Riemann Sum from Indefinite Integral

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SUMMARY

The discussion focuses on calculating the Riemann Sum for the integral \(\int _3 ^7 \left(\frac{3}{x} + 2\right) dx\) using both right and left endpoints with \(n=4\). The correct approach involves dividing the interval from \(x=3\) to \(x=7\) into four equal parts, resulting in the right endpoint sum being calculated as \((\frac{4}{4})[f(3) + f(4) + f(5) + f(6)]\) and the left endpoint sum as \((\frac{4}{4})[f(7) + f(4) + f(5) + f(6)]\). The user initially miscalculated the areas of the rectangles, leading to confusion about the correct Riemann Sum values.

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Homework Statement


Consider the integral,

\int _3 ^7 (\frac{3}{x} + 2) dx

a) Find the Riemann Sum for this integral using right endpoints and n=4.

b) Find the Riemann Sum for this integral using left endpoints and n=4.

Homework Equations


The sum,
\sum^{n = 4} (\frac{3}{x} + 2)

The graph,
b0de2dab.jpg


f(x) (white)
x = 3 (blue)
x = 7 (blue)
y = 3 (yellow)
y = f(7) (red)

The Attempt at a Solution



The right endpoint would be (7, f(7))
The product of the sum would be the area of the box, so
f(7) * 4
(this answer is in an acceptable format for my teacher)

The left endpoint would be (3,F(3))
The product of the sum would be the area of the box, so
f(3) * 4 = 12

But I'm wrong, evidently.

Thanks in advance, guys!
 
Last edited:
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The condition n = 4 in the two parts of the problem means that you are to divide the interval from x = 3 to x = 7 into four equal parts, and then find the total area of the four rectangles you are to construct for the Riemann sum, using either the "Right Endpoint" or "Left Endpoint Rule".
 
You're right! I can't believe I forgot this. Thanks dynamicsolo!

btw, That's some great advice in your signature.

For future reference,

\sum_i^4 \frac{3}{x} + 2 \rightarrow

Right
<br /> (i)(\frac{n}{7-3})[f(3) + f(4) + f(5) + f(6)]<br />

Left
<br /> (i)(\frac{n}{7-3})[f(7) + f(4) + f(5) + f(6)]<br />
 

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