Arithmetic Progression : Problems and Solutions - Part (2)

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1. If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ term of an A.P. are $a, b, c$ respectively, then show that $(a-b) r$ $+(b-c) p$ $+(c-a) q=0$.





Let the first term and the comnon difference of given A.P. be $A$ and $D$. By the proldem, $u_{p}=a$ $A+(p-1) D=a$ $u_{q}=b$ $A+(q-1) D=b$ $u_{r}=c$ $A+(r-1) D=c$ $\therefore\ (a-b) r=(p-q) D r$ $\hspace{2.2cm}=(p r-q r) D \ldots(1)$ $\quad\ (p-c) p=(q-r) D p$ $\hspace{2.2cm} =(p q-p r) D \ldots(2)$ $\quad\ (c-a) q =(r-p) D q$ $\hspace{2.2cm} =(q r-p q) D \ldots(3)$ Summing equations $(1),(2)$ and $(3)$ $(a-b) r+(b-c) p+(c-a) q=0$



2. Show that the sum of $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ term of an A.P is equal to twice the $m^{\text {th }}$ term.





Let the first tern be $a$ and the common difference be $d$ for the given A.P. $\therefore\ u_{m+n}=a+(m+n-1) d$ $\quad\ u_{m-n}=a+(m-n-1) d$ $\quad\ u_{m+n}+u_{m-n}=2 a+(2 n-1) d$ $\hspace{3.2cm} =2[a+(m-1) d]$ $\hspace{3.2cm} =2 u_{m}$



3. If $(m+1)^{\text {th }}$ term of an AP is twice the $(n+1)^{\text {th }}$ term, prove that $(3 m+1)^{\text {th }}$ term is twice the $(m+n+1)^{\text {th }}$ term.





Let the first tern be $a$ and the common difference be $d$ for the given A.P. By the problem, $u_{m+1}=2 u_{n+1}$ $a+m d=2(a+n d)$ $a+m d=2 a+2 n d$ $a=m d-2 n d$ $u_{m+n+1} =a+(m+n) d$ $\hspace{1.5cm}=m d-2 n d+m d+n d$ $\hspace{1.5cm}=2 m d-n d$ $\displaystyle u_{3 m+1} =a+3 m d$ $\hspace{1.5cm}=m d-2 n d+3 m d$ $\hspace{1.5cm}=4 m d-2 n d$ $\hspace{1.5cm}=2(2 m d-n d)$ $\hspace{1.5cm}=2 u_{m+n+1}$



4. The digits of a positive integer having three digits are in A.P. The sum of the digits is 15 and the number obtained by reversing the digits is 594 less than the original number. Find the number.





Let the hundredis digit, ten's digit and one's digit of a positive integer be $a, b$ and $c$ respectively. By the problem, $a, b, c$ are in A.P. $\therefore\ a=a$ $\quad\ b=a+d$ $\quad\ c=a+2 d$ $\quad\ a+b+c=15$ (given) $\quad\ 3 a+3 d=15$ $\therefore\ a+d=5\Rightarrow b=5$ $\therefore$ given integer $=100 a+10 b+c$ Original neumber - New formed nunber $=594$ $100 a+10 b+c-(100 c+10 b+a)=594$ $99 a-99 c=594$ $\quad\ a-c=6$ $\quad -2 d=6$ $\quad\quad d=-3$ $\therefore\ a-3=5$ $\quad\ a=8$ $\therefore\ c=2$ $\therefore$ The number is $852 .$



5. If $\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P., then prove that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in A.P.





$\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P. $\dfrac{c+a-b}{b}-\dfrac{b+c-a}{a}=\dfrac{a+b-c}{c}-\dfrac{c+a-b}{b}$ $\dfrac{c+a}{b}-1-\dfrac{b+c}{a}+1=\dfrac{a+b}{c}-1-\dfrac{c+a}{b}+1$ $\dfrac{c+a}{b}-\dfrac{b+c}{a}=\dfrac{a+b}{c}-\dfrac{c+a}{b}$ $\dfrac{a^{2}+a c-b^{2}-b c}{a b}=\dfrac{b^{2}+a b-c^{2}-a c}{b c}$ $\dfrac{a^{2}-b^{2}+a c-b c}{a}=\dfrac{b^{2}-c^{2}+a b-a c}{c}$ $\dfrac{(a-b)(a+b)+c(a-b)}{a}=\dfrac{(b-c)(b+c)+a(b-c)}{c}$ $\dfrac{(a-b)(a+b+c)}{a}=\dfrac{(b-c)(a+b+c)}{c}$ $\dfrac{a-b}{a}=\dfrac{b-c}{c}$ $\therefore\ a c-b c=a b-a c$ $\therefore\ \dfrac{a c}{a b c}-\dfrac{b c}{a b c}=\dfrac{a b}{a b c}-\dfrac{a c}{a b c}$ $\quad\ \dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b}$ $\therefore\ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ is an A.P.



6. If $a, b, c$ are in A.P., then prove that $(a-c)^{2}=4\left(b^{2}-a c\right)$.





$\quad\ a, b, c$ are in A.P. $\therefore\ b-a=c-b$ $\quad\ a+c=2 b$ $\quad\ a+c-2c=2 b-2 c$ $\quad\ a-c=2(b-c)$ $\quad\ (a-c)^{2}=4(b-c)^{2}$ $\quad\ (a-c)^{2} =4\left(b^{2}-2 b c+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-(a+c) c+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-a c-c^{2}+c^{2}\right) $ $\hspace{2.15cm}=4\left(b^{2}-a c\right)$



7. If $a, b, c$ are in A.P., then prove that $b+c, c+a, a+b$ are also in A.P.





$\quad\ a, b, c$ are in A.P. $\therefore \ b-a=c-b$ $\quad\ 2 b=c+a$ $\quad\ 2 b+c+a=c+a+c+a$ $\quad\ (b+c)+(a+b)=(c+a)+(c+a)$ $\therefore\ (c+a)-(b+c)=(a+b)-(c+a)$ $\therefore\ b+c, c+a, a+b$ are in A.P.



