Question 1:
Solution:
(a)
$\begin{array}{l}\frac{dy}{dx}=5x-\frac{5}{x^2}\\ \text{At turning point}\left(m,9\right),\frac{dy}{dx}=0.\\ 5m-\frac{5}{m^2}=0\\ \frac{5}{m^2}=5m\\ m^3=1\\ m=1\end{array}$

(b)
$\begin{array}{l}\frac{dy}{dx}=5x-\frac{5}{x^2}=5x-5x^{-2}\\ \frac{d^{2}y}{d{x}^2}=5+\frac{10}{x^3}\\ \text{When }x=1,\frac{d^{2}y}{d{x}^2}=15\text{ (> 0)}\\ \text{Thus, }\left(1,9\right)\text{ is a minimum point}\text{.}\\ \end{array}$

(c)
$\begin{array}{l}y={\displaystyle \int \left(5x-5x^{-2}\right)}dx\\ y=\frac{5x^2}{2}+\frac{5}{x}+c\\ \text{At turning point }\left(1,9\right),x=1\text{ and }y=9.\\ 9=\frac{5{\left(1\right)}^2}{2}+\frac{5}{1}+c\\ c=\frac{3}{2}\\ \text{Equation of the curve: }y=\frac{5x^2}{2}+\frac{5}{x}+\frac{3}{2}\end{array}$

Question 2:
Solution:
(a)
  $\begin{array}{l}f\text{'}\left(x\right)=6x^2-7x\\ f\left(x\right)={\displaystyle \int \left(6x^2-7x\right)}dx\\ f\left(x\right)=\frac{6x^3}{3}-\frac{7x^2}{2}+c\\ 3=2{\left(1\right)}^3-\frac{7{\left(1\right)}^2}{2}+c\text{at point}\left(1,3\right)\\ c=\frac{9}{2}\\ \therefore f\left(x\right)=2x^3-\frac{7x^2}{2}+\frac{9}{2}\end{array}$

Question 5:
In Diagram below, the straight line WY is normal to the curve   $y=\frac{1}{2}{x}^2+1$ at B (2, 4). The straight line BQ is parallel to the y–axis.


Find
(a) the value of t,
(b) the area of the shaded region,
(c) the volume generated, in terms of π, when the region bounded by the curve, the y–axis
  and the straight line y = 4 is revolved through 360° about the y-axis.
Solution:
(a)
$\begin{array}{l}y=\frac{1}{2}{x}^2+1\\ \text{Gradient of tangent, }\frac{dy}{dx}=2\left(\frac{1}{2}x\right)=x\\ \text{At point }B\text{, }\frac{dy}{dx}=2\\ \text{Gradient of normal, }{m}_2=-\frac{1}{2}\\ \frac{4-0}{2-t}=-\frac{1}{2}\\ 8=-2+t\\ t=10\end{array}$


(b)
$\begin{array}{l}\text{Area of the shaded region}\\ =\text{Area under the curve + Area of triangle }BQY\\ ={\displaystyle {\int }_0^2\left(\frac{1}{2}{x}^2+1\right)}dx+\frac{1}{2}\left(10-2\right)\left(4\right)\\ ={\left[\frac{x^3}{6}+x\right]}_0^2+16\\ =\left[\frac{8}{6}+2\right]-0+16\\ =19\frac{1}{3}{\text{ unit}}^2\end{array}$


(c)
$\begin{array}{l}\text{At }y-\text{axis, }x=0,y=\frac{1}{2}\left(0\right)+1=1\\ y=\frac{1}{2}{x}^2+1,x^2=2y-2\\ \text{Volume generated }\\ \text{= }\pi {\displaystyle \int x^{2}dy}\\ =\pi {\displaystyle {\int }_1^4\left(2y-2\right)}dy\\ =\pi {\left[y^2-2y\right]}_1^4\\ =\pi \left[\left(16-8\right)-\left(1-2\right)\right]\\ =9\pi {\text{ unit}}^3\end{array}$

