f(x) = (x³+1)/((sinx)^1/2)) ~ g(x) = 1/(x^1/2) near 0 because sinx/x = 1 near 0 (and x³+1/1 too).
Since 1/(x^1/2) is convergent near 0 then f(x) also is.
Is that right?
How do I know whether this is convergent or divergent:
Integral of (x³+1)/((sinx)^1/2) dx between 0 and pi/2
I know that this integral is convergent if Lim n->0 of Integral of (x³+1)/((sinx)^1/2)) dx between n and pi/2 exists and is not infinite (why is that?). Otherwise its divergent...
Tks for that... but what i really wanted to know is why F'(x) = B'(x)A'(B(x))... IOW a proof of the chain rule. I guess was naturally inclined to think that F'(x) would equal A'(B(x)).
Ok, i have this problem:
Calculate: Limit x->0 of ((integral of cos t² dt between 0 and x²) / (integral of e^(-t²) dt))
So, the limit of both integrals is 0 since the interval between both integrals tends to 0. I used L'Hopital then, so:
Limit x->0 F'(x)/G'(x) = (cos (x^4) *...
Ok, tks...
But why does this integral calculates this volume?
I mean... you took those integrals i made and multiplied each for its f(x) and for pi. Why does that make the volume?
Again sorry about any wrong english.
Hello, I'm new around here. I was having trouble with a problem, i thought i could look for help on the net. Anyway here's the problem:
Calculate the volume of a solid obtained by the rotation around Ox of all points (x,y) in RxR where y >= x*x, y <= square root of x and y <= 1/(8x)...