8. If $a, b, c$ are in A.P., then prove that $\dfrac{1}{b c}$, $\dfrac{1}{c a}$, $\dfrac{1}{a b}$ are also in A.P.





$\begin{aligned} & a, b, c \text{ are in A.P.}\\\\ \therefore\ &b-a =c-b\\\\ &\dfrac{b}{a b c}-\dfrac{a}{a b c}=\dfrac{c}{a b c}-\dfrac{b}{a b c}\\\\ &\dfrac{1}{a c}-\dfrac{1}{b c} =\dfrac{1}{a b}-\dfrac{1}{a c}\\\\ &\dfrac{1}{b c},\ \dfrac{1}{c a},\ \dfrac{1}{a b}\ \text{ are in A.P.} \end{aligned}$



9. If $a, b, c$ are in A.P., then prove that $(b+c-a)$,$(c+a-b)$,$(a+b-c)$ are in AP.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ &a-b=b-c \\\\ &2 a-2 b=2 b-2 c \\\\ &c+2 a-2 b-c=a+2 b-2 c-a \\\\ &c+a-b-b-c+a=a+b-c-c-a+b \\\\ &(c+a-b)-(b+c-a)=(a+b-c)-(c+a-b) \\\\ \therefore\ &b+c-a),\ (c+a-b),\ (a+b-c)\ \text { are in A.P.} \end{aligned}$



10. If $a, b, c$ are in A.P., then prove that $a^{2}(b+c)$, $b^{2}(c+a)$, $c^{2}(a+b)$ are also in A.P.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ \therefore\ &a+c=2 b \\\\ &a^{2}(b+c)+c^{2}(a+b) \\\\ =& a^{2} b+a^{2} c+c^{2} a+c^{2} b \\\\ =& a^{2} b+c a(a+c)+c^{2} b \\\\ =& a b+c a(2 b)+c^{2} b \\\\ =& a^{2} b+2 a b c+c b \\\\ =& a^{2} b +a b c+a b c+c^{2} b\\\\ =& a b(a+c)+b c(a+c) \\\\ =& a b(2 b)+b c(2 b) \\\\ =& 2 a b^{2}+2 b^{2} c \\\\ =& 2 b^{2}(c+a)\\\\ \therefore\ & 2 b^{2}(c+a)=a^{2}(b+c)+c^{2}(a+b) \\\\ \therefore\ & b^{2}(c+a)+b^{2}(c+a)=a^{2}(b+c)+c^{2}(a+b) \\\\ \therefore\ & b^{2}(c+a)-a^{2}(b+c)=c^{2}(a+b)-b^{2}(c+a) \\\\ \therefore\ & a^{2}(b+c), b^{2}(c+a), c^{2}(a+b) \text { are in A.P.} \end{aligned}$



11. If $a, b, c$ are in A.P., then prove that $b c-a^{2}$, $c a-b^{2}$, $a b-c^{2}$ are in AP.





$\begin{aligned} &a, b, c \text { are in A.P.} \\\\ \therefore\ &b-a=c-b \\\\ \therefore\ &(b-a)(a+b+c)=(c-b)(a+b+c) \\\\ &(b-a)(b+a)+(b-a) c=(c-b)(c+b)+(c-b) a \\\\ &b^{2}-a^{2}+b c-c a=c^{2}-b^{2}+c a-a b \\\\ &\left(b c-a^{2}\right)-\left(c a-b^{2}\right)=\left(c a-b^{2}\right)-\left(a b-c^{2}\right) \\\\ &\text{Multiply both sides with}\ -1,\\\\ &\left(c a-b\right)^{2}-\left(b c-a^{2}\right)=\left(a b-c^{2}\right)-\left(c a-b^{2}\right) \\\\ \therefore\ &b c-a^{2}, c a-b^{2}, a b-c^{2} \text { are in A.P.} \end{aligned}$



12. If $a, b, c$ are in A.P., then prove that $\dfrac{1}{\sqrt{b}+\sqrt{c}}$, $\dfrac{1}{\sqrt{c}+\sqrt{a}}$, $\dfrac{1}{\sqrt{a}+\sqrt{b}}$ are also in A.P.





$\begin{aligned} &a,\ b,\ c \text { are in A.P.} \\\\ \therefore\ & b-a=c-b\\\\ &(\sqrt{b})^{2}-(\sqrt{a})^{2}=(\sqrt{e})^{2}-(\sqrt{b})^{2}\\\\ &(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})=(\sqrt{c}-\sqrt{b})(\sqrt{c}+\sqrt{b})\\\\ &\dfrac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{c}}=\dfrac{\sqrt{c}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\\\\ &\dfrac{(\sqrt{b}+\sqrt{c})-(\sqrt{a}+\sqrt{c})}{\sqrt{b}+\sqrt{c}}=\dfrac{(\sqrt{a}+\sqrt{c})-(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}\\\\ &1-\dfrac{\sqrt{a}+\sqrt{c}}{\sqrt{b}+\sqrt{c}}=\dfrac{\sqrt{a}+\sqrt{c}}{\sqrt{a}+\sqrt{b}}-1\\\\ &\text { Dividing both sides with } \sqrt{a}+\sqrt{c}\\\\ &\dfrac{1}{\sqrt{a}+\sqrt{c}}-\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{1}{\sqrt{a}+\sqrt{c}}\\\\ \therefore\ &\dfrac{1}{\sqrt{b}+\sqrt{c}}, \dfrac{1}{\sqrt{a}+\sqrt{c}}, \dfrac{1}{\sqrt{a}+\sqrt{b}} \text { are in A.P } \end{aligned}$



13. If $a, b, c$ are in A.P., then prove that $a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)$, $b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)$, $c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$ are also in A.P.