Question 1:
$\begin{array}{l}\text{Find the integral of each of the following}\text{.}\\ (a){\displaystyle \int \left(\frac{3}{x^2}-\frac{5}{2x^3}+2\right)dx}\\ (b){\displaystyle \int x^2\left(x^5+2x\right)dx}\\ (c){\displaystyle \int \frac{3x^4+2x}{x^3}dx}\\ (d){\displaystyle \int \frac{(7+x)(7-x)}{x^4}dx}\\ (e){\displaystyle \int {(5x-1)}^{3}dx}\\ (f){\displaystyle \int \frac{3}{{\left(4x+7\right)}^8}dx}\end{array}$


Solution:
(a)
$\begin{array}{l}{\displaystyle \int \left(\frac{3}{x^2}-\frac{5}{2x^3}+2\right)dx}\\ ={\displaystyle \int \left(3x^{-2}-\frac{5x^{-3}}{2}+2\right)dx}\\ =-3x^{-1}+\frac{5x^{-2}}{4}+2x+c\\ =-\frac{3}{x}+\frac{5}{4x^2}+2x+c\end{array}$


(b)
$\begin{array}{l}{\displaystyle \int x^2\left(x^5+2x\right)dx}\\ ={\displaystyle \int \left(x^7+2x^3\right)}dx\\ =\frac{x^8}{8}+\frac{2x^4}{4}+c\\ =\frac{x^8}{8}+\frac{x^4}{2}+c\end{array}$


(c)
$\begin{array}{l}{\displaystyle \int \frac{3x^4+2x}{x^3}dx}\\ ={\displaystyle \int \left(\frac{3x^4}{x^3}+\frac{2x}{x^3}\right)}dx={\displaystyle \int \left(3x+2x^{-2}\right)}dx\\ =\frac{3x^2}{2}-\frac{2}{x}+c\end{array}$


(d)
$\begin{array}{l}{\displaystyle \int \frac{(7+x)(7-x)}{x^4}dx}={\displaystyle \int \left(\frac{49-x^2}{x^4}\right)}dx\\ ={\displaystyle \int \left(\frac{49}{x^4}-\frac{1}{x^2}\right)dx}={\displaystyle \int \left(49x^{-4}-x^{-2}\right)dx}\\ =\frac{49x^{-3}}{-3}+\frac{1}{x}+c\\ =-\frac{49}{3x^3}+\frac{1}{x}+c\end{array}$


(e)
$\begin{array}{l}{\displaystyle \int {(5x-1)}^{3}dx}\\ =\frac{{(5x-1)}^4}{\left(4\right)\left(5\right)}+c\\ =\frac{1}{20}{\left(5x-1\right)}^4+c\end{array}$


(f)
$\begin{array}{l}{\displaystyle \int \frac{3}{{\left(4x+7\right)}^8}dx}={\displaystyle \int 3{\left(4x+7\right)}^{-8}dx}\\ =\frac{3{\left(4x+7\right)}^{-7}}{\left(-7\right)\left(4\right)}+c\\ =-\frac{3}{28{\left(4x+7\right)}^7}+c\end{array}$

Question 11:
$\begin{array}{l}\text{Given }{\displaystyle {\int }_{-2}^{3}g(x)dx=4},\text{ and }{\displaystyle {\int }_{-2}^{3}h(x)dx=9}\text{, find the value of}\\ \text{(a) }{\displaystyle {\int }_{-2}^{3}5g(x)dx,}\\ \text{(b) }m\text{ if }{\displaystyle {\int }_{-2}^3\left[g(x)+3h\left(x\right)+4m\right]dx=12}\end{array}$

Solution:
(a)
$\begin{array}{l}{\displaystyle {\int }_{-2}^{3}5g(x)dx=5}{\displaystyle {\int }_{-2}^{3}g(x)dx}\\ =5\times 4\\ =20\end{array}$

(b)
$\begin{array}{l}{\displaystyle {\int }_{-2}^3\left[g(x)+3h\left(x\right)+4m\right]dx=12}\\ {\displaystyle {\int }_{-2}^{3}g(x)dx+3}{\displaystyle {\int }_{-2}^{3}h\left(x\right)dx+}{\displaystyle {\int }_{-2}^{3}4mdx=12}\\ 4+3\left(9\right)+4m{\left[x\right]}_{-2}^3=12\\ 4m\left[3-\left(-2\right)\right]=-19\\ 20m=-19\\ m=-\frac{19}{20}\end{array}$

