Example 1
Find the area of the shaded region.


Solution:
$\begin{array}{l}\text{Area of the shaded region}\\ ={\displaystyle {\int }_a^{b}ydx}\\ ={\displaystyle {\int }_0^4\left(6x-x^2\right)dx}\\ ={\left[\frac{6x^2}{2}-\frac{x^3}{3}\right]}_0^4\\ =\left[3{(4)}^2-\frac{{\left(4\right)}^3}{3}\right]-0\\ =26\frac{2}{3}{\text{unit}}^2\end{array}$


Example 2
Find the area of the shaded region.


Solution:
$\begin{array}{l}=\frac{1}{2}\times \text{base}\times \text{height +}{\displaystyle {\int }_7^{8}xdy}\\ =\frac{1}{2}\times \left(7\right)\left(7\right)+{\displaystyle {\int }_7^8\left(8y-y^2\right)dy}\\ =\frac{49}{2}+{\left[\frac{8y^2}{2}-\frac{y^3}{3}\right]}_7^8\\ =24\frac{1}{2}+\left[4{(8)}^2-\frac{{\left(8\right)}^3}{3}\right]-\left[4{(7)}^2-\frac{{\left(7\right)}^3}{3}\right]\\ =24\frac{1}{2}+85\frac{1}{3}-81\frac{2}{3}\\ =28\frac{1}{6}{\text{unit}}^2\end{array}$

3.4b Laws of Definite Integrals



Example:
Given that ${\displaystyle {\int }_3^{7}f(x)dx=5}$ , find the values for each of the following:

$\begin{array}{l}\text{(a)}{\displaystyle {\int }_3^{7}6f(x)dx}\\ \text{(b)}{\displaystyle {\int }_3^7[3-f(x)]dx}\\ \text{(c)}{\displaystyle {\int }_7^{3}2f(x)dx}\\ \text{(d)}{\displaystyle {\int }_3^{4}f(x)dx}+{\displaystyle {\int }_4^{5}f(x)dx+{\displaystyle {\int }_3^{7}f(x)dx}}\\ \text{(e)}{\displaystyle {\int }_3^7\frac{f(x)+7}{2}dx}\end{array}$


Solution:
$\begin{array}{l}\text{(a)}{\displaystyle {\int }_3^{7}6f(x)dx}=6{\displaystyle {\int }_3^{7}f(x)dx}\\ =6(5)=30\\ \\ \text{(b)}{\displaystyle {\int }_3^7[3-f(x)]dx={\displaystyle {\int }_3^{7}3dx-{\displaystyle {\int }_3^{7}f(x)}dx}}\\ ={\left[3x\right]}_3^7-5\\ =\left[3(7)-3(3)\right]-5\\ =7\\ \\ \text{(c)}{\displaystyle {\int }_7^{3}2f(x)dx}=-{\displaystyle {\int }_3^{7}2f(x)dx}\\ =-2{\displaystyle {\int }_3^{7}f(x)dx}\\ =-2(5)\\ =-10\\ \\ \text{(d)}{\displaystyle {\int }_3^{4}f(x)dx}+{\displaystyle {\int }_4^{5}f(x)dx+{\displaystyle {\int }_3^{7}f(x)dx}}\\ ={\displaystyle {\int }_3^{7}f(x)dx}\\ =5\\ \\ \text{(e)}{\displaystyle {\int }_3^7\frac{f(x)+7}{2}dx}={\displaystyle {\int }_3^7\left[\frac{1}{2}f(x)+\frac{7}{2}\right]dx}\\ ={\displaystyle {\int }_3^7\frac{1}{2}f(x)dx}+{\displaystyle {\int }_3^7\frac{7}{2}dx}\\ =\frac{1}{2}{\displaystyle {\int }_3^{7}f(x)dx}+{\left[\frac{7x}{2}\right]}_3^7\\ =\frac{1}{2}(5)+\left[\frac{7(7)}{2}-\frac{7(3)}{2}\right]\\ =\frac{5}{2}+14\\ =16\frac{1}{2}\end{array}$

Question 14 (3 marks):
It is given that ${\displaystyle \int \frac{5}{{\left(2x+3\right)}^n}}dx=\frac{p}{{\left(2x+3\right)}^5}+c$ , where c, n and p are constants.
Find the value of n and of p.

