Discussion questions · Section 1 of 2

The diagram shows part of the graph of $y=x^3+1$ from $x=0$ to $x=1$. The area under the curve may be approximated by the total area, $A$, of $n$ rectangles as shown. The width of each rectangle is $1/n$.

Given that $\displaystyle\sum_{r=1}^{n}r^3=\left[\frac{n(n+1)}2\right]^2$, show that $A=1+\dfrac14\left(1-\dfrac1n\right)^2$.

Hence deduce the exact value of $\displaystyle\int_0^1(x^3+1)\,dx$.

Additional printed Student lecture ask: Construct a similar diagram so that overestimation occurs.

Using partial fractions, find $\displaystyle\int_0^2\frac{9x^2+x-13}{(2x-5)(x^2+9)}\,dx$. Give your answer as $a\ln b+c\tan^{-1}d$, with rational $a,b,c,d$.

Find the exact value of $p$ such that $\displaystyle\int_0^1\frac{dx}{4-x^2}=\int_0^{1/(2p)}\frac{dx}{\sqrt{1-p^2x^2}}$.

$f(x)=xe^x$ for $0<x\le1$, and $f(x)=e(2-x)$ for $1<x\le2$. Also $f(x)=f(x+2)$ for all real $x$.

(i) Sketch $y=f(x)$ for $-2<x\le4$.

(ii) Find $\displaystyle\int_{-1}^{3}f(x)\,dx$ exactly.

(a)(i) Find $\displaystyle\int_{1/3}^{a}\frac{\ln(3x)}{x^3}\,dx$, where $a>1/3$.

(ii) Hence evaluate the same integral with upper limit $\infty$. You may assume $\ln a/a^2\to0$ as $a\to\infty$.

(b) The region under $y=3/(1+4x^2)$ between $x=-\sqrt3/2$ and $x=\sqrt3/2$ is rotated through $2\pi$ about the x-axis. Use $2x=\tan\theta$ to show that the volume is $9\pi\int_0^{\pi/3}\cos^2\theta\,d\theta$, and evaluate it exactly.

For the curve $y=x^3$, the region between the curve, $x=1$, $x=q$ and the x-axis has the same area as the region between the curve, $y=1$, $y=8$ and the y-axis, where $q>1$. Find $q$ exactly as $a^{1/b}$ for integers $a,b$.