Question 12:
$\begin{array}{l}\text{(a) Find the value of }{\displaystyle {\int }_{-1}^1{\left(3x+1\right)}^{3}dx.}\\ \text{(b) Evaluate }{\displaystyle {\int }_3^4\frac{1}{\sqrt{2x-4}}}dx.\end{array}$

Solution:
$\begin{array}{l}\text{a) }{{\displaystyle {\int }_{-1}^1{\left(3x+1\right)}^{3}dx=\left[\frac{{\left(3x+1\right)}^4}{4\left(3\right)}\right]}}_{-1}^1\\ ={\left[\frac{{\left(3x+1\right)}^4}{12}\right]}_{-1}^1\\ =\frac{1}{12}\left[4^4-{\left(-2\right)}^4\right]\\ =\frac{1}{12}\left(256-16\right)\\ =20\end{array}$

$\begin{array}{l}\text{(b) }{\displaystyle {\int }_3^4\frac{1}{\sqrt{2x-4}}}dx={\displaystyle {\int }_3^4\frac{1}{{\left(2x-4\right)}^{\frac{1}{2}}}}dx\\ ={\displaystyle {\int }_3^4{\left(2x-4\right)}^{-\frac{1}{2}}}dx\\ ={\left[\frac{{\left(2x-4\right)}^{-\frac{1}{2}+1}}{\frac{1}{2}\left(2\right)}\right]}_3^4\\ ={\left[\sqrt{2x-4}\right]}_3^4\\ =\left[\sqrt{2\left(4\right)-4}-\sqrt{2\left(3\right)-4}\right]\\ =2-\sqrt{2}\end{array}$

Question 13:
$\begin{array}{l}\text{Given that }y=\frac{x^2}{2x-1},\text{ show that}\\ \frac{dy}{dx}\text{=}\frac{2x\left(x-1\right)}{{\left(2x-1\right)}^2}\text{. Hence, evaluate }{\displaystyle {\int }_{-2}^2\frac{x\left(x-1\right)}{4{\left(2x-1\right)}^2}dx}.\end{array}$

Solution:
$\begin{array}{l}y=\frac{x^2}{2x-1}\\ \frac{dy}{dx}=\frac{\left(2x-1\right)\left(2x\right)-x\left(2\right)}{{\left(2x-1\right)}^2}\\ =\frac{4x^2-2x-2x^2}{{\left(2x-1\right)}^2}\\ =\frac{2x^2-2x}{{\left(2x-1\right)}^2}\\ =\frac{2x\left(x-1\right)}{{\left(2x-1\right)}^2}\left(\text{shown}\right)\\ \\ {\displaystyle {\int }_{-2}^2\frac{2x\left(x-1\right)}{{\left(2x-1\right)}^2}dx}={\left[\frac{x^2}{2x-1}\right]}_{-2}^2\\ \frac{1}{8}{\displaystyle {\int }_{-2}^2\frac{2x\left(x-1\right)}{{\left(2x-1\right)}^2}dx}=\frac{1}{8}{\left[\frac{x^2}{2x-1}\right]}_{-2}^2\\ \frac{1}{4}{\displaystyle {\int }_{-2}^2\frac{x\left(x-1\right)}{{\left(2x-1\right)}^2}dx}=\frac{1}{8}\left[\left(\frac{2^2}{2\left(2\right)-1}\right)-\left(\frac{{\left(-2\right)}^2}{2\left(-2\right)-1}\right)\right]\\ =\frac{1}{8}\left[\left(\frac{4}{3}\right)-\left(\frac{4}{-5}\right)\right]\\ =\frac{1}{8}\left(\frac{32}{15}\right)\\ =\frac{4}{15}\end{array}$

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

Question 5:
$\begin{array}{l}\text{Given }{\displaystyle \int \left(6x^2+1\right)dx=m{x}^3+x}+c,\\ \text{where }m\text{ and }c\text{ are constants, find}\\ \text{(a) the value of }m\text{.}\\ \text{(b) the value of }c\text{ if }{\displaystyle \int \left(6x^2+1\right)dx=13\text{ when }x=1.}\end{array}$

Solution:
(a)
$\begin{array}{l}{\displaystyle \int \left(6x^2+1\right)dx=m{x}^3+x}+c\\ \frac{6x^3}{3}+x+c=m{x}^3+x+c\\ 2x^3+x+c=m{x}^3+x+c\\ \text{Compare the both sides,}\\ \therefore m=2\end{array}$