$\begin{aligned} &a,\ b,\ c\ \text { ane in A.P.}\\\\ \therefore\ & b-a=c-b \\\\ \therefore\ & a+c=2 b \\\\ & a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ = & \dfrac{a}{b}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{c}{b} \\\\ = & \dfrac{a+c}{b}+\dfrac{a}{c}+\dfrac{c}{a} \\\\ = & \dfrac{2 b}{b}+\dfrac{a^{2}+c^{2}}{a c}\\\\ = & 2+\dfrac{a^{2}+c^{2}}{a c} \\\\ = & 2+\dfrac{(a+c)^{2}-2 a c}{a c} \\\\ = & 2+\dfrac{(a+c)^{2}}{a c}-2 \\\\ = & \dfrac{(a+c)^{2}}{a c} \\\\ = & (a+c)\left(\dfrac{a+c}{a c}\right) \\\\ = & 2 b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\\\\ \therefore\ & 2 b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)=a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ & b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)=a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \\\\ & b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)-a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)=c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-b\left(\dfrac{1}{c}+\dfrac{1}{a}\right) \\\\ \therefore\ & a\left(\dfrac{1}{b}+\dfrac{1}{c}\right), b\left(\dfrac{1}{c}+\dfrac{1}{a}\right), c\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \text { are in A.P.} \end{aligned}$



14. If $a^{2}, b^{2}, c^{2}$ are in A.P., then prove that $\dfrac{1}{b+c}$, $\dfrac{1}{c+a}$, $\dfrac{1}{a+b}$ are also in A.P.





$\begin{aligned} &a^{2}, b^{2}, c^{2} \text { are in A.P.} \\\\ \therefore\ &b^{2}-a^{2}=c^{2}-b^{2} \\\\ &(b+a)(b-a)=(c+b)(c-b) \\\\ &\dfrac{b-a}{b+c}=\dfrac{c-b}{a+b} \\\\ &\dfrac{(b+c)-(c+a)}{b+c}=\dfrac{(c+a)-(a+b)}{a+b} \\\\ &1-\dfrac{c+a}{b+c}=\dfrac{c+a}{a+b}-1\\\\ &\text { Dividing both sides with } c+a \\\\ &\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a} \\\\ \therefore\ &\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b} \text { are in A.P.} \end{aligned}$



15. If $a^{2}$, $b^{2}$, $c^{2}$ are in A.P., then prove that $\dfrac{a}{b+c}$, $\dfrac{b}{c+a}$, $\dfrac{c}{a+b}$ are also in A.P.





$\begin{aligned} &a^{2}, b^{2}, c^{2} \text { are in A.P.} \\\\ \therefore\ &b^{2}-a^{2}=c^{2}-b^{2} \\\\ &(b+a)(b-a)=(c+b)(c-b) \\\\ &\dfrac{b-a}{b+c}=\dfrac{c-b}{a+b} \\\\ &\dfrac{(b+c)-(c+a)}{b+c}=\dfrac{(c+a)-(a+b)}{a+b} \\\\ &1-\dfrac{c+a}{b+c}=\dfrac{c+a}{a+b}-1\\\\ &\text { Dividing both sides with } c+a \\\\ &\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a} \\\\ &\text { Multiplying both sides with } a+b+c,\\\\ &\dfrac{a+b+c}{c+a}-\dfrac{a+b+c}{b+c}=\dfrac{a+b+c}{a+b}-\dfrac{a+b+c}{c+a}\\\\ &\dfrac{c+a}{c+a}+\dfrac{b}{c+a}-\dfrac{a}{b+c}-\dfrac{b+c}{b+c}=\dfrac{a+b}{a+b}+\dfrac{c}{a+b}-\dfrac{c+a}{c+a}-\dfrac{b}{c+a}\\\\ &1+\dfrac{b}{c+a}-\dfrac{a}{b+c}-1=1+\dfrac{c}{a+b}-1-\dfrac{b}{c+a}\\\\ &\dfrac{b}{c+a}-\dfrac{a}{b+c}=\dfrac{c}{a+b}-\dfrac{b}{c+a}\\\\ &\dfrac{a}{b+c}, \dfrac{b}{c+a}, \dfrac{c}{a+b} \text { are in A.P. } \end{aligned}$



16. If the $m^{\text {th }}$ term of an A.P. is $\dfrac{1}{n}$ and $n^{\text {th }}$ term is $\dfrac{1}{m}$, then show that $u_{m n}=1$.





Let the first term and the common difference of the given A.P. be $a$ and $d$ respectively. $u_{m=} \dfrac{1}{n} $ $a+(m-1) d=\dfrac{1}{n}---(1) $ $u_{n}=\dfrac{1}{m} $ $a+(n-1) d=\dfrac{1}{m}---(2) $ $(1)-(2) \Rightarrow(m-n) d=\dfrac{1}{n}-\dfrac{1}{m}$ $(m-n) d=\dfrac{m-n}{m n}$ $d=\dfrac{1}{m n}$ $a+(m-1) \dfrac{1}{m n}=\dfrac{1}{n}$ $a=\dfrac{1}{n}-\dfrac{m-1}{m n}$ $\quad=\dfrac{m-m+1}{m n}$ $\quad=\dfrac{1}{m n}$ $u_{m n} =a+(m n-1) d $ $\quad\quad=\dfrac{1}{m n}+(m n-1) \dfrac{1}{m n} $ $\quad\quad=\dfrac{1}{m n}+1-\dfrac{1}{m n} $ $\quad\quad=1$



17. If the $p^{\text {th }}$ term of an A.P. is $q$ and the $q^{\text {th }}$ term is $p$, find its $n^{\text {th }}$ term in terms of $p, q$ and $n$.





$\begin{aligned} &\text{Let the first term}=a\\\\ &\text{the common difference}=d\\\\ &u_{p}=q \\\\ &a+(p-1) d=q---(1) \\\\ &u_{q}=p \\\\ &a+(q-1) d=p---(2) \\\\ &(1)-(2) \\\\ &(p-q) d=q-p \\\\ &(p-q) d=-(p-q)\\\\ &\therefore d=-1 \\\\ &\therefore a+(p-1)(-1)=q \\\\ &\begin{aligned} u_{n} &=a+(n-1) d \\\\ &=p+q-1+(n-1)(-1) \\\\ &=p+q-n \end{aligned} \end{aligned}$



18. If $\log _{10} 2, \log _{10}\left(2^{x}-1\right)$ and $\log _{10}\left(2^{x}+3\right)$ are three consecutive terms of an A.P., find the value of $x$.





