Integration as Area

In this question, you may assume that, if a continuous function takes both positive and negative values in an interval, then it takes the value \(0\) at some point in that interval.

1. The function \(\f\) is continuous and \(\f(x)\) is non-zero for some value of \(x\) in the interval \(0\le x \le 1\). Prove by contradiction, or otherwise, that if \[ \int_0^1 \f(x) \d x = 0\,, \] then \(\f(x)\) takes both positive and negative values in the interval \(0\le x\le 1\).
2. The function \(\g\) is continuous and \[ \int_0^1 \g(x) \, \d x = 1\,, \quad \int_0^1 x\g(x) \, \d x = \alpha\, , \quad \int_0^1 x^2\g(x) \, \d x = \alpha^2\,. \tag{\(*\)} \] Show, by considering \[ \int_0^1 (x - \alpha)^2 \g(x) \, \d x \,, \] that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\). Find a function of the form \(\g(x) = a+bx\) that satisfies the conditions \((*)\) and verify that \(\g(x)=0\) for some value of \(x\) in the interval \(0\le x \le 1\).
3. The function \(\h\) has a continuous derivative \(\h'\) and \[ \h(0) = 0\,, \quad \h(1) = 1\,, \quad \int_0^1 \h(x) \, \d x = \beta\,, \quad \int_0^1 x \h(x) \, \d x = \tfrac{1}{2}\beta (2 - \beta) \,. \] Use the result in part (ii) to show that \(\h^\prime(x)=0\) for some value of \(x\) in the interval \(0\le x\le 1\).

A spherical loaf of bread is cut into parallel slices of equal thickness. Show that, after any number of the slices have been eaten, the area of crust remaining is proportional to the number of slices remaining. A European ruling decrees that a parallel-sliced spherical loaf can only be referred to as `crusty' if the ratio of volume \(V\) (in cubic metres) of bread remaining to area \(A\) (in square metres) of crust remaining after any number of slices have been eaten satisfies \(V/A<1\). Show that the radius of a crusty parallel-sliced spherical loaf must be less than \(2\frac23\) metres. [{\sl The area \(A\) and volume \(V\) formed by rotating a curve in the \(x\)--\(y\) plane round the \(x\)-axis from \(x=-a\) to \(x=-a+t\) are given by \[ A= 2\pi\int_{-a}^{-a+t} { y}\left( 1+ \Big(\frac{\d {y}}{\d x}\Big)^2\right)^{\frac12} \d x\;, \ \ \ \ \ \ \ \ \ \ \ V= \pi \int_{-a}^{-a+t} {y}^2 \d x \;. \ \ ] \] }

For any number \(x\), the largest integer less than or equal to \(x\) is denoted by \([x]\). For example, \([3.7]=3\) and \([4]=4\). Sketch the graph of \(y=[x]\) for \(0\le x<5\) and evaluate \[ \int_0^5 [x]\;\d x. \] Sketch the graph of \(y=[\e^{x}]\) for \(0\le x< \ln n\), where \(n\) is an integer, and show that \[ \int_{0}^{\ln n}[\e^{x}]\, \d x =n\ln n - \ln (n!). \]

Show Solution

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\begin{align*} \int_0^5 [x]\;\d x &= 0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 1 \\ &= 10 \end{align*}

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\begin{align*} \int_{0}^{\ln n}[\e^{x}]\, \d x &= \sum_{k=1}^{n-1} \int_{\ln k}^{\ln (k+1)}[\e^{x}]\, \d x \\ &= \sum_{k=1}^{n-1} k \l \ln (k+1) - \ln (k) \r\\ &= \sum_{k=1}^{n-1} \l( (k+1) \l \ln (k+1) - \ln (k) \r - \ln(k+1) \r \\ &= \sum_{k=1}^{n-1} (k+1) \ln (k+1) - \sum_{k=1}^{n-1} k \ln (k) - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - 1 \ln 1 - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - \ln \l \prod_{k=1}^{n-1} (k+1)\r \\ &= n \ln n - \ln (n!) \end{align*}

Find constants \(a_{1}\), \(a_{2}\), \(u_{1}\) and \(u_{2}\) such that, whenever \({\mathrm P}\) is a cubic polynomial, \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t =a_{1}{\mathrm P}(u_{1})+a_{2}{\mathrm P}(u_{2}).\]

If \({\rm f}(t)\ge {\rm g}(t)\) for \(a\le t\le b\), explain very briefly why \(\displaystyle \int_a^b {\rm f}(t) \d t \ge \int_a^b {\rm g}(t) \d t\). Prove that if \(p>q>0\) and \(x\ge1\) then $$\frac{x^p-1}{ p}\ge\frac{x^q-1}{ q}.$$ Show that this inequality also holds when \(p>q>0\) and \(0\le x\le1\). Prove that, if \(p>q>0\) and \(x\ge0\), then $$\frac{1}{ p}\left(\frac{x^p}{ p+1}-1\right)\ge \frac{1}{q}\left(\frac{x^q}{ q+1}-1\right).$$