(b)

$\begin{array}{l}{\displaystyle \int \left(6x^2+1\right)dx=13\text{ when }x=1.}\\ 2{\left(1\right)}^3+1+c=13\\ 3+c=13\\ c=10\end{array}$

Question 6:
$\begin{array}{l}\text{It is given that }{\displaystyle {\int }_5^{k}g(x)dx=6},\text{ and }{\displaystyle {\int }_5^k\left[g\left(x\right)+2\right]dx}=14,\\ \text{find the value of }k.\end{array}$

Solution:
$\begin{array}{l}{\displaystyle {\int }_5^k\left[g\left(x\right)+2\right]dx}=14\\ {\displaystyle {\int }_5^{k}g\left(x\right)dx}+{\displaystyle {\int }_5^{k}2dx}=14\\ 6+{\left[2x\right]}_5^k=14\\ 2\left(k-5\right)=8\\ k-5=4\\ k=9\end{array}$

Question 7:
$\text{Given }{\displaystyle {\int }_k^2(4x+7)dx=28},\text{ calculate the possible value of }k.$

Solution:
$\begin{array}{l}{\displaystyle {\int }_k^2(4x+7)dx=28}\\ {\left[2x^2+7x\right]}_k^2=28\\ 8+14-\left(2k^2+7k\right)=28\\ 22-2k^2-7k=28\\ 2k^2+7k+6=0\\ \left(2k+3\right)\left(k+2\right)=0\\ k=-\frac{3}{2}\text{ or }k=-2\end{array}$

Question 8:
$\text{Given }y=\frac{5x}{x^2+1}\text{ and }\frac{dy}{dx}=g\left(x\right),\text{ find the value of }{\displaystyle {\int }_0^{3}2g\left(x\right)dx.}$

Solution:
$\begin{array}{l}\text{Since}\frac{dy}{dx}=g\left(x\right),\text{ thus }y={\displaystyle \int g\left(x\right)}dx\\ {\displaystyle {\int }_0^{3}2g\left(x\right)dx=2{\displaystyle {\int }_0^{3}g\left(x\right)dx}}\\ =2{\left[y\right]}_0^3\\ =2{\left[\frac{5x}{x^2+1}\right]}_0^3\\ =2\left[\frac{5\left(3\right)}{3^2+1}-0\right]\\ =2\left(\frac{15}{10}\right)\\ =3\end{array}$

Question 9:
$\text{Find }{\displaystyle {\int }_5^k\left(x+1\right)dx,\text{ in terms of }k.}$

Solution:
$\begin{array}{l}{{\displaystyle {\int }_5^k\left(x+1\right)dx=\left[\frac{x^2}{2}+x\right]}}_5^k\\ =\left(\frac{k^2}{2}+k\right)-\left(\frac{5^2}{2}+5\right)\\ =\frac{k^2+2k}{2}-\frac{35}{2}\\ =\frac{k^2+2k-35}{2}\end{array}$

Question 10:
$\begin{array}{l}\text{Given that}={\displaystyle {\int }_2^{5}g(x)dx=-2}\text{. Find}\\ \text{(a) the value of }{\displaystyle {\int }_5^{2}g(x)dx,}\\ \text{(b) the value of }m\text{ if }{\displaystyle {\int }_2^5\left[g(x)+m\left(x\right)\right]dx=19}\end{array}$

Solution:
$\begin{array}{l}\text{(a) }{\displaystyle {\int }_5^{2}g(x)dx=}-{\displaystyle {\int }_2^{5}g(x)dx}\\ =-\left(-2\right)\\ =2\end{array}$

$\begin{array}{l}\text{(b) }{\displaystyle {\int }_2^5\left[g(x)+m\left(x\right)\right]dx=19}\\ {\displaystyle {\int }_2^{5}g(x)dx+m}{\displaystyle {\int }_2^{5}xdx=19}\\ -2+m{\left[\frac{x^2}{2}\right]}_2^5=19\\ \frac{m}{2}{\left[x^2\right]}_2^5=21\\ \frac{m}{2}\left[25-4\right]=21\\ 21m=42\\ m=2\end{array}$

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).

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