$\begin{aligned} &\log _{10} 2, \log _{10}\left(2^{x}-1\right) \text { and } \log _{10}\left(2^{x}+3\right) \text { are in A.P.} \\\\ \therefore\ &\log _{10}\left(2^{x}-1\right)-\log _{10} 2=\log _{10}\left(2^{x}+3\right)-\log _{10}\left(2^{x}-1\right) \\\\ &\log _{10}\left(\frac{2^{x}-1}{2}\right)=\log _{10}\left(\frac{2^{x}+3}{2^{x}-1}\right) \\\\ &\frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1} \\\\ &\left(2^{x}-1\right)^{2}=2 \cdot 2^{x}+6 \\\\ &\left(2^{x}\right)^{2}-22^{x}+1=2 \cdot 2^{x}+6\\\\ &\left(2^{x}\right)^{2}-4 \cdot 2^{x}-5=0 \\\\ &\left(2^{x}+1\right)\left(2^{x}-5\right)=0 \\\\ & \text{For}\ 2^{x}=-1, \text { which is not possible for every } x \in \mathbb{R}.\\\\ &\text{For}\ 2^{x}=5, \\\\ \therefore\ & x=\log _{2} 5 \end{aligned}$

Arithmetic Progression : Problems and Solutions - Part (1)

Arithmetic Progression

An arithmetic progression is a sequence in which the difference between two consecutive terms is a constant. ကိန်းစဉ်တစ်ခု၏ နီးစပ် (ကပ်လျက်) ကိန်းနှစ်လုံး ခြားနားခြင်းသည် ကိန်းသေဖြစ်လျှင် ၎င်းကိန်းစဉ်ကို arithmetic progression ဟုခေါ်သည်။ That constant is called the common difference of the progression. If $u_{1}, u_{2}, u_{3}, \ldots u_{n-1}, u_{n}$ is an A.P., then $u_{2}-u_{1}=u_{3}-u_{2}=\ldots=u_{n}-u_{n-1}=$ constant $u_{n}-u_{n-1}=d$ and $u_{n}=u_{n-1}+d$ where $\boldsymbol{d}$ is called the common difference. The $n^{\text {th }}$ term of an $A . P$ is given by $\begin{array}{|l|}\hline u_{n}=a+(n-1)d \\ \hline\end{array}$ where $u_{n}=n^{\text {th }}$ term, $a=$ first term (or) $u_{1}$, $d=$ common difference $n=$ number of terms

Exercises

1. In each of the following A.P., find
   (a) the common difference (b) the $10^{\text {th }}$ term (c) the $n^{\text {th }}$ term.
   (i) $1,3,5,7, \ldots$
   (ii) $10,9,8,7, \ldots$
   (iii) $1,2 \dfrac{1}{2}, 4,5 \dfrac{1}{2}, \ldots$
   (iv) $20,18,16,14, \ldots$
   (v)$-25,-20,-15,-10, \ldots$
   (vi) $-\dfrac{1}{8}$,$-\dfrac{1}{4}$,$-\dfrac{3}{8}$,$-\dfrac{1}{2}$, $\ldots$





(i) $1,3,5,7, \ldots$ is an A.P. $\quad$ (a) $d=3-1=2$ $\quad$ (b) $a=1, d=2$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =1+9(2)$ $\hspace{2cm}=19$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=1+(n-1) 2$ $\hspace{2cm} =1+2 n-2$ $\hspace{2cm}=2 n-1$ (ii) $10,9,8,7, \ldots$ is an A.P. $\quad$ (a) $d=9-10=-1$ $\quad$ (b) $a=10, d=-1$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =10+9(-1)$ $\hspace{2cm}=1$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=10+(n-1) (-1)$ $\hspace{2cm} =10-n+1$ $\hspace{2cm}=11-n$ (iii) $1,2\dfrac{1}{2},4, 5\dfrac{1}{2}, \ldots$ is an A.P. $\quad$ (a) $d=2\dfrac{1}{2}-1=1\dfrac{1}{2}=\dfrac{3}{2}$ $\quad$ (b) $a=1, d=\dfrac{3}{2}$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =1+9\left(\dfrac{3}{2}\right)$ $\hspace{2cm}=\dfrac{29}{2}$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=1+(n-1)\left(\dfrac{3}{2}\right)$ $\hspace{2cm} =1+\dfrac{3n}{2}-\dfrac{3}{2}$ $\hspace{2cm}=\dfrac{1}{2}(3n-1)$ (iv) $20,18,18,14, \ldots$ is an A.P. $\quad$ (a) $d=18-20=-2$ $\quad$ (b) $a=20, d=-2$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =20+9(-2)$ $\hspace{2cm}=2$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=20+(n-1) (-2)$ $\hspace{2cm} =20-2 n+2$ $\hspace{2cm}=22-2n$ (v) $-25,-20,-15,-10, \ldots$ is an A.P. $\quad$ (a) $d=-20-(-25)=5$ $\quad$ (b) $a=-25, d=5$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =-25+9(5)$ $\hspace{2cm}=20$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=-25+(n-1)5$ $\hspace{2cm} =-25+5n -5$ $\hspace{2cm}=5n-30$ (vi) $-\dfrac{1}{8},-\dfrac{1}{4},-\dfrac{3}{8},-\dfrac{1}{2}, \ldots$ is an A.P. $\quad$ (a) $d=-\dfrac{1}{4}-\left(-\dfrac{1}{8}\right)=- \dfrac{1}{8}$ $\quad$ (b) $a=-\dfrac{1}{8}, d=-\dfrac{1}{8}$ $\hspace{1.2cm} u_{10} =a+9 d$ $\hspace{2cm} =-\dfrac{1}{8}+9\left(-\dfrac{1}{8}\right)$ $\hspace{2cm}=-\dfrac{5}{4}$ $\quad$ (c) $ u_{n}\hspace{.3cm} =a+(n-1) d$ $\hspace{2cm}=-\dfrac{1}{8}+(n-1)\left(-\dfrac{1}{8}\right)$ $\hspace{2cm} =-\dfrac{1}{8}-\dfrac{n}{8}+\dfrac{1}{8}$ $\hspace{2cm}=\dfrac{n}{8}$



2. The $5^{\text {th }}$ term of an arithmetic progression is 10 while the $15^{\text {th }}$ term is 40 . Write down the first 5 terms of the A.P.





