Question 3:
Solution:
(a)
$\begin{array}{l}\text{Given gradient function of a curve }\frac{dy}{dx}=6x^2-12x\\ \text{The equation of the curve,}\\ y={\displaystyle \int \left(6x^2-12x\right)}dx\\ y=\frac{6x^3}{3}-\frac{12x^2}{2}+c\\ y=2x^3-6x^2+c\\ -14=2{\left(2\right)}^3-6{\left(2\right)}^2+c\text{, at point }P\left(2,-14\right)\\ -14=-8+c\\ c=-6\\ y=2x^3-6x^2-6\end{array}$


(b)
$\begin{array}{l}\frac{dy}{dx}=6x^2-12x\\ \text{At turning points, }\frac{dy}{dx}=0\\ 6x^2-12x=0\\ 6x\left(x-2\right)=0\\ x=0\text{, }x=2\\ \\ x=0,y=2{\left(0\right)}^3-6{\left(0\right)}^2-6=-6\\ x=2,y=2{\left(2\right)}^3-6{\left(2\right)}^2-6=-14\\ \\ \frac{d^{2}y}{d{x}^2}=12x-12\\ \\ \text{When }x=0\text{, }\frac{d^{2}y}{d{x}^2}=12\left(0\right)-12=-12<0\\ \left(0,-6\right)\text{ is a maximum point}\text{.}\\ \\ \text{When }x=2\text{, }\frac{d^{2}y}{d{x}^2}=12\left(2\right)-12=12>0\\ \left(2,-14\right)\text{ is a minimum point}\text{.}\end{array}$

3.5 Integration as the Summation of Areas

(A) Area of the region between a Curve and the x-axis.

3.2 Integration by Substitution


$\begin{array}{l}\text{Let }u=ax+b\\ \text{Thus, }\frac{du}{dx}=a\\ \therefore dx=\frac{du}{a}\end{array}$

$\begin{array}{l}{{\displaystyle \int \left(3x+5\right)}}^{3}dx.\\ \text{Let }u=3x+5\\ \frac{du}{dx}=3\\ dx=\frac{du}{3}\\ \\ {{\displaystyle \int \left(3x+5\right)}}^{3}dx\\ ={\displaystyle \int u^3}\frac{du}{3}\leftarrow \boxed{\begin{array}{l}\text{substitute }3x+5=u\\ \text{and }dx=\frac{du}{3}\end{array}}\\ =\frac{1}{3}{\displaystyle \int u^3}du\\ =\frac{1}{3}\left(\frac{u^4}{4}\right)+c\\ =\frac{1}{3}\left(\frac{{\left(3x+5\right)}^4}{4}\right)+c\leftarrow \boxed{\begin{array}{l}\text{ substitute back}\\ u=3x+5\end{array}}\\ =\frac{{\left(3x+5\right)}^4}{12}+c\end{array}$

(B) Using Formula method
$\begin{array}{l}{\displaystyle \int {\left(ax+b\right)}^n}=\frac{{\left(ax+b\right)}^{n+1}}{\left(n+1\right)a}+c\\ \\ \text{Hence,}\\ {\displaystyle \int {\left(3x+5\right)}^{3}dx}=\frac{{\left(3x+5\right)}^4}{4\left(3\right)}+c\\ =\frac{{\left(3x+5\right)}^4}{12}+c\end{array}$

Example 2 (Formula method):

  $\begin{array}{l}{\displaystyle \int {\left(ax+b\right)}^n}=\frac{{\left(ax+b\right)}^{n+1}}{\left(n+1\right)a}+c\\ \\ \text{Hence,}\\ {\displaystyle \int {\left(3x+5\right)}^{3}dx}=\frac{{\left(3x+5\right)}^4}{4\left(3\right)}+c\\ =\frac{{\left(3x+5\right)}^4}{12}+c\end{array}$

$\begin{array}{l}{\displaystyle \int adx}=ax+C\\ \text{Example}\\ {\displaystyle \int 2dx}=2x+C\end{array}$

Type 2:

$\begin{array}{l}{\displaystyle \int a{x}^{n}dx}=\frac{a{x}^{n+1}}{n+1}+C\\ \\ \\ Example\text{1}\\ {\displaystyle \int 2x^{3}dx}=\frac{2x^4}{4}+C=\frac{x^4}{2}+C\\ \\ \\ Example\text{2}\\ {\displaystyle \int \frac{2}{3x^5}dx}={\displaystyle \int \frac{2}{3}{x}^{-5}dx=\frac{2}{3}(\frac{x^{-4}}{-4})+C}\\ =\frac{\cancel{2}}{3}(\frac{x^{-4}}{\cancel{-4}})+C=\frac{x^{-4}}{-6}+C\end{array}$

Type 3:

$\begin{array}{l}{\displaystyle \int \left(u+v\right)dx={\displaystyle \int udx\pm {\displaystyle \int vdx}}}\\ u\text{ and }v\text{ are functions in }x\\ \\ Example\text{ 1}\\ {\displaystyle \int 3x^2+2xdx={\displaystyle \int 3x^{2}dx+{\displaystyle \int 2xdx}}}\\ =\frac{3x^3}{3}+\frac{2x^2}{2}+C\\ =\frac{\cancel{3}{x}^3}{\cancel{3}}+\frac{\cancel{2}{x}^2}{\cancel{2}}+C\\ =x^3+x^2+C\end{array}$

$\begin{array}{l}y={\displaystyle \int \left(2x+8\right)}\\ y=\frac{2x^2}{2}+8x+C\end{array}$



Example 2:

Therefore minimum point = (2, 3).

