MHB Jessica's question at Yahoo Answers regarding approximate integration

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The discussion focuses on approximating the area under the curve of the function f(x) = 9/x over the interval [1, 3] using four subintervals with right endpoints. The areas of the rectangles formed by these subintervals are calculated, resulting in an approximate integral value of 8.55. The conversation also highlights that increasing the number of subintervals can improve the approximation, leading to a more accurate limit as n approaches infinity. The exact integral value is established as 9ln(3), approximately 9.89. The thread emphasizes the importance of using more subintervals for better accuracy in definite integral approximations.
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Here is the question:

Area of the definite integral?

Find an approximation of the area of the region R under the graph of the function f on the interval [1, 3]. Use n = 4 subintervals. Choose the representative points to be the right endpoints of the subintervals.
f(x) = 9/x

I have posted a link there to this topic so the OP can see my work.
 
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Re: Jessica's quation at Yahoo! Answers regarding approximate integration

Hello Jessica,

Let's look at a plot of the curve and the 4 rectangles the sum of whose areas we are to use to get an approximate value for the definite integral $$\int_1^3\frac{9}{x}\,dx$$:

View attachment 924

Rectangles have an area $A$ given by $A=bh$ where $b$ in the measure of the base and $h$ is the measure of the height. For each of these rectangles the base is $$\frac{3-1}{4}=\frac{1}{2}$$.

The red rectangle has an area of:

$$A_1=\frac{1}{2}\cdot\frac{9}{\frac{3}{2}}=3$$

The green rectangle has an area of:

$$A_2=\frac{1}{2}\cdot\frac{9}{2}=\frac{9}{4}$$

The blue rectangle has an area of:

$$A_3=\frac{1}{2}\cdot\frac{9}{\frac{5}{2}}=\frac{9}{5}$$

The orange rectangle has an area of:

$$A_4=\frac{1}{2}\cdot\frac{9}{3}=\frac{3}{2}$$

And so we may state:

$$\int_1^3\frac{9}{x}\,dx\approx3+\frac{9}{4}+\frac{9}{5}+\frac{3}{2}=\frac{171}{20}=8.55$$

We can improve the approximation by taking more sub-intervals. Let's let $n$ be the number of these regular partitions, and using the right-end-points, we may state the area of the $k$th rectangle as:

$$\Delta A=\frac{3-1}{n}\cdot\frac{9}{1+k\cdot\frac{2}{n}}=\frac{18}{n+2k}$$

And so we may state:

$$\int_1^3\frac{9}{x}\,dx\approx18\sum_{k=1}^n\frac{1}{n+2k}$$

$$\int_1^3\frac{9}{x}\,dx=18\lim_{n\to\infty}\left( \sum_{k=1}^n\frac{1}{n+2k} \right)$$

Now, since we know:

$$\int_1^3\frac{9}{x}\,dx=9\ln(3)\approx9.887510598013$$

We may state:

$$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{n+2k} \right)=\ln\left(\sqrt{3} \right)$$
 

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