Luas rantau berlorek, L
$\begin{array}{l}={\displaystyle {\int }_a^{b}ydx}\\ ={\displaystyle {\int }_0^4\left(6x-x^2\right)dx}\\ ={\left[\frac{6x^2}{2}-\frac{x^3}{3}\right]}_0^4\\ =\left[3{(4)}^2-\frac{{(4)}^3}{3}\right]-0\\ =26\frac{2}{3}{\text{unit}}^2\end{array}$

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

Soalan 14 (4 markah):
Rajah menunjukkan lengkung y = g(x). Garis lurus ialah tangen kepada lengkung itu.
Rajah

Diberi g’(x) = –4x + 8, cari persamaan lengkung itu.


Penyelesaian:
$\begin{array}{l}\text{Diberi }g\text{'}\left(x\right)=-4x+8\\ \text{Titik maximum apabila }g\text{'}\left(x\right)=0\\ -4x+8=0\\ 4x=8\\ x=2\\ \text{Maka, titik maximum ialah }\left(2,11\right).\\ \\ g\text{'}\left(x\right)=-4x+8\\ {\displaystyle \int g\text{'}\left(x\right)}={\displaystyle \int \left(-4x+8\right)}dx\\ g\left(x\right)=\frac{-4x^2}{2}+8x+c\\ g\left(x\right)=-2x^2+8x+c\\ \\ \text{Gantikan }\left(2,11\right)\text{ ke dalam }g\left(x\right):\\ 11=-2{\left(2\right)}^2+8\left(2\right)+c\\ c=3\\ \\ \text{Maka, persamaan lengkung ialah}\\ g\left(x\right)=-2x^2+8x+3\end{array}$

Soalan 15 (3 markah):
Diberi bahawa ${\displaystyle \int \frac{5}{{\left(2x+3\right)}^n}}dx=\frac{p}{{\left(2x+3\right)}^5}+c$ , dengan keadaan c, n dan p ialah pemalar.
Cari nilai n dan nilai p.

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

Soalan 10 (10 markah):
Rajah menunjukkan lengkung y = 2x² – 18 dan garis lurus PQ yang merupakan tangen kepada lengkung itu pada titik K.

Diberi bahawa kecerunan garis lurus PQ ialah 4.
(a) Cari koordinat titik K
(b) Hitung luas rantau berlorek.
(c) Apabila rantau yang dibatasi oleh lengkung, paksi-x dan garis lurus y = h diputarkan melalui 180^o pada paksi-y, isi padu kisaran ialah 65π unit³.  Cari nilai h.


Penyelesaian: 
(a)
$\begin{array}{l}y=2x^2-18\\ \frac{dy}{dx}=4x\\ \\ \text{Kecerunan garis lurus }PQ=4\\ 4x=4\\ x=1\\ \\ \text{Apabila }x=1,\\ y=2{\left(1\right)}^2-18=-16\\ \\ \text{Koordinat titik }K=\left(1,-16\right).\end{array}$


(b)
$\begin{array}{l}\text{Pada paksi-}x\text{, }y=0\\ \therefore 2x^2-18=0\\ 2x^2=18\\ x^2=9\\ x=\pm 3\\ \text{Lengkung itu memotong paksi-}x\text{ di titik }\left(-3,0\right)\text{ dan titik }\left(3,0\right)\text{.}\\ \\ \text{Luas rantau berlorek}\\ \text{= Luas }\Delta -\text{Luas rantau yang dibatasi oleh lengkung}\\ =\frac{1}{2}\left(5-1\right)\left(16\right)-{\displaystyle {\int }_1^{3}ydx}\\ =32-{\displaystyle {\int }_1^3\left(2x^2-18\right)dx}\\ =32-\left|{\left[\frac{2x^3}{3}-18x\right]}_1^3\right|\\ =32-\left|\left(\frac{2{\left(3\right)}^3}{3}-18\left(3\right)\right)-\left(\frac{2{\left(1\right)}^3}{3}-18\left(1\right)\right)\right|\\ =32-\left|\left(18-54-\frac{2}{3}+18\right)\right|\\ =32-\left|-18\frac{2}{3}\right|\\ =32-18\frac{2}{3}\\ =13\frac{1}{3}{\text{ unit}}^2\end{array}$


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

Soalan 9 (10 markah):
Rajah menunjukkan garis lurus 4y = x – 2 menyentuh lengkung x = y² + 6 pada titik P.

Rajah
Cari
(a) koordinat P,
(b) luas kawasan berlorek,
(c) isi padu kisaran, dalam sebutan π, apabila rantau yang dibatasi oleh lengkung dan garis lurus x = 8 dikisarkan melalui 180^o pada paksi-x.


