure that it is correct, can anyone help me out Probability of getting a straight flush for diamonds (or hand 2) 11C5 = 55440 because there are 1712304 possibilities of 5 community cards turning up...

Alright so i did some additional work to get some probability values and this is what i have done so far cards to help hand 1 win are A,2,3,4,5,6,7,8,9,10,J,Q 12/13 = 92% cards to help hand 2 win are A,2,3,4,5,6,7,8,9,10,J,Q,K all diamonds (5 diamonds turning up) 13/13 = 100%...

Homework Statement You will assess the probability of a certain hand winning after the starting hands are dealt. You only need to consider two players at a time. (At no point are we considering betting or folding playing a part in winning) Starting Points -Try and watch some games of Texas...

me again, started doing Q4, but am not entirely sure where to go next, can anyone point me in the right direction? so for x, volume about a solid is V= \int \pi y2 dy between x=b and x=a which gives - V = \int \pi x2 dx = \pi (x3/3) between x= b and x= a V = \pi(b3/3) - (a3/3)...

and i was working on doing it for between bn, an and b and a. \int xn dx = xn+1/n+1 between x=b and x=a and \int y1/n dy = y1/n + 1(1/n + 1) between y= bn and y= an for y, i got nbn+1/(n+1) - nan+1/(n+1) but for x i don't know where to go after bn+1/n - an+1/n any advice...

alright so an update on things that i have done and the ratio to go with it it should say \int xn dx = xn+1/n+1 between x=0 and x=1 =1/n+1 with \int y1/n dy = y1/n+1/(1/n +1) between y=0 and y=1 =n/n+1 which gives a ratio of 1/n+1:n/n+1 or 1:n

ok this is the work i have done so far y=x2 \int x2 dx = x3/3 betwen x=1 and x=0 gives dy/dx = 1/3 1-1/3 = 2/3 so the ratio between area A and B is 1/3 : 2/3 then again with this time with y=x3 \int x3 dx = x4/4 between x=1 and x=0 gives dy/dx = 1/4 1-1/4 = 3/4 so the ratio is...

Need to investgate the ratio of the areas formed when y=xn is graphed between two arbirary parameters x=a and x=b such that a<b 1. Given the funtion y=x2, consider the region formed by this function from x=0 to x=1 and the x-axis. Label this area B. Label the region form y=0 to y=1 and the...