Show that $\displaystyle\int_{n}^{n+1} \ln x\, dx = (n+1)\ln(n+1) - n\ln n - 1.$ (ii) The diagram shows the graph of $y = \ln x$ between $x = n$ and $x = n+1$. The area of the shaded region represents the error when the value of the integral in part (i) is approximated by using a single trapezium. Show that the area of the shaded region is $(n + \tfrac{1}{2})\ln\bigl(1 + \tfrac{1}{n}\bigr) - 1.$ (iii) Use a series expansion to show that if $n$ is large enough for $\dfrac{1}{n^3}$ and higher powers of $\dfrac{1}{n}$ to be neglected, then the area in part (ii) is approximately equal to $\dfrac{k}{n^2}$, where $k$ is a constant to be determined.

1. Apply integration by parts to evaluate the indefinite integral of $\ln x$.

Let $u = \ln x \implies \frac{du}{dx} = \frac{1}{x}$
Let $\frac{dv}{dx} = 1 \implies v = x$

$\int \ln x \, dx = x \ln x - \int x \left(\frac{1}{x}\right) \, dx = x \ln x - x$

2. Evaluate the definite integral from $n$ to $n+1$.

$\int_{n}^{n+1} \ln x \, dx = [x \ln x - x]_n^{n+1}$
$= \bigl((n+1)\ln(n+1) - (n+1)\bigr) - \bigl(n\ln n - n\bigr)$
$= (n+1)\ln(n+1) - n\ln n - n - 1 + n$
$= (n+1)\ln(n+1) - n\ln n - 1 \quad \text{(Shown)}$

3. Calculate the area of the single trapezium used to approximate the integral.

$\text{Area of trapezium} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$= \frac{1}{2} (\ln n + \ln(n+1))(n+1 - n)$
$= \frac{1}{2} \ln n + \frac{1}{2} \ln(n+1)$

4. Set up the expression for the area of the shaded region.

The shaded region is the exact area under the curve minus the area of the trapezium approximation.

$\text{Area of shaded region} = \int_{n}^{n+1} \ln x \, dx - \text{Area of trapezium}$
$= \bigl((n+1)\ln(n+1) - n\ln n - 1\bigr) - \left(\frac{1}{2} \ln n + \frac{1}{2} \ln(n+1)\right)$

5. Simplify the expression using logarithm properties to match the required result.

$= \left(n + 1 - \frac{1}{2}\right)\ln(n+1) - \left(n + \frac{1}{2}\right)\ln n - 1$
$= \left(n + \frac{1}{2}\right)\ln(n+1) - \left(n + \frac{1}{2}\right)\ln n - 1$
$= \left(n + \frac{1}{2}\right)[\ln(n+1) - \ln n] - 1$
$= \left(n + \frac{1}{2}\right)\ln\left(\frac{n+1}{n}\right) - 1$
$= \left(n + \frac{1}{2}\right)\ln\left(1 + \frac{1}{n}\right) - 1 \quad \text{(Shown)}$

6. Expand $\ln\left(1 + \frac{1}{n}\right)$ using the standard Maclaurin series expansion.

Using the standard expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$, substitute $x = \frac{1}{n}$:

$\ln\left(1 + \frac{1}{n}\right) = \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \dots$

(Note: It is crucial to expand up to $\frac{1}{n^3}$ because it will be multiplied by $n$, producing a $\frac{1}{n^2}$ term.)

7. Multiply the series by $\left(n + \frac{1}{2}\right)$ and neglect terms of $\frac{1}{n^3}$ and higher.

$\left(n + \frac{1}{2}\right)\ln\left(1 + \frac{1}{n}\right) \approx \left(n + \frac{1}{2}\right)\left(\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3}\right)$
$= n\left(\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3}\right) + \frac{1}{2}\left(\frac{1}{n} - \frac{1}{2n^2}\right) \quad \text{(neglecting higher powers)}$
$= \left(1 - \frac{1}{2n} + \frac{1}{3n^2}\right) + \left(\frac{1}{2n} - \frac{1}{4n^2}\right)$
$= 1 + \left(-\frac{1}{2n} + \frac{1}{2n}\right) + \left(\frac{1}{3n^2} - \frac{1}{4n^2}\right)$
$= 1 + \frac{1}{12n^2}$

8. Compute the approximate area and determine $k$.

Substitute back into the full area expression:

$\text{Area} = \left(n + \frac{1}{2}\right)\ln\left(1 + \frac{1}{n}\right) - 1$
$\approx \left(1 + \frac{1}{12n^2}\right) - 1$
$= \frac{1}{12n^2}$

Comparing this with $\frac{k}{n^2}$, we obtain:
$k = \frac{1}{12}$