Recent content by Spartan029

wow that's awesome. now i feel retarded haha. but seriosuly... thanks gabba!

\int_0^1 f'(x)dx = f(1) - f(0) ??

Homework Statement Let f be twice differentiable with f(0)=6, f(1)=5, f'(1)=2 Evaluate the integral \int_0^{1}x f''(x) dx Homework Equations \int uv' dx = uv = \int u'v dx The Attempt at a Solution u = x and v' = f''(x) so u' = 1dx and v = f'(x) so xf'(x) -...

geez this problem is pwning me lol. so we go... g(x)x^2 - \int_0^{10} 2xg(x)dx (2.3)(0) - 2(0)(2.3) = 0, for x=0 (3.1)(4) - 2(2)(3.1) = 0, for x=2 (4.1)(16) - 2(4)(4.1) = 54.4, for x=4 (5.5)(36) - 2(6)(5.5) = 132, for x=6 ...and so on... calculate area under (connected) points (0,0)...

oh wait how do i work in that g(x)x^{2} part

okay awesome! thanks for helping me out!

\int_0^{10} 2xg(x)dx when x= 0, 2xg(x) = 0 x=2, 2xg(x) = 2(2)(3.1) = 12.4 x=4, 2xg(x) = ... = 32.8 x=6, 2xg(x) = ... = 66 x=8, 2xg(x) = ... = 94.4 x=10, 2xg(x) = ... = 122 connect thesse and estimate area under from 0 to 10? makes sense, but is there any other way to solve the problem?

Homework Statement Estimate \int_{0}^{10} f(x) g'(x) dx for f(x) = x^{2} and g has the values in the following table. \begin{array}{l | c|c|c|c|c|c |} \hline \hline g&0&2&4&6&8&10\\ \hline g(x)&2.3&3.1&4.1&5.5&5.9&6.1\\ \hline \end{array}...

cool. thanks for helping me redbelly!

Thats where I am confused. is it 7 units south and 20 east, both starting at the flagpole (origin)? or 7 south from the 50 meter mark, and 20 east from the 70m mark? the wording on this problem is pwning me. lol

am i allowed to bump this? :)

I see the point 1 to point 2 thing now, arrow points northwest. the one i drew before posting here went in the opposite direction (northeast) my point 2 was 7 inches right above where my 5 inch (east) mark was, and i drew an arrow from the flagpole to that. oops. thanks for clarifying that...

I'm stuck at part b on this problem. I've been working at it for an hour at least! lol Homework Statement At one instant a bicyclist is 50 m due east of a park's flagpole, going due south with a speed of 7 m/s. Then, 24 s later, the cyclist is 70 m due north of the flagpole, going due...

lol, yea i messed up there, thanks for pointing it out. I figured out how to solve the problem thanks for helping me!

Hello everyone. I am stumped here with this problem, i feel like it should be fairly simple but i can't seem to figure it out. Homework Statement If the acceleration for a given object is given by the function: a(t) = +(3 m/s^3) · t (Note: units are included in the eqn, so if [t]=s...