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Find the area using Riemann Sums for $f(x)=1/x$ between $x=3$ and $x=5$ using a right sum with $2$ rectangles of equal width

$n=2, a=3, b=5$. I keep getting 1/4 as my answer but my review says its 9/20. Please help

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  • $\begingroup$ How did you get $1/4$? $\endgroup$ Nov 15, 2019 at 16:19

3 Answers 3

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Yes, the answer is $\frac9{20}$. After all:

  • The first rectangle has its vertices at $(3,0)$, $(4,0)$, $\left(4,\frac14\right)$, and $\left(3,\frac14\right)$; therefore, its area is $\frac14$.
  • The second rectangle has its vertices at $(4,0)$, $(5,0)$, $\left(5,\frac15\right)$, and $\left(4,\frac15\right)$; therefore, its area is $\frac15$.

And $\frac14+\frac15=\frac9{20}$.

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The width of the rectangles are width $\frac{5-3}{2}=1$. The right Riemann sum would give you $$f(4)\times1+f(5)\times1=\frac14+\frac15=\boxed{\frac9{20}}.$$

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Between $x = 3$ and $x = 5$ with two equal width rectangles means one rectangle goes from $x = 3$ to $x = 4$, and the other from $x = 4$ to $x = 5$.

The first rectangle, going from $x = 3$ to $x = 4$ has height $\frac14$, as that is the function value at the right side of this interval. As the width is $1$, the area is $\frac14$.

The second rectangle, going from $x = 4$ to $x = 5$, has height $\frac15$, as that's the function value at the right side of the interval. Thus its area is $\frac15$.

In total, the areas of these two rectangles is $\frac14 + \frac15 = \frac{9}{20}$.

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