$u_{5}=10$ $a+4 d=10 \ldots (1)$ $u_{15}=40$ $a+14 d=40 \ldots (2)$ $(2)-(1)\Rightarrow 10 d=30$ $\therefore \ d=3$ Substituting $d=3$ in equation (1),we get $\quad\ a=-2$ $\therefore\ u_{1}=-2$ $\quad\ u_{2}=u_{1}+d=-2+3=1$ $\quad\ u_{3}=u_{2}+d=1+3=4$ $\quad\ u_{4}=u_{3}+d=4+3=7$ $\quad\ u_{5}=u_{4}+d=7+3=10$ $\therefore$ The first 5 terms are $-2,1,4,7,10$.



3. The $5^{\text {th }}$ and $10^{\text {th }}$ terms of an A.P. are 8 and $-7$ respectively. Find the $100^{\mathrm{th}}$ and $500^{\text {th }}$ terms of the A.P.





In an $A P$, $u_{5}=8$ $a+4 d=8 \ldots(1)$ $u_{10}=-7$ $a+9 d=-7 \ldots(2)$ $(2)-(1) \Rightarrow 5 d=-15$ $\hspace{2.6cm}d=-3$ $\therefore\ a+ 4(-3)=8 $ $\quad\ a= 20 $ $u_{100} =a+99 d $ $\quad\quad =20+99(-3) $ $\quad\quad =-277 $ $u_{500} =a+499 d $ $\quad\quad =20+499(-3) $ $\quad\quad =-1477$



4. The sixth term of an A.P. is 32 while the tenth term is 48 . Find the common difference and the $21^{\text {st }}$ term.





In an $A P$, $u_{6}=32$ $a+5 d=32 \ldots(1)$ $u_{10}=48$ $a+9 d=48 \ldots(2)$ $(2)-(1) \Rightarrow 4 d=16$ $\hspace{2.6cm}d=4$ $\therefore\ a+ 5(4)=32 $ $\quad\ a= 12 $ $u_{21} =a+20 d $ $\quad\ =12+20(4) $ $\quad\ =92 $



5. Which term of the A.P. $6,13,20,27, \ldots$ is $111 ?$





$6,13,20,27, \ldots,$ is an $A P .$ $\therefore\ a=6$ $d=13-6=7$ Let $u_{n}=111$ $a+(n-1) d=111$ $6+(n-1) 7=111$ $(n-1) 7=105$ $n-1=15$ $n=16$ $\therefore u_{16}=117$



6. If $u_{1}=6$ and $u_{30}=-52$ in an A.P., find the common difference.





$u_{1} =6$ $\therefore\ a =6$ $u_{30} =-52$ $a+29 d =-52$ $6+29 d=-52$ $29 d =-58$ $d =-2$



7. In an A.P., $u_{1}=3$ and $u_{7}=39$. Find (a) the first five terms of the A.P. (b) the $20^{\text {th }}$ term of the A.P.





$u_{1}=3$ $\therefore\ a=3$ $u_{7}=39$ $a+6 d=39$ $3+6 d=39$ $6 d=36$ $\therefore d=6$ $u_{1} =3 $ $u_{2} =u_{1}+d=3+6=9 $ $u_{3} =u_{2}+d=9+6=15 $ $u_{4} =u_{3}+d=15+6=21 $ $u_{5} =u_{4}+d=21+6=27 $ $u_{20} =a+19 d $ $\quad\ =3+19 \times 6 $ $\quad\ =117$



8. The four angles of a quadrilateral are in A.P. Given that the value of the largest is three times the value of the smallest angle, find the values of all four angles.





Let the four angles be $\alpha, \beta, \gamma$, and $\delta$. By the problem, $\alpha, \beta, r, \delta$ is an $A P .$ $\therefore\ \beta=\alpha+d$ $r=\alpha+2 d$ $\delta=\alpha+3 d$ where $d$ is a common difference. Since $\alpha+\beta+\gamma+\delta=360^{\circ}$ $4 \alpha+6 d=360^{\circ}$ $2 \alpha+3 d=180^{\circ}\ldots(1)$ By the prolblem, $\delta=3 \alpha$ $\alpha+3 d=3 \alpha$ $\therefore\ 2 \alpha-3 d=0 \ldots(2)$ $(1)+(2) \Rightarrow 4 \alpha =90^{\circ} $ $\alpha =45^{\circ} $ $(1)-(2) \Rightarrow 6 d =90^{\circ} $ $d =30^{\circ} $ $\therefore\ \alpha=45^{\circ} $ $\beta=45^{\circ}+30^{\circ} =75^{\circ} $ $\gamma =45^{\circ}+60^{\circ}=105^{\circ} $ $\delta=45^{\circ}+90^{\circ} =135^{\circ}$



9. If the $n^{\text {th }}$ term of an A.P. $2,3 \dfrac{7}{8}, 5 \dfrac{3}{4}, \ldots$ is equal to the $n^{\text {th }}$ term of an A.P. $187$ , $184 \dfrac{1}{4}$, $181 \dfrac{1}{2}$, $\ldots$, find $n$.





