Limit of an integral function

In this video, I showed how to find the limit of an integral function using FTC

[devotion]

L'Hospital isn't necessary here. It is much easier to apply the Mean value theorem here. This gives you sqrt(1+x^3) for an x between 2 and 2+h, and then let h go to Zero and this also gives you the result.

[skleon]

Define F(t) such that F'(t) = f(t) = sqrt(1+t^3)
L = lim_{h -> 0} (1/h) \int_2^{2+h) f(t) dt = lim_{h -> 0} (1/h) (F(2+h) - F(2))
By the definition of the derivative
L = F'(2) = f(2) = sqrt(1+2^3) = sqrt(9) = 3

[catherinegrimes]

I did this in my head. For an infinitesimal h the area given by the integral would by SQRT(1 + t^3)h that is evaluated at t = 2. This gives 3h and when divided by the denominator of h the answer is 3.
Have you thought about calculating the area of a circle of radius r by setting y = SQRT(r^2 - x^2)?
The area of the circle is 4 times the INTEGRAL of y dx with limits 0 to r.
You expand SQRT(r^2 - x^2) using the binomial theorem to produce a series.
Integrate all terms in the series and factor the r^2 term out of the series.
Sum the series using Excel or something similar and this will give PI/4.
Multiply PI/4 by the 4 outside of the integral and the final answer is PI r^2
I worked this out years ago and have never seen it anywhere. It shows an example of simple integration, use of the binomial theorem, how PI can be represented as a series and how it can be calculated using software like Excel.
I think that it would make a good example in a video.

[JourneyThroughMath]

When I saw the way the problem was written, I knew FTC would come in play, but I missed using L'Hopitals rule. I think its the early morning

[kilianklaiber]

You merely need to use the fundamental theorem of calculus. The limit of said integral is just the derivative of the integral function at t is equal to two.

[johnnolen]

Is it necessary to invoke L' Hopital here? The limit as written appears to be the definition of F '(2), which by the FTC is sqrt[1 + 2^3] = sqrt[9] = 3 ◼

[nothingbutmathproofs]

Suppose the int (sqrt(1+t^3))dt = F(t). Then you have (F(2+h)-F(2))/h as h goes to 0. That answer if F'(2), where F'(t) = sqrt(1+t^3). So F'(2) = 3

[leonard]

I did it a bit simpler (in my opinion). I just defined the function under the integral as f'(t) = sqrt(1 + t³) (you'll see why I defined it as f' and not just f)
So if you solve the Integral, only the Integral is equal to f(2 + h) - f(2). If you then multiply it with 1/h it's (f(2 + h) - f(2)) / h. This is just the differential quotient. Thus the solution to this is just f'(2), so you just have to plug in a 2 into the previously defined function, so f'(2) = sqrt(1 + 2³) = sqrt(9) = 3.

[xgx]

This is the derivative F'(2) of the function F(x)=int_0^x f(t) dt with f(t)=sqrt(1+t^2). By Newton-Leibnitz theorem, F'(2)=f(2)=3.
Another short solution is via the mean value theorem (see comment below).
Finally, any physicist or engineer will simply observe that the integral is h*sqrt(1+2^2)+O(h^2).

Prime Newtons specializes on trivial results presented with unnecessary verbiage using the most clumsy techniques possible. Invoking L'Hospital's rule in this example is pure incompetence.

[anglaismoyen]

Kind of weird to think you can't do 1/h times the integral because it's infinity times zero, but putting the integral as the numerator over h is allowed even though it's the same thing. I get why, it's just interesting.

[muadzmubarak]

Can you explain to me why the derivative of (2+h) can appear

[lukaskamin]

Sorry what do FTC 1 and 2 stand for. I wasn't studying this(another country). BTw is it the same across all the English speaking countries or it is limited to USA(or wherever you reside?)?

[davefried]

wouldn’t the answer be +3 and -3 since the sqrt of 9 has two answers?