Explain diagrammatically, or otherwise, why \[ \frac{\mathrm{d}}{\mathrm{d}x}\int_{a}^{x}\mathrm{f}(t)\,\mathrm{d}t=\mathrm{f}(x). \] Show that, if \[ \mathrm{f}(x)=\int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t+1, \] then \(\mathrm{f}(x)=\mathrm{e}^{x}.\) What is the solution of \[ \mathrm{f}(x)=\int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t? \] Given that \[ \int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t=\int_{x}^{1}t^{2}\mathrm{f}(t)\,\mathrm{d}t+x-\frac{x^{5}}{5}+C, \] find \(\mathrm{f}(x)\) and show that \(C=-2/15.\)

Criticise each step of the following arguments. You should correct the arguments where necessary and possible, and say (with justification) whether you think the conclusion are true even though the argument is incorrect.

1. The function \(g\) defined by \[ \mathrm{g}(x)=\frac{2x^{3}+3}{x^{4}+4} \] satisfies \(\mathrm{g}'(x)=0\) only for \(x=0\) or \(x=\pm1.\) Hence the stationary values are given by \(x=0\), \(\mathrm{g}(x)=\frac{3}{4}\) and \(x=\pm1,\) \(\mathrm{g}(x)=1.\) Since \(\frac{3}{4}<1,\) there is a minimum at \(x=0\) and maxima at \(x=\pm1.\) Thus we must have \(\frac{3}{4}\leqslant\mathrm{g}(x)\leqslant1\) for all \(x\).
2. \({\displaystyle \int(1-x)^{-3}\,\mathrm{d}x=-3(1-x)^{-4}}\quad\) and so \(\quad{\displaystyle \int_{-1}^{3}(1-x)^{-3}\,\mathrm{d}x=0.}\)

The function \(\mathrm{f}\) satisfies the condition \(\mathrm{f}'(x)>0\) for \(a\leqslant x\leqslant b\), and \(\mathrm{g}\) is the inverse of \(\mathrm{f}.\) By making a suitable change of variable, prove that \[ \int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x=b\beta-a\alpha-\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y, \] where \(\alpha=\mathrm{f}(a)\) and \(\beta=\mathrm{f}(b)\). Interpret this formula geometrically, in the case where \(\alpha\) and \(a\) are both positive. Prove similarly and interpret (for \(\alpha>0\) and \(a>0\)) the formula \[ 2\pi\int_{a}^{b}x\mathrm{f}(x)\,\mathrm{d}x=\pi(b^{2}\beta-a^{2}\alpha)-\pi\int_{\alpha}^{\beta}\left[\mathrm{g}(y)\right]^{2}\,\mathrm{d}y. \]

Show Solution

Let \(u = f(x)\) then \(\frac{\d u}{\d x} = f'(x)\) and \begin{align*} \int_a^b f(x) \d x &\underbrace{=}_{\text{IBP}} \left [ xf(x) \right]_a^b - \int_a^b x f'(x) \d x \\ &\underbrace{=}_{u = f(x)} b \beta - a \alpha - \int_{u = f(a) = \alpha}^{u = f(b) = \beta} g(u) \d u \\ &= b \beta - a \alpha - \int_{\alpha}^{\beta} g(u) \d u \end{align*}

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\[ \underbrace{\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x}_{\text{red area}}=\underbrace{b\beta}_{\text{whole area}}-\underbrace{a\alpha}_{\text{area in green}}-\underbrace{\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y}_{\text{area in blue}}, \] \begin{align*} 2\pi \int_a^b x f(x) \d x &\underbrace{=}_{\text{IBP}}\pi \left [ x^2 f(x) \right]_a^b - \pi \int_a^b x^2 f'(x) \d x \\ &\underbrace{=}_{x = g(u)} \pi (b^2 \beta - a^2 \alpha) - \pi \int_{u = f(a) = \alpha}^{u = f(b) = \beta} [g(u)]^2 \d u \\ &= \pi(b^2 \beta - a^2 \alpha) - \pi \int_\alpha^\beta [g(u)]^2 \d u \end{align*} This is the volume outside the function in the volume of revolution about the \(y\) axis between \( \alpha\) and \(\beta\).

1. Prove that \[ \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5) \] and deduce that \[ \sum_{r=1}^{n}r^{5}<\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5). \]
2. Prove that, if \(n>1,\) \[ \sum_{r=0}^{n-1}r^{5}>\tfrac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n. \]
3. Let \(\mathrm{f}\) be an increasing function. If the limits \[ \lim_{n\rightarrow\infty}\sum_{r=0}^{n-1}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right)\qquad\mbox{ and }\qquad\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right) \] both exist and are equal, the definite integral \({\displaystyle \int_{0}^{a}\mathrm{f}(x)\,\mathrm{d}x}\) is defined to be their common value. Using this definition, prove that \[ \int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6. \]