Understanding the Limit of a Sum in the Integral of f(x)=x

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In summary, the conversation discusses how to solve for f(x) when given that its input is a function from a to b. The first step is to find the limit as h approaches infinity of f(a+i(b-a))/h. This is then simplified to f(a+i(b-a))/h+i. Next, the limit on f(a+i(b-a))/h+i is found as a function of h and i. Finally, the RHS is shown to be (b+a)/2.
  • #1
ManishR
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[tex]\intop_{a}^{b}f(x)dx=\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}f\left(a+i\frac{(b-a)}{h}\right)\frac{(b-a)}{h}[/tex]

if f(x) = x

[tex]\intop_{a}^{b}f(x)dx=(b-a)\left[\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(a+i\frac{(b-a)}{h})\frac{1}{h}\right][/tex]

[tex]\Rightarrow\intop_{a}^{b}f(x)dx=(b-a)\left[\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(\frac{a}{h}+i\frac{(b-a)}{h^{2}})\right][/tex]

[tex]\Rightarrow\intop_{a}^{b}f(x)dx=(b-a)\left[\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(\frac{a}{h}+i\frac{(b-a)}{h^{2}})\right][/tex]

[tex]\Rightarrow\intop_{a}^{b}f(x)dx=(b-a)\left[a+\left(\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(i\frac{(b-a)}{h^{2}})\right)\right][/tex]

[tex]\Rightarrow\intop_{a}^{b}f(x)dx=(b-a)\left[a+(b-a)\left(\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(\frac{i}{h^{2}})\right)\right][/tex]

let

[tex]\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}(\frac{i}{h^{2}})=k[/tex]

[tex]\Rightarrow\intop_{a}^{b}f(x)dx=(b-a)\left[a+(b-a)k\right][/tex]

if

[tex]\intop_{a}^{b}f(x)dx=b^{2}-a^{2}[/tex]

[tex]\Rightarrow b^{2}-a^{2}=(b-a)\left[a+(b-a)k\right][/tex]

[tex]\Rightarrow k=\frac{b}{b-a}[/tex]

but k cannot be function of anything.

so what's wrong here ?
 
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  • #2
Hi ManishR! :smile:

i] ∑ i/h2 = (1/h2) ∑ i

ii] ∫ab x dx is not b2 - a2 :wink:
 
  • #3
You goofed going from the 4th to 5th line when you pulled the "a" out of the limit. You have to equalize the denominators, collect on i and solve the summations. Then you'll get the required 1/2 (b2-a2).
 
  • #4
tiny-tim said:
ii] ∫ab x dx is not b2 - a2 :wink:

[tex]\int_{a}^{b}f(x)dx=\frac{1}{2}(b^{2}-a^{2})[/tex]

[tex]\Rightarrow\frac{1}{2}(b^{2}-a^{2})=(b-a)[a+(b-a)k][/tex]

[tex]\Rightarrow k=\frac{1}{2}[/tex]

ok everything is alright
thanks tiny-tim
 
Last edited:
  • #5
dimitri151 said:
You goofed going from the 4th to 5th line when you pulled the "a" out of the limit. You have to equalize the denominators, collect on i and solve the summations. Then you'll get the required 1/2 (b2-a2).

i don't see any problem in 5th step. would you care to elaborate.
 
  • #6
It's technically not incorrect. It just looked like that's why you got confused. You should have finished your algebra and then did your summations. How could you know the sum on 1 and not on i?
 
  • #7
dimitri151 said:
It's technically not incorrect. It just looked like that's why you got confused.

again how ?

are you saying

[tex]\sum_{i=0}^{h}(f(h)+g(i))=h.f(h)+\sum_{i=0}^{h}g(i)[/tex]

is wrong ?
 
  • #8
ManishR said:
again how ?

are you saying

[tex]\sum_{i=0}^{h}(f(h)+g(i))=h.f(h)+\sum_{i=0}^{h}g(i)[/tex]

is wrong ?

It IS wrong.

Should be (h+1) on the right.
 
  • #9
l'Hôpital said:
It IS wrong.

Should be (h+1) on the right.

sorry i forgot to account for 0th term,

RHS = (h+1)f(h) + ,,
 
  • #10
[tex]\int_{a}^{b}xdx=(b-a)\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}\frac{f(a-i\frac{(b-a)}{h})}{h}[/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+(b-a)\left(\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}\frac{i}{h^{2}}\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+(b-a)\left(\underset{h\rightarrow\infty}{lim}\frac{1}{h^{2}}\sum_{0}^{h}i\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+(b-a)\left(\underset{h\rightarrow\infty}{lim}\frac{1}{h^{2}}\frac{h(h+1)}{2}\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+\frac{(b-a)}{2}\left(\underset{h\rightarrow\infty}{lim}\frac{h+1}{h}\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+\frac{(b-a)}{2}\left(\underset{h\rightarrow\infty}{lim}\frac{1+\frac{1}{h}}{1}\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+\frac{(b-a)}{2}\left(1+\underset{h\rightarrow\infty}{lim}\frac{1}{h}\right)\right][/tex]

[tex]\int_{a}^{b}xdx=(b-a)\left[a+\frac{(b-a)}{2}\right][/tex]

[tex]\int_{a}^{b}xdx=\frac{\left(b^{2}-a^{2}\right)}{2}[/tex]
 
  • #11
Perfect! :smile:

(nice LaTeX too, btw)

(except personally, i'd have gone straight from lim h(h+1)/2h2 to 1/2, and I'd have put in an extra line just before the end, with (b+a)/2 :wink:)
 

1. What is the problem with the integral of f(x)=x?

The problem with the integral of f(x)=x is that it does not have a definite solution. This is because the function f(x)=x does not have a defined antiderivative. As a result, the integral cannot be evaluated using traditional integration techniques.

2. Can the integral of f(x)=x be solved using other methods?

Yes, the integral of f(x)=x can be solved using numerical methods such as the trapezoidal rule or Simpson's rule. These methods approximate the integral by dividing the area under the curve into smaller, simpler shapes and summing their areas. While this method may not give an exact solution, it can provide a close approximation.

3. What is the significance of the problem in the integral of f(x)=x?

The problem in the integral of f(x)=x highlights the limitations of traditional integration techniques and the importance of numerical methods in solving complex integrals. It also serves as a reminder that not all functions have a defined antiderivative, and some integrals may not have a definite solution.

4. Are there any real-world applications of the integral of f(x)=x?

Yes, the integral of f(x)=x has many real-world applications in physics, engineering, and economics. For example, it can be used to calculate the displacement of an object with a changing velocity, the area under a velocity-time graph, or the total cost of a product with a changing price.

5. Can the integral of f(x)=x be solved using calculus?

No, the integral of f(x)=x cannot be solved using traditional methods of calculus, such as integration by substitution or integration by parts. This is because these methods rely on the existence of an antiderivative, which does not exist for the function f(x)=x.

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