Definite Integral by Definition

OmniNewton
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Homework Statement


Let A be the area of the region that lies under the graph of f(x) = 2x 2 + 5 between x = 0 and x = 4. Find an expression for A using n rectangles. Then evaluate this expression.

Homework Equations


Answer is 188/3
h= (4/n)

The Attempt at a Solution


2943aeaecd3d29cb25c8faaa4fe7e4d4.png
[/B]
The problem I am having is I do not know how to simplify this limit further after factoring out the h.
 
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##f(a+h)## is not ##5+h## but ##2(a+h)^2+5## !?

Aren't you supposed to find ##A(n)## ?
 
BvU said:
##f(a+h)## is not ##5+h## but ##2(a+h)^2+5## !?

Aren't you supposed to find ##A(n)## ?

You are right about 2(a+h)^2+5. So you would suggest to convert this limit all to "n" and evaluate the limit as n ---> infinity?
 
If you have faithfully rendered the problem statement, the answer is an expression. Something like ##\displaystyle A = \sum_{i=1}^n \ ... \ ##

The way I read it you want to work out e.g. the middle riemann sum

300px-MidRiemann2.svg.png
 
OmniNewton said:

Homework Statement


Let A be the area of the region that lies under the graph of f(x) = 2x 2 + 5 between x = 0 and x = 4. Find an expression for A using n rectangles. Then evaluate this expression.

Homework Equations


Answer is 188/3
h= (4/n)

The Attempt at a Solution


2943aeaecd3d29cb25c8faaa4fe7e4d4.png
[/B]
The problem I am having is I do not know how to simplify this limit further after factoring out the h.

I cannot read your formula for f(x). Is it ##f(x) = 2x^2 + 5##? ##f(x) = (2x)^2 + 5##? Something else? If it really does involve the second power (and if you don't want to use either LaTeX or the formula template in the top green panel, then you need to use "^" instead, as in 2 x^2 + 5 or (2x)^2 + 5.
 
Well, the book answer helps you out, Ray :wink:
 
BvU said:
Well, the book answer helps you out, Ray :wink:

Well, of course I can guess what the OP means, but the whole point is that I should not need to. He/she needs to adhere to the standard notation; that's why it was invented.
 
Agree.
But it looks as if Newt is out of the picture ?
 

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