Bounds using rectangles

\includegraphics{figure_3} The diagram shows the curve with equation \(y = 1 - x^2\) for \(0 \leq x \leq 1\), together with a set of \(n\) rectangles of width \(\frac{1}{n}\).

1. By considering the sum of the areas of the rectangles, show that $$\int_0^1 (1 - x^2) \, dx < \frac{4n^2 + 3n - 1}{6n^2}.$$ [4]
2. Use a similar method to find, in terms of \(n\), a lower bound for \(\int_0^1 (1 - x^2) \, dx\). [4]

Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O. \includegraphics{figure_11}

1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section. [4]
2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. [2]

The curve of the roof may be modelled by \(y = -0.013x^3 + 0.16x^2 - 0.082x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.

1. Use integration to find the area of the cross-section according to this model. [4]
2. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\). [2]

\includegraphics{figure_4} The diagram shows the curve \(y = e^{-\frac{1}{x}}\) for \(0 < x \leq 1\). A set of \((n - 1)\) rectangles is drawn under the curve as shown.

1. Explain why a lower bound for \(\int_0^1 e^{-\frac{1}{x}} dx\) can be expressed as $$\frac{1}{n}\left(e^{-n} + e^{-\frac{n}{2}} + e^{-\frac{n}{3}} + \ldots + e^{-\frac{n}{n-1}}\right).$$ [2]
2. Using a set of \(n\) rectangles, write down a similar expression for an upper bound for \(\int_0^1 e^{-\frac{1}{x}} dx\). [2]
3. Evaluate these bounds in the case \(n = 4\), giving your answers correct to 3 significant figures. [2]
4. When \(n > N\), the difference between the upper and lower bounds is less than 0.001. By expressing this difference in terms of \(n\), find the least possible value of \(N\). [3]