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.
You may also use the squeeze theorem: Replace sqrt(1+t^3) by sqrt(1+2^3) and sqrt(1+(2+h)^3); the value of sqrt(1+t^3) lies between these two values. Limits of the "new" functions can be calculated easily because a constant is inside the integral. Because the two "new" functions have the same limit (3), and the value of the originial function always is in between the two "new" ones, the limit of the function "in the middle" must also be 3.
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 ◼
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.
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.
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.
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?)?
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.
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.
@@davefried If we can choose √9 to be either −3 or 3, then we can also choose √9 + √9 to be 0. But that implies that √9 = 0, which is definitely not true. It's just much easier to say √9 = 3 and −√9 = −3.
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.
You may also use the squeeze theorem: Replace sqrt(1+t^3) by sqrt(1+2^3) and sqrt(1+(2+h)^3); the value of sqrt(1+t^3) lies between these two values. Limits of the "new" functions can be calculated easily because a constant is inside the integral. Because the two "new" functions have the same limit (3), and the value of the originial function always is in between the two "new" ones, the limit of the function "in the middle" must also be 3.
Definition of differential!!
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 ◼
Not everyone will see that. Good observation.
Yes that was my thougths. The límit will be the derivative by definition. Thats means the result will be f(2) thats |3|
Great explanation sir, thank you.
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
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.
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.
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
Brilliant 🎉😊
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.
Nice 😊👍
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
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?)?
Me encanta las matemáticas
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.
Can you explain to me why the derivative of (2+h) can appear
He wrote int f(t) as F(2+h)-F(2), what do you mean by it can appear?
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.
wouldn’t the answer be +3 and -3 since the sqrt of 9 has two answers?
9 has two square roots, ±√9 = ±3
But in this case we are only dealing with the positive square root, √9 = 3
So, 3 is the only answer.
@@jumpman8282why
@@davefried If we can choose √9 to be either −3 or 3, then we can also choose √9 + √9 to be 0.
But that implies that √9 = 0, which is definitely not true.
It's just much easier to say √9 = 3 and −√9 = −3.
@@davefriedlimits cannot have multiple answers so you take the higher one, this can be proven through more calculations or take it as is.
Detta F la primitiva abbiamo (F(2+h)-F(2))/) /h che per h tendente a zero è F'(2) ossia sqrt(1+2^3)
asnwer=1/2 isit
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