Solution:
$\begin{array}{l}{\displaystyle \int \frac{5}{{\left(2x+3\right)}^n}}dx={\displaystyle \int 5{\left(2x+3\right)}^{-n}}dx\\ =\frac{5{\left(2x+3\right)}^{-n+1}}{\left(-n+1\right)\times 2}+c\\ =\frac{5}{2\left(1-n\right)}\times \frac{1}{{\left(2x+3\right)}^{n-1}}+c\\ =\frac{5}{2\left(1-n\right){\left(2x+3\right)}^{n-1}}+c\\ \\ \text{Compare }\frac{5}{2\left(1-n\right){\left(2x+3\right)}^{n-1}}\\ \text{with }\frac{p}{{\left(2x+3\right)}^5}\\ \\ n-1=5\\ n=6\\ \\ \frac{5}{2\left(1-n\right)}=p\\ \frac{5}{2\left(1-6\right)}=p\\ \frac{5}{2\left(-5\right)}=p\\ p=-\frac{1}{2}\end{array}$

Question 15 (4 marks):
Diagram shows the curve y = g(x). The straight line is a tangent to the curve.
Diagram

Given g’(x) = –4x + 8, find the equation of the curve.


Solution:
$\begin{array}{l}\text{Given }g\text{'}\left(x\right)=-4x+8\\ \text{Maximum point when }g\text{'}\left(x\right)=0\\ -4x+8=0\\ 4x=8\\ x=2\\ \text{Thus, maximum point is }\left(2,11\right).\\ \\ g\text{'}\left(x\right)=-4x+8\\ {\displaystyle \int g\text{'}\left(x\right)}={\displaystyle \int \left(-4x+8\right)}dx\\ g\left(x\right)=\frac{-4x^2}{2}+8x+c\\ g\left(x\right)=-2x^2+8x+c\\ \\ \text{Substitute }\left(2,11\right)\text{ into }g\left(x\right):\\ 11=-2{\left(2\right)}^2+8\left(2\right)+c\\ c=3\\ \\ \text{Thus, the equation of curve is}\\ g\left(x\right)=-2x^2+8x+3\end{array}$

Question 10 (10 marks):
Diagram shows a curve y = 2x² – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.

(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180^o about the y-axis, the volume generated is 65π unit³.
Find the value of h.

Solution: 
(a)
$\begin{array}{l}y=2x^2-18\\ \frac{dy}{dx}=4x\\ \\ \text{Gradient of straight line }PQ=4\\ 4x=4\\ x=1\\ \\ \text{When }x=1,\\ y=2{\left(1\right)}^2-18=-16\\ \\ \text{Coordinates of }K=\left(1,-16\right).\end{array}$


(b)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \therefore 2x^2-18=0\\ 2x^2=18\\ x^2=9\\ x=\pm 3\\ \text{The curve cuts the }x\text{-axis at }\left(-3,0\right)\text{ and }\left(3,0\right)\text{.}\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(5-1\right)\left(16\right)-{\displaystyle {\int }_1^{3}ydx}\\ =32-{\displaystyle {\int }_1^3\left(2x^2-18\right)dx}\\ =32-\left|{\left[\frac{2x^3}{3}-18x\right]}_1^3\right|\\ =32-\left|\left(\frac{2{\left(3\right)}^3}{3}-18\left(3\right)\right)-\left(\frac{2{\left(1\right)}^3}{3}-18\left(1\right)\right)\right|\\ =32-\left|\left(18-54-\frac{2}{3}+18\right)\right|\\ =32-\left|-18\frac{2}{3}\right|\\ =32-18\frac{2}{3}\\ =13\frac{1}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume generated}=65\pi \\ \pi {\displaystyle {\int }_h^0{x}^{2}dy}=65\pi \\ \pi {\displaystyle {\int }_h^0\left(\frac{y}{2}+9\right)dx}=65\pi \leftarrow \boxed{\begin{array}{l}y=2x^2-18\\ x^2=\frac{y}{2}+9\end{array}}\\ {\left[\frac{y^2}{4}+9y\right]}_h^0=65\\ 0-\left(\frac{h^2}{4}+9h\right)=65\\ -\frac{h^2}{4}-9h=65\\ h^2+36h+260=0\\ \left(h+10\right)\left(h+26\right)=0\\ h=-10\text{ or }h=-26\left(\text{rejected}\right)\\ \text{Thus, }h=-10\end{array}$

Question 9 (10 marks):
Diagram shows the straight line 4y = x – 2 touches the curve x = y² + 6 at point P.



Find
(a) the coordinates of P,
(b) the area of the shaded region,
(c) the volume of revolution, in terms of π, when the region bounded by the curve and the straight line x = 8 is revolved through 180^o about the x-axis.