Penyelesaian:
(a)
$\begin{array}{l}4y=x-\mathrm{2.........}(1)\\ x=y^2+\mathrm{6.........}(2)\\ \\ \text{Gantikan (2) ke dalam (1):}\\ 4y=\left(y^2+6\right)-2\\ y^2-4y+4=0\\ \left(y-2\right)\left(y-2\right)=0\\ y-2=0\\ y=2\\ \\ \text{Gantikan }y=2\text{ ke dalam (2):}\\ x={\left(2\right)}^2+6\\ x=10\\ \text{Oleh itu, }P=\left(10,2\right).\end{array}$

(b)
$\begin{array}{l}\text{Pada paksi-x, }y=0\\ \therefore 4y=x-2\\ 0=x-2\\ x=2\\ \\ \text{Luas kawasan berlorek}\\ \text{= Luas segi tiga}-\text{Luas rantau yang dibatasi oleh lengkung}\\ =\frac{1}{2}\left(10-2\right)\left(2\right)-{\displaystyle {\int }_6^{10}ydx}\\ =8-{\displaystyle {\int }_6^{10}\sqrt{x-6}dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y=\sqrt{x-6}\end{array}}\\ =8-{\displaystyle {\int }_6^{10}{\left(x-6\right)}^{\frac{1}{2}}dx}\\ =8-{\left[\frac{{\left(x-6\right)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_6^{10}\\ =8-{\left[\frac{2{\left(x-6\right)}^{\frac{3}{2}}}{3}\right]}_6^{10}\\ =8-\left[\frac{2{\left(10-6\right)}^{\frac{3}{2}}}{3}-\frac{2{\left(6-6\right)}^{\frac{3}{2}}}{3}\right]\\ =8-\frac{16}{3}\\ =\frac{8}{3}{\text{ unit}}^2\end{array}$

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

Soalan 8:
Rajah di bawah menunjukkan lengkung $y=\frac{4}{x^2}$ dan garis lurus y = mx + c. Garis lurus y = mx + c ialah tangen kepada lengkung pada (2, 1).
(a) Cari nilai m dan nilai c.

(b) Hitung luas kawasan berlorek.

(c) Diberi bahawa isi padu kisaran apabila rantau yang dibatasi oleh lengkung, paksi-x, garis lurus x = 2 dan x = h diputarkan melalui 360^o pada paksi-x ialah $\frac{38\pi }{81}{\text{ unit}}^3.$
Cari nilai h, dengan keadaan h > 2.


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


(b)
$\begin{array}{l}\text{Pada paksi-}x\text{, }y=0\\ \text{Dari garis lurus }y=-x+3,x=3\\ \\ \text{Luas kawasan berlorek}\\ =\text{Luas bawah lengkung}-\text{Luas segi tiga}\\ ={\displaystyle {\int }_2^{4}ydx-}\frac{1}{2}\times 1\times 1\\ ={\displaystyle {\int }_2^4\left(4x^{-2}\right)dx}-\frac{1}{2}\\ ={\left[\frac{4x^{-1}}{-1}\right]}_2^4-\frac{1}{2}\\ ={\left[-\frac{4}{x}\right]}_2^4-\frac{1}{2}\\ =\left[-\frac{4}{4}-\left(-\frac{4}{2}\right)\right]-\frac{1}{2}\\ =\frac{1}{2}{\text{ unit}}^2\end{array}$


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

Soalan 7:
Rajah di bawah menunjukkan suatu lengkung $y=\frac{1}{4}{x}^2+3$ yang menyilang suatu garis lurus y = x + 6 pada titik A.

(a) Cari koordinat A.
(b) ) hitung
(i) luas rantau berlorek M,
(ii) isipadu kisaran, dalam sebutan π, apabila rantau berlorek N diputarkan melalui 360^o pada paksi-y.


