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(D) The area $A$ is given by the integral $\int_{1}^{10} \frac{1}{x} dx = [\ln x]_{1}^{10} = \ln 10$.Consider the sum $B = \sum_{n=1}^{9} \frac{1}{n}

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$C < A < B$ and $B - A < A - C$

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(D) The area $A$ is given by the integral $\int_{1}^{10} \frac{1}{x} dx = [\ln x]_{1}^{10} = \ln 10$.Consider the sum $B = \sum_{n=1}^{9} \frac{1}{n} = 1 + \frac{1}{2} + \dots + \frac{1}{9}$. This represents the upper Riemann sum for the function $f(x) = \frac{1}{x}$ on the interval $[1, 10]$ with subintervals of width $1$. Since $f(x)$ is a decreasing function,the upper sum is greater than the area under the curve,so $B > A$.Consider the sum $C = \sum_{n=2}^{10} \frac{1}{n} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{10}$. This represents the lower Riemann sum for the function $f(x) = \frac{1}{x}$ on the interval $[1, 10]$ with subintervals of width $1$. Since $f(x)$ is decreasing,the lower sum is less than the area under the curve,so $C Thus,we have $C To compare $B-A$ and $A-C$,note that $B-A = \sum_{n=1}^{9} \frac{1}{n} - \int_{1}^{10} \frac{1}{x} dx = \sum_{n=1}^{9} (\frac{1}{n} - \int_{n}^{n+1} \frac{1}{x} dx)$.Similarly,$A-C = \int_{1}^{10} \frac{1}{x} dx - \sum_{n=2}^{10} \frac{1}{n} = \sum_{n=1}^{9} (\int_{n}^{n+1} \frac{1}{x} dx - \frac{1}{n+1})$.Since the function $f(x) = \frac{1}{x}$ is convex,the area under the curve is closer to the lower sum than the upper sum,implying $A-C Therefore,$C < A < B$ and $B - A < A - C$.

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

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