Basic Q on definite integrals and areas under the curve

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SUMMARY

The discussion centers on approximating the area under the curve of the function f(x)=x² over the interval [0,b] using Riemann sums. The user seeks clarification on determining the heights of rectangles when dividing the interval into n equal subintervals. It is established that for upper sums, the heights of the rectangles can be calculated using the function values at the right endpoints, specifically f(xi)=(ib/n)² for i=1 to n. The conversation emphasizes that using maximum, minimum, or average heights will yield similar results in the limit as n approaches infinity.

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  • Understanding of Riemann sums
  • Familiarity with the function f(x)=x²
  • Knowledge of limits in calculus
  • Basic concepts of integration
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  • Learn about the Fundamental Theorem of Calculus
  • Explore the differences between upper and lower sums
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Students studying calculus, particularly those learning about integration and Riemann sums, as well as self-learners seeking to deepen their understanding of area approximation techniques.

JOhnJDC
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Homework Statement


Consider the function f(x)=x2 on the interval [0,b]. Let n be a large positive integer equal to the number of rectangles that we will use to approximate the area under the curve f(x)=x2. If we divide the interval [0,b] by n equal subintervals by means of n-1 equally spaced points, then we have point x1=b/n, x2=2b/n, ..., xn-1=(n-1)b/n.

From this, I understand that the base of each rectangle will be b/n. What I'm not clear on is how to determine the heights of each rectangle. My book says that if we use the upper sums (rectangles that reach just above the curve), then the heights are f(x1)=(b/n)2, f(x2)=(2b/n)2, ..., f(xn)=(nb/n)2.

Can someone please explain? I can't consult a professor or other students because I'm studying calculus on my own. Thanks.
 
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Hi JOhnJDC! :smile:

You're approximating, so you can take the maximum height, the minimum height, or an average height.

If you take the minimum height, you use the height at n, if you take the maximum height, you use the height at n+1. :wink:

(and in the limit, of course, it makes no difference anyway)
 
Your parabola is increasing on [0,b]. So if you look at the two vertical sides of any rectangle, the side on the right will hit x2 at the highest point. So the rectangle based on [(i-1)/n,i/n] will be tallest if you use the right end on the function: (i/n)2 for its height, and the shortest if you use the left end:((i-1)/n)2.
 
Thank you both. I see it now.
 

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