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

[ManishR]

\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}

if f(x) = x

\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]

\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]

\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]

\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]

\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]

let

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

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

if

\intop_{a}^{b}f(x)dx=b^{2}-a^{2}

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

\Rightarrow k=\frac{b}{b-a}

but k cannot be function of anything.

so what's wrong here ?

 

[tiny-tim]

Hi ManishR!

i] ∑ i/h² = (1/h²) ∑ i

ii] ∫_a^b x dx is not b² - a²

 

[dimitri151]

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).

 

[ManishR]

ii] ∫_a^b x dx is not b² - a²

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

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

\Rightarrow k=\frac{1}{2}

ok everything is alright
thanks tiny-tim

 

[ManishR]

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.

 

[dimitri151]

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?

 

[ManishR]

It's technically not incorrect. It just looked like that's why you got confused.

again how ?

are you saying

\sum_{i=0}^{h}(f(h)+g(i))=h.f(h)+\sum_{i=0}^{h}g(i)

is wrong ?

 

[l'Hôpital]

again how ?

are you saying

\sum_{i=0}^{h}(f(h)+g(i))=h.f(h)+\sum_{i=0}^{h}g(i)

is wrong ?

It IS wrong.

Should be (h+1) on the right.

 

[ManishR]

It IS wrong.

Should be (h+1) on the right.

sorry i forgot to account for 0th term,

RHS = (h+1)f(h) + ,,

 

[ManishR]

\int_{a}^{b}xdx=(b-a)\underset{h\rightarrow\infty}{lim}\sum_{0}^{h}\frac{f(a-i\frac{(b-a)}{h})}{h}

\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]

\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]

\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]

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

\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]

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

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

\int_{a}^{b}xdx=\frac{\left(b^{2}-a^{2}\right)}{2}

 

[tiny-tim]

Perfect!

(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 )