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Jessica's question at Yahoo! Answers regarding approximate integration
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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 \(\displaystyle \int_1^3\frac{9}{x}\,dx\):
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 \(\displaystyle \frac{3-1}{4}=\frac{1}{2}\).
The red rectangle has an area of:
\(\displaystyle A_1=\frac{1}{2}\cdot\frac{9}{\frac{3}{2}}=3\)
The green rectangle has an area of:
\(\displaystyle A_2=\frac{1}{2}\cdot\frac{9}{2}=\frac{9}{4}\)
The blue rectangle has an area of:
\(\displaystyle A_3=\frac{1}{2}\cdot\frac{9}{\frac{5}{2}}=\frac{9}{5}\)
The orange rectangle has an area of:
\(\displaystyle A_4=\frac{1}{2}\cdot\frac{9}{3}=\frac{3}{2}\)
And so we may state:
\(\displaystyle \int_1^3\frac{9}{x}\,dx\approx3+\frac{9}{4}+\frac{9}{5}+\frac{3}{2}=\frac{171}{20}=8.55\)
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:
\(\displaystyle \Delta A=\frac{3-1}{n}\cdot\frac{9}{1+k\cdot\frac{2}{n}}=\frac{18}{n+2k}\)
And so we may state:
\(\displaystyle \int_1^3\frac{9}{x}\,dx\approx18\sum_{k=1}^n\frac{1}{n+2k}\)
\(\displaystyle \int_1^3\frac{9}{x}\,dx=18\lim_{n\to\infty}\left( \sum_{k=1}^n\frac{1}{n+2k} \right)\)
Now, since we know:
\(\displaystyle \int_1^3\frac{9}{x}\,dx=9\ln(3)\approx9.887510598013\)
We may state:
\(\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{n+2k} \right)=\ln\left(\sqrt{3} \right)\)
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 \(\displaystyle \int_1^3\frac{9}{x}\,dx\):
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 \(\displaystyle \frac{3-1}{4}=\frac{1}{2}\).
The red rectangle has an area of:
\(\displaystyle A_1=\frac{1}{2}\cdot\frac{9}{\frac{3}{2}}=3\)
The green rectangle has an area of:
\(\displaystyle A_2=\frac{1}{2}\cdot\frac{9}{2}=\frac{9}{4}\)
The blue rectangle has an area of:
\(\displaystyle A_3=\frac{1}{2}\cdot\frac{9}{\frac{5}{2}}=\frac{9}{5}\)
The orange rectangle has an area of:
\(\displaystyle A_4=\frac{1}{2}\cdot\frac{9}{3}=\frac{3}{2}\)
And so we may state:
\(\displaystyle \int_1^3\frac{9}{x}\,dx\approx3+\frac{9}{4}+\frac{9}{5}+\frac{3}{2}=\frac{171}{20}=8.55\)
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:
\(\displaystyle \Delta A=\frac{3-1}{n}\cdot\frac{9}{1+k\cdot\frac{2}{n}}=\frac{18}{n+2k}\)
And so we may state:
\(\displaystyle \int_1^3\frac{9}{x}\,dx\approx18\sum_{k=1}^n\frac{1}{n+2k}\)
\(\displaystyle \int_1^3\frac{9}{x}\,dx=18\lim_{n\to\infty}\left( \sum_{k=1}^n\frac{1}{n+2k} \right)\)
Now, since we know:
\(\displaystyle \int_1^3\frac{9}{x}\,dx=9\ln(3)\approx9.887510598013\)
We may state:
\(\displaystyle \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{n+2k} \right)=\ln\left(\sqrt{3} \right)\)