Recent content by ARaslan

Remember that the integral sign is a glorified plus sign. ds is an infinitesimal arc length. So when you take the integral of ds, you are adding up all the tiny infinitesimal lengths. the integral of ds without limits is pointless, because if you wanted to restate that in words, it would go like...

Well technically it does not; however a line integral is defined as an integral which is taken around a curve. By definition line integrals must be calculated along a certain path, so it does not make sense to give coordinates in the integral itself.

Yes, those limits given are the exact same limits given by the line segments. My guess is that those limits are given for problem b. For problems A and C, ignore the given limits and focus on the limits given by the line segment

Yes you are right, it does not make sense to say that there's a discontinuity at infinity. I just meant that we can not treat as a normal limit of integration and we are going to have to take the limit as that limit goes to infinity. Thank you from pointing that out. And yup, i must gave...

∫(3x+2)(2x+1)^0.5dx let f'(x) = (2x + 1)^0.5 ==> f(x) = ((2x + 1)^1.5)/ 3 g(x) = 2x + 2 ==> g'(x) = 2 = (2x + 2)((2x + 1)^1.5) / 3 - 2/3∫(2x + 1)^1.5 dx = (2x + 2)((2x + 1)^1.5) / 3 - 1/3(2x + 1)^2.5 + c = 1/3((2x + 2)(2x+1)^1.5 - (2x + 1)^ 2.5) + c

The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.

There are two discontinuities here: x = 0 and x = infinity so rewrite as lim a→0 (lim b→∞ (∫dx/(√x e*√x) from a to b)) u = √x 2du = dx/√x lim a→0 (lim b→∞ (∫e^-u du from a to b)) = lim a→0 (lim b→∞ -e^-b + e^-a) = 0 +1 = 1

a) A vector parallel to the line segment from (0,1) to (1,2) is <1,1>. Therefore the vector equation for the line segment is r(t) = <0,1> + t<1,1> = <t, t +1>. So the parameter of the line integral is : x =t y = t+1 0 <= t <= 1. dy/dt = 1 dx/dt = 1 so the integral becomes: ∫(t^2 - t - 1) +...