$2,3 \dfrac{7}{8}, 5 \dfrac{3}{4}, \ldots$ is an A.P. $a=2, d=3 \dfrac{7}{8}-2=1 \dfrac{7}{8}=\dfrac{15}{8}$ $u_{n}=a+(n-1) d$ $u_{n}=2+(n-1) \dfrac{15}{8}$ $187,184 \dfrac{1}{4}, 18-\dfrac{1}{2} \ldots$ is an A.P $a=187$ $d=184 \dfrac{1}{4}-187=-2 \dfrac{3}{4}=-\dfrac{11}{4}$ $u_{n}=a+(n-1) d$ $\quad\ =187+(n-1)\left(-\dfrac{11}{4}\right)$ $\quad\ =187-(n-1) \dfrac{n}{4}$ By the problem, $2+(n-1) \dfrac{15}{8}=187-(n-1) \dfrac{n}{4}$ $(n-1) \dfrac{15}{8}+(n-1) \dfrac{11}{4}=185$ $(n-1)\left(\dfrac{15}{8}+\dfrac{11}{4}\right)=185$ $(n-1) \dfrac{37}{8}=185$ $n-1=40$ $n=41$



10. Show that $\dfrac{1}{1+x}, \dfrac{1}{1-x^{2}}, \dfrac{1}{1-x}$ are three consecutive terms of an A.P.





$\begin{aligned} \frac{1}{1-x^{2}}-\frac{1}{1+x} &=\frac{1}{1-x^{2}}-\frac{1}{1+x} \times \frac{1-x}{1-x^{2}} \\\\ &=\frac{1}{1-x^{2}}-\frac{1-x}{1-x^{2}} \\\\ &=\frac{x}{1-x^{2}} \\\\ \frac{1}{1-x}-\frac{1}{1-x^{2}}&=\frac{1}{1-x} \times \frac{1+x}{1+x}-\frac{1}{1-x^{2}} \\\\ &=\frac{1+x}{1-x^{2}}-\frac{1}{1-x^{2}} \\\\ &=\frac{x}{1-x^{2}}\\\\ \therefore\ \frac{1}{1-x^{2}}-\frac{1}{1+x}&=\frac{1}{1-x}-\frac{1}{1-x^{2}} \\\\ \therefore\ \frac{1}{1+x},\ \frac{1}{1-x^{2}},\ & \frac{1}{1-x} \text { is an A.P.} \end{aligned}$



11. The first three terms of an A.P. are $4 p^{2}-10,8 p$ and $4 p+3$ respectively. Find the two possible values of $p .$ If $p$ is positive and that the $n^{\text {th }}$ term of the progression is $-93$, find the value of $n$.





$4 p^{2}-10,8 p, 4 p+3$ are the first three terms of an A.P. $\therefore 8 p-\left(4 p^{2}-10\right)=4 P+3-8 p$ $4 p^{2}-12 p-7=0$ $(2 p+1)(2 p-7)=0$ $\quad p=-\dfrac{1}{2}$ or $p=\dfrac{7}{2}$ If $p>0, \quad p=\dfrac{7}{2}$ $\therefore a=4\left(\dfrac{7}{2}\right)^{2}-10=39$ $u_{2}=8\left(\dfrac{7}{2}\right)=28$ $d=28-39=-11$ $u_{n}=-93$ $a+(n-1) d=-93$ $39+(n-1)(-11)=-93$ $(n-1)(-11)=-132$ $n-1=12$ $n=13$



12. The $5^{\text {th }}$ term and $8^{\text {th }}$ terms of an A.P. are $x$ and $y$ respectively. Show that the $20^{\text {th }}$ term is $5 y-4 x$.





$\quad u_{5}=x$ $\quad a+4 d=x$ $\quad u_{8}=y$ $\quad a+7 d=y$ $\quad 5 y-4 x$ $=5 a+35 d-4 a-16 d$ $=a+19 d$ $=u_{20}$ $\therefore u_{20}=5 y-4 x$



13. Given that $\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are three consecutive terms of an A.P., show also that $a^{2}, b^{2}$ and $c^{2}$ are three consecutive terms of an A.P.





$\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are three consecutive terms of an A.P. $\therefore\ \dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a}$ $\quad\ \dfrac{b-a}{(c+a)(b+c)}=\dfrac{c-b}{(a+b)(c+a)}$ $\quad\ \dfrac{b-a}{b+c}=\dfrac{c-b}{a+b}$ $\therefore\ b^{2}-a^{2}=c^{2}-b^{2}$ $\therefore\ a^2, b^2, c^2$ are three consecutive terms of an A.P.



14. A certain A.P. has 25 terms. The last three terms are $\dfrac{1}{x-4}, \dfrac{1}{x-1}$ and $\dfrac{1}{x}$, Calculate the value $x$, and the middle term of that progression.





$ \dfrac{1}{x-1}-\dfrac{1}{x-4}=\dfrac{1}{x}-\dfrac{1}{x-1} $ $ \dfrac{x-4-x+1}{(x-1)(x-4)}=\dfrac{x-1-x}{x(x-1)} $ $ \dfrac{-3}{x-4}=\dfrac{-1}{x} $ $ \therefore\ 3 x=x-4 $ $ \quad\ 2 x=-4 $ $ \quad\ x=-2 $ $\therefore\ u_{23}=\dfrac{1}{x-4}=-\dfrac{1}{6} $ $ \quad\ u_{24}=\dfrac{1}{x-1}=-\dfrac{1}{3} $ $ \quad\ u_{25}=\dfrac{1}{x}=-\dfrac{1}{2}$ $\quad\ d=-\dfrac{1}{3}+\dfrac{1}{6}=-\dfrac{1}{6}$ $\quad\ u_{23}=-\dfrac{1}{6}$ $\quad\ a+22 d=-\dfrac{1}{6}$ $\therefore\ a=\dfrac{21}{6} $ $\therefore\ \text { middle term } =u_{13} $ $\hspace{3.2cm}=a+12 d $ $\hspace{3.2cm}=\dfrac{21}{6}+12\left(-\dfrac{1}{6}\right) $ $\hspace{3.2cm}=\dfrac{3}{2}$ $\hspace{3.5cm}\mathrm{OR}$ $\quad\ \text { middle term } =\dfrac{u_{1}+u_{25}}{2}$ $\hspace{3.2cm}=\dfrac{\dfrac{21}{6}-\dfrac{1}{2}}{2}$ $\hspace{3.2cm}=\dfrac{3}{2} $



15. Given that $x^{2},(8 x+1)$ and $(7 x+2)$, where $x \neq 0$, are the $2^{\text {nd }}, 4^{\text {th }}$ and $6^{\text {th }}$ terms respectively of an A.P. Find the value of $x$, the common difference and the first term.