Solution:
(a)
$\begin{array}{l}4y=x-\mathrm{2.........}(1)\\ x=y^2+\mathrm{6.........}(2)\\ \\ \text{Substitute (2) into (1):}\\ 4y=\left(y^2+6\right)-2\\ y^2-4y+4=0\\ \left(y-2\right)\left(y-2\right)=0\\ y-2=0\\ y=2\\ \\ \text{Substitute }y=2\text{ into (2):}\\ x={\left(2\right)}^2+6\\ x=10\\ \text{Thus, }P=\left(10,2\right).\end{array}$


(b)
$\begin{array}{l}\text{At x-axis, }y=0\\ \therefore 4y=x-2\\ 0=x-2\\ x=2\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(10-2\right)\left(2\right)-{\displaystyle {\int }_6^{10}ydx}\\ =8-{\displaystyle {\int }_6^{10}\sqrt{x-6}dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y=\sqrt{x-6}\end{array}}\\ =8-{\displaystyle {\int }_6^{10}{\left(x-6\right)}^{\frac{1}{2}}dx}\\ =8-{\left[\frac{{\left(x-6\right)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_6^{10}\\ =8-{\left[\frac{2{\left(x-6\right)}^{\frac{3}{2}}}{3}\right]}_6^{10}\\ =8-\left[\frac{2{\left(10-6\right)}^{\frac{3}{2}}}{3}-\frac{2{\left(6-6\right)}^{\frac{3}{2}}}{3}\right]\\ =8-\frac{16}{3}\\ =\frac{8}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume of revolution}\\ =\pi {\displaystyle {\int }_6^8{y}^{2}dx}\\ =\pi {\displaystyle {\int }_6^8\left(x-6\right)dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y^2=x-6\end{array}}\\ =\pi {\left[\frac{x^2}{2}-6x\right]}_6^8\\ =\pi \left[\left(32-48\right)-\left(18-36\right)\right]\\ =2\pi {\text{ units}}^3\end{array}$

Question 8:
Diagram below shows the curve $y=\frac{4}{x^2}$ and the straight line y = mx + c. The straight line y = mx + c is a tangent to the curve at (2, 1).

(a) Find the value of m and of c.

(b) Calculate the area of the shaded region.

(c) It is given that the volume generated when the region bounded by the curve, the x–axis and the straight lines x = 2 and x = h is revolved through 360^o about the x-axis is $\frac{38\pi }{81}{\text{ unit}}^3.$
Find the value of h, such that h > 2.

Solution:
(a)
$\begin{array}{l}y=\frac{4}{x^2}=4x^{-2}\\ \frac{dy}{dx}=-8x^{-3}\\ =-\frac{8}{x^3}\\ \text{At }x=2,\\ \frac{dy}{dx}=-\frac{8}{2^3}\\ =-1\\ \\ \text{Equation of tangent:}\\ y-y_1=m\left(x-x_1\right)\\ y-1=-1\left(x-2\right)\\ y=-x+2+1\\ y=-x+3\\ \\ m=-1,c=3\end{array}$


(b)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \text{From the straight line }y=-x+3,x=3\\ \\ \text{Area of the shaded region}\\ =\text{Area under the curve}-\text{Area of triangle}\\ ={\displaystyle {\int }_2^{4}ydx-}\frac{1}{2}\times 1\times 1\\ ={\displaystyle {\int }_2^4\left(4x^{-2}\right)dx}-\frac{1}{2}\\ ={\left[\frac{4x^{-1}}{-1}\right]}_2^4-\frac{1}{2}\\ ={\left[-\frac{4}{x}\right]}_2^4-\frac{1}{2}\\ =\left[-\frac{4}{4}-\left(-\frac{4}{2}\right)\right]-\frac{1}{2}\\ =\frac{1}{2}{\text{ unit}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume generated}=\frac{38\pi }{81}\\ \pi {\displaystyle {\int }_2^h{y}^{2}dx}=\frac{38\pi }{81}\\ {\displaystyle {\int }_2^h{\left(\frac{4}{x^2}\right)}^{2}d}x=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(\frac{16}{x^4}\right)dx}=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(16x^{-4}\right)dx}=\frac{38}{81}\\ {\left[\frac{16x^{-3}}{-3}\right]}_2^h=\frac{38}{81}\\ {\left[-\frac{16}{3x^3}\right]}_2^h=\frac{38}{81}\\ -\frac{16}{3h^3}-\left(-\frac{16}{3{\left(2\right)}^3}\right)=\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{24}-\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{81}\\ 3h^3=81\\ h^3=27\\ h=3\end{array}$

Question 7:
Diagram below shows a curve $y=\frac{1}{4}{x}^2+3$ which intersects the straight line y = x + 6 at point A.


