Exam-Style Problem

9231 P21 - Nov 2024 - Q6 - 12 marks

5956

The diagram shows the curve with equation \(y=\left(\frac{1}{2}\right)^{x}\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac{1}{N}\).
(a) By considering the sum of the areas of these rectangles, show that \(\int_{0}^{1}\left(\frac{1}{2}\right)^{x} \mathrm{~d} x\gt L_{N}\), where
\(L_{N}=\frac{1}{2 N\left(2^{\frac{1}{N}}-1\right)} .\)
(b) Use a similar method to find, in terms of \(N\), an upper bound \(U_{N}\) for \(\int_{0}^{1}\left(\frac{1}{2}\right)^{x} \mathrm{~d} x\).
(c) Find the least value of \(N\) such that \(U_{N}-L_{N} \leqslant 10^{-3}\).

(d) Given that \(\int_{0}^{1}\left(\frac{1}{2}\right)^{x} \mathrm{~d} x=\frac{1}{2 \ln 2}\), use the value of \(N\) found in part (c) to find upper and lower bounds for \(\ln 2\).

Solution Checked by expert

Answer: (a) \(\displaystyle \int_0^1\left(\frac12\right)^x\,dx \gt L_N\), where \(\displaystyle L_N=\frac{1}{2N\left(2^{1/N}-1\right)}\).

(b) \(\displaystyle U_N=\frac{2^{1/N}}{2N\left(2^{1/N}-1\right)}\), so \(\displaystyle \int_0^1\left(\frac12\right)^x\,dx \lt U_N\).

(c) \(\displaystyle U_N-L_N=\frac{1}{2N}\). Hence \(\frac{1}{2N}\le 10^{-3}\Rightarrow N\ge 500\). The least value is \(500\).

(d) Using \(N=500\),

\(\displaystyle L_{500}\lt \frac{1}{2\ln 2}\lt U_{500}\Rightarrow \frac{1}{2U_{500}}\lt \ln 2\lt \frac{1}{2L_{500}}\).

So \(\displaystyle 0.693\lt \ln 2\lt 0.694\).

The curve \(y=\left(\frac12\right)^x\) is decreasing on \([0,1]\). Each rectangle has width \(\frac1N\).

(a) Using right-end ordinates gives rectangles below the curve, so

\(\displaystyle \int_0^1\left(\frac12\right)^x\,dx \gt \frac1N\left[\left(\frac12\right)^{1/N}+\left(\frac12\right)^{2/N}+\cdots+\left(\frac12\right)^{N/N}\right].\)

Hence

\(\displaystyle L_N=\frac1N\sum_{n=1}^N\left(\frac12\right)^{n/N}.\)

Let \(r=\left(\frac12\right)^{1/N}\). Then this is a geometric series:

\(\displaystyle \sum_{n=1}^N r^n=\frac{r(r^N-1)}{r-1}.\)

Since \(r^N=\frac12\),

\(\displaystyle L_N=\frac1N\cdot\frac{r\left(\frac12-1\right)}{r-1}=\frac{r}{2N(1-r)}.\)

Now substitute \(r=2^{-1/N}\):

\(\displaystyle L_N=\frac{2^{-1/N}}{2N\left(1-2^{-1/N}\right)}=\frac{1}{2N\left(2^{1/N}-1\right)}.\)

Therefore

\(\displaystyle \int_0^1\left(\frac12\right)^x\,dx \gt L_N=\frac{1}{2N\left(2^{1/N}-1\right)}.\)

(b) Using left-end ordinates gives rectangles above the curve, so

\(\displaystyle \int_0^1\left(\frac12\right)^x\,dx \lt \frac1N\left[1+\left(\frac12\right)^{1/N}+\cdots+\left(\frac12\right)^{(N-1)/N}\right].\)

Thus

\(\displaystyle U_N=\frac1N\sum_{n=0}^{N-1}\left(\frac12\right)^{n/N}.\)

Again with \(r=2^{-1/N}\),

\(\displaystyle \sum_{n=0}^{N-1}r^n=\frac{r^N-1}{r-1},\)

\(\displaystyle U_N=\frac1N\cdot\frac{\frac12-1}{r-1}=\frac{1}{2N(1-r)}.\)

Hence

\(\displaystyle U_N=\frac{1}{2N\left(1-2^{-1/N}\right)}=\frac{2^{1/N}}{2N\left(2^{1/N}-1\right)}.\)

(c)

\(\displaystyle U_N-L_N=\frac{2^{1/N}}{2N\left(2^{1/N}-1\right)}-\frac{1}{2N\left(2^{1/N}-1\right)}=\frac{1}{2N}.\)

We require

\(\displaystyle \frac{1}{2N}\le 10^{-3}.\)

So \(2N\ge 1000\), giving the least integer value

\(\displaystyle N=500.\)

(d) Given

\(\displaystyle \int_0^1\left(\frac12\right)^x\,dx=\frac{1}{2\ln 2},\)

we have

\(\displaystyle L_N\lt \frac{1}{2\ln 2}\lt U_N.\)

Since all quantities are positive, taking reciprocals gives

\(\displaystyle \frac{1}{2U_N}\lt \ln 2\lt \frac{1}{2L_N}.\)

Using \(N=500\),

\(\displaystyle 0.693\lt \ln 2\lt 0.694.\)