$x^{2}, 8 x+1,7 x+2$ are $2^{\text{nd}}, 4^{\text{th}}$ and $6^{\text{th}}$ terms of an A.P. $\therefore\ 8 x+1-x^{2}=7 x+2-(8 x+1)$ $\quad\ x^{2}-9 x=0$ $\quad\ x(x-9)=0$ Since $x \neq 0, x-9=0,$ $\therefore\ x=9 .$ $\therefore\ u_{2}=81$ $\quad\ a+d=81$ ---(1) $\quad\ u_{4}=73$ $\quad\ a+3 d=73$ --- (2) $(2)-(1) \Rightarrow 2 d=-8$ $\hspace{2.6cm} d=-4$ $\therefore\ a-4=81 $ $\quad\ a=85$



16. If $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in A.P., express $b$ in terms of $a$ and $c$.





$\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ ane in A.P. $\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b}$ $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$ $\dfrac{2}{b}=\dfrac{a+c}{a c}$ $b=\dfrac{2 a c}{a+c}$



17. If $m$ times the $m^{\text {th }}$ term of an A.P. is equal to $n$ times the $n^{\text {th }}$ term where $m \neq n$ find $(m+n)^{\text {th }}$ term of the progression.





Let $a$ and $d$ be the first term and the common difference of given A.P. By the problem, $\quad\ m u_{m}=n u_{n}$ $\quad\ m(a+(m-1) d)=n(a+(n-1) d)$ $\quad\ m a+\left(m^{2}-n\right) d=n a+\left(n^{2}-n\right) d$ $\quad\ m a-n a+\left(m^{2}-n^{2}-m+n\right) d=0$ $\quad\ (m-n) a+[(m+n)(m-n)-(m-n)] d=0$ $\quad\ (m-n)[a+(m+n-1) d]=0$ $\quad\ $ Since $m \neq n, m-n \neq 0$ $\therefore\ a+(m+n-1) d=0$ $\therefore\ u_{m+n}=0$



18. If $a^{2}+2 b c, b^{2}+2 a c, c^{2}+2 a b$ are in A.P., show that $\dfrac{1}{b-c}$, $\dfrac{1}{c-a}$, $\dfrac{1}{a-b}$ are in A.P.





$a^{2}+2 b c, b^{2}+2 a c, c^{2}+2 a b$ are in A.P. $\therefore\ b^{2}+2 a c-a^{2}-2 b c=c^{2}+2 a b-b^{2}-2 a c$ $\quad\ b^{2}-a^{2}+a c(a-b)=c^{2}-b^{2}+2 a(b-c)$ $\quad\ (b-a)(b+a)-2 c(b-a)=(c-b)(c+b)-2 a(c-b)$ $\quad\ (b-a)(b+a-2 c)=(c-b)(c+b-2 a)$ $\quad\ -(a-b)(a+b-2 c)=-(b-c)(b+c-2 a)$ $\quad\ (a-b)(2 c-a-b)=(b-c)(2 a-b-c)$ $\quad\ (a-b)[(c-a)+(c-b)]=(b-c)[(a-b)+(a-c)]$ $\quad\ (a-b)[(c-a)-(b-c)]=(b-c)[(a-b)-(c-a)]$ $\quad\ (a-b)(c-a)-(a-b)(b-c)=(b-c)(a-b)-(b-c)(c-a)$ Dividing both sides with $(a-b)(b-c)(c-a)$, $\quad\ \dfrac{1}{b-c}-\dfrac{1}{c-a}=\dfrac{1}{c-a}-\dfrac{1}{a-b}$ $\quad\ \dfrac{1}{c-a}-\dfrac{1}{b-c}=\dfrac{1}{a-b}-\dfrac{1}{c-a}$ $\therefore\ \dfrac{1}{b-c}, \dfrac{1}{c-a}, \dfrac{1}{a-b}$ are in A.P.

Definite Integral : Area Between Curves

Integration နှင့် ပတ်သက်ပြီး အောက်ပါ သင်ခန်းစာ များကို ရေးသားခဲ့ပါသည်။ ယခုသင်ခန်းစာကို ဖော်ပြပါသင်ခန်းစာများနှင့် တွဲဖက်လေ့လာရမည် ဖြစ်ပါသည်။

1. AREA UNDER A CURVE
2. ANTI-DERIVATIVE
3. PRACTICE PROBLEMS FOR INDEFINITE INTEGRATION
4. INTEGRATION OF $\dfrac{1}{x}$ AND $\dfrac{1}{ax+b}$ EXPONENTIAL FUNCTION
5. INTEGRATION BY SUBSTITUTION
6. INDEFINITE INTEGRATION

Problems

1. The diagram shows parts of the line $y=3 x+10$ and the curve $y=x^{3}-5 x^{2}+3 x+10$. The line and the curve both pass through the point $A$ on the $y$ -axis. The curve has a maximum at the point $B$ and a minimum at the point $C$. The line through $C$, parallel to the $y$ -axis, intersects the line $y=3 x+10$ at the point $D$.
   • Show that the line $A D$ is a tangent to the curve at $A$.
   • Find the $x$ -coordinate of $B$ and of $C$.
   • Find the area of the shaded region $A B C D$, showing all your working.





Line: $y=3 x+10$ When the line cuts $y$ -axis, $x=0$ $\therefore y=10$ Hence, the coordinate of the point $A$ is $(0,10)$ Curve: $y=x^{3}-5 x^{2}+3 x+10$ $y^{\prime}=3 x^{2}-10 x+3$ At $x=0, y^{\prime}=3$ The equation of tangent to the curve at $(0,10)$ is $y-10=3 x$ $y=3 x+10$ At turning points, $y^{\prime}=0$ $\therefore \quad 3 x^{2}-10 x+3=0$ $(3 x-1)(x-3)=0$ When $x=\dfrac{1}{3}, y=\dfrac{283}{27}$ When $x=3, y=1$ $y^{\prime \prime}=9 x-10$ When $x=\dfrac{1}{3}, y^{\prime \prime}=-7<0$ When $x=3, y^{\prime \prime}=17>0$ $\therefore \quad B=\left(\dfrac{1}{3}, \dfrac{283}{27}\right)$ and $C=(3,1)$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{0}^{3}\left[3 x+10-\left(x^{3}-5 x^{2}+3 x+10\right)\right]\ d x$ $=\displaystyle\int_{0}^{3}\left(-x^{3}+5 x^{2}\right)\ d x$ $=\left[-\dfrac{x^{4}}{4}+\dfrac{5 x^{3}}{3}\right]_{0}^{3}$ $=45-\dfrac{8-1}{4}$ $=\dfrac{180-81}{4}$ $=\dfrac{99}{4}$




2. The graph of $y=x^{2}-6 x+10$ cuts the $y$ -axis at $A$. The graphs of $y=x^{2}-6 x+10$ and $y=x+10$ cut one another at $A$ and $B$. The line $B C$ is perpendicular to the $x$ -axis. Calculate the area of the shaded region enclosed by the curve and the line $A B$, showing all your working.