(a) Find the coordinates of A.
(b) Calculate
(i) the area of the shaded region M,
(ii) the volume generated, in terms of π, when the shaded region N is revolved 360^o about the y-axis.

Solution:
(a)
$\begin{array}{l}y=\frac{1}{4}{x}^2+\mathrm{3..........}\left(1\right)\\ y=x+\mathrm{6..........}\left(2\right)\\ \text{Substitute (2) into (1),}\\ x+6=\frac{1}{4}{x}^2+3\\ 4x+24=x^2+12\\ x^2-4x-12=0\\ \left(x+2\right)\left(x-6\right)=0\\ x=-2\text{ or }x=6\left(\text{rejected}\right)\\ \\ \text{When }x=-2\\ y=-2+6=4\\ \text{Therefore, }A=\left(-2,4\right).\end{array}$


(b)(i)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \text{From }y=x+6,x=-6\\ \\ \text{Area of region }M\\ =\text{Area of triangle}+\text{Area under the curve}\\ =\frac{1}{2}\times \left(6-2\right)\times 4+{\displaystyle {\int }_{-2}^{0}ydx}\\ =8+{\displaystyle {\int }_{-2}^0\left(\frac{1}{4}{x}^2+3\right)dx}\\ =8+{\left[\frac{x^3}{4\left(3\right)}+3x\right]}_{-2}^0\\ =8+\left[0-\left(\frac{{\left(-2\right)}^3}{12}+3\left(-2\right)\right)\right]\\ =8+\left[0-\left(-\frac{8}{12}-6\right)\right]\\ =8+\left[0-\left(-\frac{20}{3}\right)\right]\\ =14\frac{2}{3}{\text{ unit}}^2\end{array}$


(b)(ii)
$\begin{array}{l}\text{At }y\text{-axis, }x=0,\\ y=\frac{1}{4}\left(0\right)+3\\ y=3\\ \\ y=\frac{1}{4}{x}^2+3\\ 4y=x^2+12\\ x^2=4y-12\\ \\ \text{Volume of }N\\ \text{= }\pi {\displaystyle {\int }_3^4{x}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_3^4\left(4y-12\right)dy}\\ \text{= }\pi {\displaystyle {\int }_3^4\left(2y^2-12y\right)dy}\\ =\pi {\left[\left(2y^2-12y\right)\right]}_3^4\\ =\pi \left[\left(2{\left(4\right)}^2-12\left(4\right)\right)-\left(2{\left(3\right)}^2-12\left(3\right)\right)\right]\\ =\pi \left(-16+18\right)\\ =2\pi {\text{ unit}}^3\end{array}$

Solution:

$\begin{array}{l}x=y^2-1\to \left(1\right)\\ 3y=2x\\ x=\frac{3}{2}y\to \left(2\right)\\ \text{Substitute (2) into (1),}\\ \frac{3}{2}y=y^2-1\\ 2y^2-3y-2=0\\ \left(2y+1\right)\left(y-2\right)=0\\ y=-\frac{1}{2}\text{ or }y=2\end{array}$


$\begin{array}{l}\text{When }y=2,x=\frac{3}{2}\left(2\right)=3,Q=\left(3,2\right)\\ \\ I_1\left(\text{Volume of cone}\right)\\ =\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(3\right)}^2\left(2\right)\\ =6\pi {\text{ unit}}^3\\ \\ I_2\left(\text{Volume of the curve}\right)\\ \text{= }\pi {\displaystyle {\int }_1^2{x}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_1^2{\left(y^2-1\right)}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_1^2\left(y^4-2y^2+1\right)dy}\\ =\pi {\left[\frac{y^5}{5}-\frac{2y^3}{3}+y\right]}_1^2\\ =\pi \left[\left(\frac{2^5}{5}-\frac{2{\left(2\right)}^3}{3}+2\right)-\left(\frac{1^5}{5}-\frac{2{\left(1\right)}^3}{3}+1\right)\right]\\ =\pi \left(\frac{46}{15}-\frac{8}{15}\right)\\ =\frac{38}{15}\pi {\text{ unit}}^3\\ \\ \therefore \text{ Volume generated }=I_1-I_2\\ =6\pi -\frac{38}{15}\pi \\ =\frac{52}{15}\pi {\text{ unit}}^3\end{array}$