Penyelesaian:
(a)
$\begin{array}{l}y=\frac{1}{4}{x}^2+\mathrm{3..........}\left(1\right)\\ y=x+\mathrm{6..........}\left(2\right)\\ \text{Gantikan (2) ke dalam (1),}\\ x+6=\frac{1}{4}{x}^2+3\\ 4x+24=x^2+12\\ x^2-4x-12=0\\ \left(x+2\right)\left(x-6\right)=0\\ x=-2\text{ or }x=6\left(\text{ditolak}\right)\\ \\ \text{Apabila }x=-2\\ y=-2+6=4\\ \text{Oleh itu, }A=\left(-2,4\right).\end{array}$


(b)(i)
$\begin{array}{l}\text{Pada paksi-}x\text{, }y=0\\ \text{Dari }y=x+6,x=-6\\ \\ \text{Luas kawasan berlorek }M\\ =\text{Luas segi tiga}+\text{Luas di bawah lengkung}\\ =\frac{1}{2}\times \left(6-2\right)\times 4+{\displaystyle {\int }_{-2}^{0}ydx}\\ =8+{\displaystyle {\int }_{-2}^0\left(\frac{1}{4}{x}^2+3\right)dx}\\ =8+{\left[\frac{x^3}{4\left(3\right)}+3x\right]}_{-2}^0\\ =8+\left[0-\left(\frac{{\left(-2\right)}^3}{12}+3\left(-2\right)\right)\right]\\ =8+\left[0-\left(-\frac{8}{12}-6\right)\right]\\ =8+\left[0-\left(-\frac{20}{3}\right)\right]\\ =14\frac{2}{3}{\text{ unit}}^2\end{array}$


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

Soalan 11:
$\begin{array}{l}\text{Diberi}={\displaystyle {\int }_2^{5}g(x)dx=-2}\text{. Cari}\\ \text{(a) nilai bagi }{\displaystyle {\int }_5^{2}g(x)dx,}\\ \text{(b) nilai bagi }m\text{ jika }{\displaystyle {\int }_2^5\left[g(x)+m\left(x\right)\right]dx=19}\end{array}$

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

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

Soalan 12:
$\begin{array}{l}\text{a) Cari nilai bagi }{\displaystyle {\int }_{-1}^1{\left(3x+1\right)}^{3}dx.}\\ \text{(b) Nilaikan }{\displaystyle {\int }_3^4\frac{1}{\sqrt{2x-4}}}dx.\end{array}$

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

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

Soalan 13:
$\begin{array}{l}\text{Diberi }y=\frac{x^2}{2x-1},\text{ tunjukkan}\\ \frac{dy}{dx}\text{=}\frac{2x\left(x-1\right)}{{\left(2x-1\right)}^2}\text{. Seterusnya, nilaikan }{\displaystyle {\int }_{-2}^2\frac{x\left(x-1\right)}{4{\left(2x-1\right)}^2}dx}.\end{array}$

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

Soalan 8:
$\begin{array}{l}\text{Diberi }{\displaystyle {\int }_{-2}^{3}g(x)dx=4},\text{ dan }{\displaystyle {\int }_{-2}^{3}h(x)dx=9}\text{, cari nilai bagi}\\ \text{(a) }{\displaystyle {\int }_{-2}^{3}5g(x)dx,}\\ \text{(b) }m\text{ jika }{\displaystyle {\int }_{-2}^3\left[g(x)+3h\left(x\right)+4m\right]dx=12}\end{array}$

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

Soalan 9:
$\text{Diberi }y=\frac{5x}{x^2+1}\text{ dan }\frac{dy}{dx}=g\left(x\right),\text{ cari nilai bagi }{\displaystyle {\int }_0^{3}2g\left(x\right)dx.}$

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

Soalan 10:
$\text{Cari }{\displaystyle {\int }_5^k\left(x+1\right)dx,\text{ dalam sebutan }k.}$

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

Soalan 5:
$\begin{array}{l}\text{Diberi }{\displaystyle \int \left(6x^2+1\right)dx=m{x}^3+x}+c,\\ \text{dengan keadaan }m\text{ dan }c\text{ ialah pemalar, cari}\\ \text{(a) nilai }m\text{.}\\ \text{(b) nilai }c\text{ jika }{\displaystyle \int \left(6x^2+1\right)dx=13\text{ apabila }x=1.}\end{array}$

Penyelesaian:
(a)
$\begin{array}{l}{\displaystyle \int \left(6x^2+1\right)dx=m{x}^3+x}+c\\ \frac{6x^3}{3}+x+c=m{x}^3+x+c\\ 2x^3+x+c=m{x}^3+x+c\\ \text{Banding kedua-dua belah,}\\ \text{Maka, }m=2\end{array}$

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

Soalan 6:
$\begin{array}{l}\text{Diberi bahawa }{\displaystyle {\int }_5^{k}g(x)dx=6},\text{ dan }{\displaystyle {\int }_5^k\left[g\left(x\right)+2\right]dx}=14,\\ \text{cari nilai }k.\end{array}$

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

Soalan 7:
$\text{Diberi }{\displaystyle {\int }_k^2(4x+7)dx=28},\text{ hitung nilai yang mungkin bagi }k.$

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