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I am currently working on the problem below and I am in need of help.

Consider the definite integral $\int_{1}^{2}\frac{1}{t}dt$.

(a)By the dividing the interval $1\leq t\leq 2$ into $n$ equal parts and choosing appropriate sample points, show that $$\sum_{j=1}^{n}\frac{1}{n+j}< \int_{1}^{2}\frac{1}{t}dt< \sum_{j=0}^{n-1}\frac{1}{n+j}$$ (b)How large should $n$ be to approximate $\int_{1}^{2}\frac{1}{t}dt$ with an error of at most $5(10^{-6}) $ using one of the sums in part (a)?

Hint: What is the difference between the underestimate and over estimate?

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I presume you know what Riemann sums are? –  J. M. Nov 24 '11 at 11:36
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2 Answers

The "appropriate" sample points are: $1+1/n, 1+2/n,1+3/n,\ldots,2$ for the sum on the left and $1 , 1+1/n,1+2/n,\ldots,1-(n-1)/n$ for the sum on the right.

The function $f(t)={1\over t}$ is decreasing over $[1,2]$.

To obtain a lower estimate of the integral, we take the sum of the areas of the rectangles $$A_1, A_2, \ldots, A_n,$$ where $A_i$ is the rectangle whose base is $[1+(i-1)/n, 1+i/n]$ and whose height is $f(1+i/n)={1\over 1+{i\over n}}$ ($f$ evaluated at the right endpoint of $A_i$). This gives the underestimate $$ L=\sum_{i=1}^n\underbrace{ {1\over 1+{i\over n}}}_{{\rm height\ of\ }A_i }\cdot \underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i}. $$

So $L$ is the sum on the left in your post.

By taking the height of $A_i$ to be $f(1+(i-1)i/n)={1\over 1+{(i-1)\over n}}$ ($f$ evaluated at the left endpoint of $A_i$), we obtain the over estimate: $$U=\sum_{i=1}^n\underbrace{ {1\over 1+{i-1\over n}}}_{{\rm height\ of\ }A_i }\cdot \underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i-1} =\sum_{i=0}^{n-1} {1\over n+i } . $$

So $U$ is the sum on the right in your post.

For part $b)$, using the above, if you estimate the integral with one of these sums, the error $E$ will satisfy: $$ |E|\le U-L=\sum_{i=0}^{n-1} {1\over n+i }-\sum_{i=1}^n {1\over n+i}={1\over n}-{1\over 2n}={1\over 2n} $$ So you just need to find the smallest value of $n$ that satisfies ${1\over 2n}\le5\cdot10^{-6}$.

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For part (a), we divide the interval $[1,2]$ into $n$ intervals which only interests at boundary points, i.e. $[1+\frac{j-1}{n},1+\frac{j}{n}]$ where $1\leq j\leq n$. Note that $\frac{1}{t}$ is an decreasing function, which implies that for $1\leq j\leq n$ $$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt$$ where $$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt=\frac{1}{1+\frac{j}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j}{n}}\cdot\frac{1}{n}=\frac{1}{n+j}$$ and $$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt=\frac{1}{1+\frac{j-1}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j-1}{n}}\cdot\frac{1}{n}=\frac{1}{n+j-1}.$$ That is $$\frac{1}{n+j}<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\frac{1}{n+j-1}\mbox{ for }1\leq j\leq n.$$ Now summming up the above inequalities give the result: $$\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=1}^n\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt=\int_1^2\frac{1}{t}dt<\sum_{j=1}^n\frac{1}{n+j-1}=\sum_{j=0}^{n-1}\frac{1}{n+j}.$$

For part (b), it follows from part (a) that $$0<\int_1^2\frac{1}{t}dt-\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=0}^{n-1}\frac{1}{n+j}-\sum_{j=1}^n\frac{1}{n+j}=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}.$$ To approximate $\int_1^2\frac{1}{t}dt$ with an error of at most $5(10^{−6})$ using the sum $\sum_{j=1}^n\frac{1}{n+j}$ in part (a), we must have $$\frac{1}{2n}\leq 5(10^{−6}),$$ that is $n\geq 10^5$.

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