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## Main Question or Discussion Point

The definite integral of a function ##f(x)## from ##a## to ##b## as the limit of a sum is:

$$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h(f(a)+f(a+h)+.. ..+f(a+(n-2)h)+f(a+(n-1)h))$$

where ##h=\frac{b-a}{n}##. So, replacing ##h## with ##\frac{b-a}{n}## gives:

$$\lim_{n\rightarrow \infty}(b-a)\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+.. ..+f\left(a+(n-2)\frac{b-a}{n}\right)+f\left(a+(n-1)\frac{b-a}{n}\right)}{n}$$

This seems to be of the form ##\frac{\infty}{\infty}## because the numerator is an infinite sum and the denominator also tends to infinity, so I think L'Hospital's rule can be applied. So, applying L'Hospital's rule gives:

$$\int_a^bf(x)dx=\lim_{n\rightarrow \infty}\frac{(b-a)^2}{n^2}\left(-f'\left(a+\frac{b-a}{n}\right)-2f'\left(a+2\frac{(b-a)}{n}\right)-3f'\left(a+3\frac{(b-a)}{n}\right)-.....+3f'\left(a+(n-3)\frac{(b-a)}{n}\right)+2f'\left(a+(n-2)\frac{(b-a)}{n}\right)+f'\left(a+(n-1)\frac{(b-a)}{n}\right)\right)$$

Replacing ##\frac{(b-a)}{n}## with ##h## gives:

$$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h^2(-f'(a+h)-2f'(a+2h)-3f'(a+3h)-.....+3f'(a+(n-3)h)+2f'(a+(n-2)h)+f'(a+(n-1)h))$$

Replacing ##f'(x)## with ##f(x)## and hence ##f(x)## with ##\int f(x)dx## gives:

$$\int_a^b\left(\int f(x)dx\right)dx=\lim_{h\rightarrow 0}h^2(-f(a+h)-2f(a+2h)-3f(a+3h)-.....+3f(a+(n-3)h)+2f(a+(n-2)h)+f(a+(n-1)h))$$

Does the above series make any sense to you? I'm asking this because the starting terms of the series are negative and the first term is multiplied by 1, the second term by 2, the third term by 3, etc,i.e the multipliers are increasing by 1, but the last terms of the series are positive and the multipliers are decreasing by 1 instead as we move forward in the series. Is it something like some transition from negative terms to positive terms takes place in the middle or something?

Or Is applying L'Hospital's rule not allowed in this case?

$$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h(f(a)+f(a+h)+.. ..+f(a+(n-2)h)+f(a+(n-1)h))$$

where ##h=\frac{b-a}{n}##. So, replacing ##h## with ##\frac{b-a}{n}## gives:

$$\lim_{n\rightarrow \infty}(b-a)\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+.. ..+f\left(a+(n-2)\frac{b-a}{n}\right)+f\left(a+(n-1)\frac{b-a}{n}\right)}{n}$$

This seems to be of the form ##\frac{\infty}{\infty}## because the numerator is an infinite sum and the denominator also tends to infinity, so I think L'Hospital's rule can be applied. So, applying L'Hospital's rule gives:

$$\int_a^bf(x)dx=\lim_{n\rightarrow \infty}\frac{(b-a)^2}{n^2}\left(-f'\left(a+\frac{b-a}{n}\right)-2f'\left(a+2\frac{(b-a)}{n}\right)-3f'\left(a+3\frac{(b-a)}{n}\right)-.....+3f'\left(a+(n-3)\frac{(b-a)}{n}\right)+2f'\left(a+(n-2)\frac{(b-a)}{n}\right)+f'\left(a+(n-1)\frac{(b-a)}{n}\right)\right)$$

Replacing ##\frac{(b-a)}{n}## with ##h## gives:

$$\int_a^bf(x)dx=\lim_{h\rightarrow 0}h^2(-f'(a+h)-2f'(a+2h)-3f'(a+3h)-.....+3f'(a+(n-3)h)+2f'(a+(n-2)h)+f'(a+(n-1)h))$$

Replacing ##f'(x)## with ##f(x)## and hence ##f(x)## with ##\int f(x)dx## gives:

$$\int_a^b\left(\int f(x)dx\right)dx=\lim_{h\rightarrow 0}h^2(-f(a+h)-2f(a+2h)-3f(a+3h)-.....+3f(a+(n-3)h)+2f(a+(n-2)h)+f(a+(n-1)h))$$

Does the above series make any sense to you? I'm asking this because the starting terms of the series are negative and the first term is multiplied by 1, the second term by 2, the third term by 3, etc,i.e the multipliers are increasing by 1, but the last terms of the series are positive and the multipliers are decreasing by 1 instead as we move forward in the series. Is it something like some transition from negative terms to positive terms takes place in the middle or something?

Or Is applying L'Hospital's rule not allowed in this case?

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