Review the following Itegration of trigomometic functions: $\int\cos x dx = \sin x + C$ $\int \sin x dx = -\cos x + C$ $\int \sec^2 x dx = tan x + C$ $\int \csc^2 x dx = - \cot x + C$ $\int \sec x \tan x dx = \sec x + C$ $\int \csc x \cot x dx = -\csc x + C$ $\int \sec x dx = \ln |\sec x + \tan x| +C$ $\int\csc x dx = -\ln |\csc x + \cot x| + C$..

The equation $y=\sqrt{1-(x-1)^2}$ represents the semicircle above the x-axis with centre $(1,0)$ and radius 1 unit. By considering an appropriate region of this semicircle, find $\int_1^2 \sqrt{1-(x-1)^2}dx$ in terms of $\pi$.

Let $y=f(x)$ where $f(x)$ is defined and continuous on the interval $[a, b]$. Denote the anti-derivative of $f(x)$ by $F(x)$, that is $\frac{d}{dx}F(x) = f(x)$. Then $\int_a^b f(x) dx = F(b) - F(a)$ and $\int_b^a f(x) dx =-\int_a^bf(x) dx$. You can use the graphic calculator (GC) to evaluate definite integrals. By using the substitution $x=\cos\theta$, or otherwise, find in terms of $p$, where $\frac{1}{2} < p < 1$, an expression for $\int_p^{\frac{1}{2}} \frac{1}{x^2\sqrt{1-x^2}} dx$.

Let $y=f(x)$ where $f(x)$ is defined and continuous on the inteval $[a,b]$. Area under the curve is the limit of sums of the area of rectangles, that is, $\int_a^b f(x) dx = \lim_{\delta x \rightarrow \infty} \sum y \delta x$. Divide the interval $[a,b]$ into $n$ equal sub-intervervals, each of width $\delta x = \frac{b-a}{n}$. The diagram shows that the graph of $y=\frac{1}{x}$ in the interval $[1,2]$ with $n$ rectangles of equal width. Show that the total area of $n$ rectangles, $S_n=\sum_{r=1}^n\frac{1}{n+r}$. Hence deduce the exact value of $\lim_{n\rightarrow \infty} S_{n}$.

The area under the curve of $f(x)$ from $x=a$ to $x=b$ is $\int_a^b f(x) dx$, integrate with respect to x. The area under the curve of $g(y)$ from $y=c$ to $y=d$ is $\int_c^d g(y) dy$, integrate with respect to y. Note that $\int_a^b f(x) dx$ is negative when the curve of $y=f(x)$ is below the x-axis. So ara uner the curve is $\int_a^b |f(x)| dx$ or $|\int_a^b f(x) dx|$. Sketch the graph of $y=x(x-1)(x-2)$ and hence find the area of region enclosed by the graph and the x-axis from $x=0$ to $x=2$.

When the area uner part of a curve is rotated about a straight line, the solid formed is called a solid of revolution. Such a solid is always symmetrical about the axis of rotation. The Region $R$ is bounded by the curve $y=\sin x$, $y=\frac{1}{2}$ and the y-axis. Find the volume of the solid of revolution formed when the region $R$ is rotated through $360^o$ about the x-axis, giving your answer in exact form.

A portion of the curve $y^2 = ax$, where $a$ is a positive constant, is rotated about the y-axis to form the curved surface of an open vase. The vase has a horizontal circular base of radius $ar^2$ and a horizontal rim of radius $4ar^2$. Show that the depth of the vase is $ar$. The region $S$ is bounded by the same portion of the curve $y^2=ax$, $x=ar^2$, $x=4ar^2$ and the x-axis. Given that the volume of the solid formed by rotating $S$ about the y-axis through an angle of $2\pi$ radians is $\frac{31\pi}{10}$, find the depth of the vase in terms of $r$ only.