Definite Integral by Definition

In summary: And I was trying to get some discussion going.)In summary, the problem is to find the area, A, of the region under the graph of f(x) = 2x^2 + 5 between x = 0 and x = 4 by using n rectangles and then evaluating the expression. The book answer is 188/3, with h = 4/n. The attempt at a solution involved difficulty simplifying the limit after factoring out the h. It is suggested to convert the limit to "n" and evaluate as n approaches infinity.
  • #1
OmniNewton
105
5

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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  • #2
##f(a+h)## is not ##5+h## but ##2(a+h)^2+5## !?

Aren't you supposed to find ##A(n)## ?
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
Well, the book answer helps you out, Ray :wink:
 
  • #7
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.
 
  • #8
Agree.
But it looks as if Newt is out of the picture ?
 

FAQ: Definite Integral by Definition

What is a definite integral by definition?

A definite integral by definition is a mathematical concept used to find the exact area under a curve. It is represented by ∫ and has a lower and upper limit, also known as the limits of integration.

How is a definite integral by definition calculated?

To calculate a definite integral by definition, you must first divide the area under the curve into smaller rectangles or trapezoids. Then, you find the area of each shape and add them together to get an approximation of the total area. As the number of shapes increases, the approximation gets closer to the exact value.

What is the difference between a definite integral by definition and a definite integral by the Fundamental Theorem of Calculus?

A definite integral by definition involves using geometric shapes to approximate the area under a curve, while a definite integral by the Fundamental Theorem of Calculus involves using the antiderivative of a function to find the exact area. The latter is typically faster and more accurate, but it can only be used for certain functions.

What is the significance of the limits of integration in a definite integral by definition?

The limits of integration determine the range over which the integral is being calculated. They specify the starting and ending points on the x-axis, which ultimately affects the value of the integral. Changing the limits of integration can result in a different area under the curve.

Can a definite integral by definition be negative?

Yes, a definite integral by definition can be negative. This occurs when the curve dips below the x-axis, resulting in a negative area. The negative value simply indicates that the area under the curve in that region is being subtracted from the total area.

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