The diagram shows the parts of the graph of $y = \frac{2}{1+x}$ between $x = 1$ and $x = 2$. P

Question

The diagram shows the parts of the graph of $y = \frac{2}{1+x}$ between $x = 1$ and $x = 2$. Prove that the total area of the four rectangles, each of equal width, as shown in the diagram below, may be expressed as $\sum_{r=1}^{4} \frac{2}{8+r}$.[2]

In general, when there are $n$ rectangles, each of equal width, under the curve between $x = 1$ and $x = 2$ (instead of just $4$), find an expression, in the form $\sum_{r=1}^{n} \mathrm{f}(r)$, for their total area.

Hence show that $\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right)$.

Deduce that $\sum_{r=1}^{n} \frac{2}{2n+r} < 2 \ln \left(\frac{3}{2}\right) < \sum_{r=1}^{n} \frac{2}{2n+r-1}$.

Accepted Answer

Four right-endpoint rectangles have area $\sum_{r=1}^4\frac2{8+r}$. For $n$ rectangles, $A_{R,n}=\sum_{r=1}^n\frac2{2n+r}$ and $\sum_{r=1}^n\frac2{2n+r}<2\ln\frac32<\sum_{r=1}^n\frac2{2n+r-1}$.

PJC P1 Q9

PJC P1 Q9

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