MHB Riemann Sum Definite Integral Question

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To evaluate the integral of (x-2) from -2 to 2 using the definition of a definite integral, one can choose n equally spaced subintervals and use the right endpoints as sample points. Each rectangle's width is determined by w = 4/n, and the height of each rectangle can be calculated by substituting the right endpoint into the function (x-2). By calculating the sum of the areas of these rectangles and taking the limit as n approaches infinity, the definite integral can be evaluated. This method allows for the approximation of the area under the curve without needing a specific value for n. Ultimately, this approach leads to the correct evaluation of the integral.
ISITIEIW
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So the question is Evaluate (x-2)dx as the integral goes from -2 to 2 using the definition of a definite integral, choosing your sample points to be the right endpoints of the subintervals…

Ok, so i understand how to do this problem if it gave me an actual number of interval like n=6 but it doesn't and I'm not sure how to go about actually solving it without an actual number for n.

Thanks :)
 
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ISITIEIW said:
So the question is Evaluate (x-2)dx as the integral goes from -2 to 2 using the definition of a definite integral, choosing your sample points to be the right endpoints of the subintervals…

Ok, so i understand how to do this problem if it gave me an actual number of interval like n=6 but it doesn't and I'm not sure how to go about actually solving it without an actual number for n.

Thanks

Welcome to MHB, ISITIEIW! :)

I suggest you pick n equally spaced narrow rectangles numbered i = 1, ..., n.
To visualize it, you can start with n=6.
Then each rectangle will have width w=4/n.
For each rectangle we can pick an arbitrary coordinate $x_i$ between its left side and its right side. Say we pick the center, what would $x_i$ be then?
Substitute that $x_i$ in (x-2) and you get the height of each rectangle.

Can you calculate the sum of the areas of the rectangles?
And then let n go to infinity?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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