$\begin{array}{l}\frac{dy}{dx}=2x-4\\ y={\displaystyle \int \left(2x-4\right)dx}\\ y=\frac{2x^2}{2}-4x+C\\ y=x^2-4x+C\end{array}$

  $\begin{array}{l}\text{(a)}{\displaystyle {\int }_{-1}^0\left(3x^2-2x+5\right)dx}\\ \text{(b)}{\displaystyle {\int }_0^2{\left(2x+1\right)}^{3}dx}\end{array}$

$\begin{array}{l}{\displaystyle \int \frac{4}{{\left(1+x\right)}^4}dx=m{\left(1+x\right)}^n}+c\\ {\displaystyle \int 4}{\left(1+x\right)}^{-4}dx=m{\left(1+x\right)}^n+c\\ \frac{4{\left(1+x\right)}^{-3}}{-3\left(1\right)}+c=m{\left(1+x\right)}^n+c\\ -\frac{4}{3}{\left(1+x\right)}^{-3}+c=m{\left(1+x\right)}^n+c\\ m=-\frac{4}{3},n=-3\end{array}$

$\begin{array}{l}\text{Given }{\displaystyle {\int }_{-1}^{2}2g(x)dx=4}\text{, and }{\displaystyle {\int }_{-1}^2\left[mx+3g\left(x\right)\right]dx}=15.\\ \text{Find the value of constant }m.\end{array}$
$\begin{array}{l}{\displaystyle {\int }_{-1}^2\left[mx+3g\left(x\right)\right]dx}=15\\ {\displaystyle {\int }_{-1}^{2}mxdx}+{\displaystyle {\int }_{-1}^{2}3g\left(x\right)dx}=15\\ {\left[\frac{m{x}^2}{2}\right]}_{-1}^2+3{\displaystyle {\int }_{-1}^{2}g\left(x\right)dx}=15\\ \left[\frac{m{\left(2\right)}^2}{2}-\frac{m{\left(-1\right)}^2}{2}\right]+\frac{3}{2}{\displaystyle {\int }_{-1}^{2}2g\left(x\right)dx}=15\\ 2m-\frac{1}{2}m+\frac{3}{2}\left(4\right)=15\leftarrow \boxed{\text{given}{\displaystyle {\int }_{-1}^{2}2g(x)dx=4}}\\ \frac{3}{2}m+6=15\\ \frac{3}{2}m=9\\ m=9\times \frac{2}{3}\\ m=6\end{array}$

$\begin{array}{l}\text{Given}\frac{d}{dx}\left(\frac{2x}{3-x}\right)=g\left(x\right)\\ {\displaystyle \int g\left(x\right)dx=}\frac{2x}{3-x}\\ \text{Thus,}\\ {\displaystyle {\int }_1^{2}g(x)dx}={\left[\frac{2x}{3-x}\right]}_1^2\\ =\frac{2\left(2\right)}{3-2}-\frac{2\left(1\right)}{3-1}\\ =4-1\\ =3\end{array}$

Question 6:
Diagram below shows part of the curve   $y=\frac{2}{{\left(3x-2\right)}^2}$  which passes through B (1, 2).


(a) Find the equation of the tangent to the curve at the point B.
(b) A region is bounded by the curve, the x-axis and the straight lines x = 2 and x = 3.
(i) Find the area of the region.
(ii) The region is revolved through 360° about the x–axis. Find the volume generated, in terms of p.
Solution:
(a)
$\begin{array}{l}y=\frac{2}{{\left(3x-2\right)}^2}=2{\left(3x-2\right)}^{-2}\\ \frac{dy}{dx}=-4{\left(3x-2\right)}^{-3}\left(3\right)\\ \frac{dy}{dx}=\frac{-12}{{\left(3x-2\right)}^3}\\ \frac{dy}{dx}=\frac{-12}{{\left(3\left(1\right)-2\right)}^3}\text{,}x=1\\ \frac{dy}{dx}=-12\\ y-2=-12\left(x-1\right)\\ y-2=-12x+12\\ y=-12x+14\end{array}$


(b)(i)
$\begin{array}{l}\text{Area =}{\displaystyle {\int }_2^{3}ydx}\\ ={\displaystyle {\int }_2^3\frac{2}{{\left(3x-2\right)}^2}dx=}{\displaystyle {\int }_2^{3}2}{\left(3x-2\right)}^{-2}dx\\ ={\left[\frac{2{\left(3x-2\right)}^{-1}}{-1\left(3\right)}\right]}_2^3\\ ={\left[-\frac{2}{3\left(3x-2\right)}\right]}_2^3\\ =\left[-\frac{2}{3\left[3\left(3\right)-2\right]}\right]-\left[-\frac{2}{3\left[3\left(2\right)-2\right]}\right]\\ =-\frac{2}{21}+\frac{1}{6}\\ =\frac{1}{14}{\text{unit}}^2\end{array}$


(b)(ii)
$\begin{array}{l}\text{Volume generated}\\ \text{=}\pi {\displaystyle \int y^{2}dx}\\ =\pi {\displaystyle {\int }_2^3\frac{4}{{\left(3x-2\right)}^4}}dx=\pi {\displaystyle {\int }_2^3\text{4}{\left(3x-2\right)}^{-4}}dx\\ =\pi {\left[\frac{\text{4}{\left(3x-2\right)}^{-3}}{-3\left(3\right)}\right]}_2^3=\pi {\left[-\frac{\text{4}}{9{\left(3x-2\right)}^3}\right]}_2^3\\ =\pi \left[-\frac{\text{4}}{9{\left[3\left(3\right)-2\right]}^3}\right]-\left[-\frac{\text{4}}{9{\left[3\left(2\right)-2\right]}^3}\right]\\ =\pi \left(-\frac{4}{3087}+\frac{4}{576}\right)\\ =\frac{31}{5488}\pi {\text{unit}}^3\end{array}$