Line: $y_{1}=x+10$ Curve: $y=x^{2}-6 x+10$ When the line intersect the curves, $y_{1}=y_{2}$ $\therefore\ x^{2}-6 x+10=x+10$ $x^{2}-7 x=0$ $x(x-7)=0$ $x=0$ or $x=7$ When $x=0, y=10$ When $x=7, y=17$ $\therefore A=(0,10)$ and $B=(7,17) .$ $\quad$ Area of shaded region $=\displaystyle\int_{0}^{7}\left(y_{1}-y_{2}\right)\ d x$ $=\displaystyle\int_{0}^{7}\left(7 x-x^{2}\right)\ d x$ $=\left[\dfrac{7 x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{0}^{7}$ $=\dfrac{7^{3}(3-2)}{6}=\dfrac{343}{6}$




3. The diagram shows the graph of $y=\sqrt{4+x}$, which meets the $y$ -axis at the point $A$ and the line $x=5$ at the point $B$. Find the area of the region enclosed by the curve and the straight line $A B$.





Curve: $y_{1}=\sqrt{4+x}$ When $x=0, \quad y=\sqrt{4}=2$ When $x=5, \quad y=\sqrt{9}=3$ $\therefore A=(0,2)$ and $B=(5,3)$ $\therefore$ Equation of $A B$ is $y_{2}=\dfrac{3-2}{5-0} x+2$ $\quad =\dfrac{x}{5}+2$ $\begin{aligned} & \text { Area of required region } \\\\ =& \displaystyle\int_{0}^{5}\left(y_{1}-y_{2}\right)\ d x \\\\ =& \displaystyle\int_{0}^{5}\left(\sqrt{4+x}-\dfrac{x}{5}-2\right)\ d x \\\\ =& \displaystyle\int_{0}^{5}(4+x)^{1/2}\ d x-\displaystyle\int_{0}^{5} \dfrac{x}{5} d x-\displaystyle\int_{0}^{5} 2\ d x\\\\ =& \displaystyle\int_{0}^{5}(4+x)^{1 / 2}\ d(4+x)-\displaystyle\int_{0}^{5} \dfrac{x}{5}\ d x-\displaystyle\int_{0}^{5} 2\ d x[\because d(4+x)=d x] \\\\ =& \left[\dfrac{2}{3}(4+x)^{3 / 2}\right]_{0}^{5}-\left[\dfrac{x^{2}}{10}\right]_{0}^{5}-[2 x]_{0}^{5} \\\\ =& \dfrac{2}{3}\left[9^{3 / 2}-4^{3 / 2}\right]-\dfrac{25}{10}-10 \\\\ =& \dfrac{2}{3}[27-8]-\dfrac{5}{2}-10 \\\\ =& \dfrac{38}{3}-\dfrac{25}{2}=\dfrac{1}{6} \end{aligned}$




4. The diagram shows part of the graph of $y=1-2 \cos 3 x$, which crosses the $x$ -axis at the point $A$ and has a maximum at the point $B$.
   • Find the coordinates of $A$.
   • Find the coordinates of $B$.
   • Showing all your working, find the area of the shaded region bounded by the curve, the $x$ -axis and the perpendicular from $B$ to the $x$ -axis.





$y=1-2 \cos 3 x$ When $y=0,$ $1-2 \cos 3 x=0$ $\cos 3 x=\dfrac{1}{2}$ By the diagram, $3 x=\dfrac{\pi}{3}$ $x=\dfrac{\pi}{9}$ $\therefore \quad A=\left(\dfrac{\pi}{9}, 0\right)$ Since $-1 \le \cos 3 x \le 1$ The maximum value of $y=3$, when $\cos 3 x=-1$ By the diagran, $3 x=\pi$ $x=\dfrac{\pi}{3}$ $\therefore B=\left(\dfrac{\pi}{3}, 3\right)$ $\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}}(1-2 \cos 3 x) \ d x \\\\ =& \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}} 1 \ d x-\dfrac{2}{3} \displaystyle\int_{\frac{\pi}{9}}^{\frac{\pi}{3}} \cos 3 x\ d(3 x) \\\\ =&[x]_{\frac{\pi}{9}}^{\frac{\pi}{3}}-\left[\dfrac{2}{3} \sin 3 x\right]_{\frac{\pi}{9}}^{\frac{\pi}{3}}\\\\ =&\left(\dfrac{\pi}{3}-\dfrac{\pi}{9}\right)-\dfrac{2}{3}\left[\sin \pi-\sin \dfrac{\pi}{3}\right] \\\\ =&\dfrac{2 \pi}{9}-\dfrac{2}{3}\left[0-\dfrac{\sqrt{3}}{2}\right] \\\\ =&\dfrac{2 \pi}{9}+\dfrac{\sqrt{3}}{3} \end{aligned}$




5. The diagram shows part of the curve $y=3 x-x^{\frac{3}{2}}$ and the lines $y=3 x$ and $2 y=27-3 x$. The curve and the line $y=3 x$ meet the $x$ -axis at $O$ and the curve and the line $2 y=27-3 x$ meet the $x$ -axis at $A$.
   • Find the coordinates of $A$.
   • Verify that the coordinates of $B$ are $(3,9)$
   • Find the area of the shaded region.





Line1: $y_{1}=3 x$ Line2: $2 y_{2}=27-3 x$ When line 2 cuts $x$ -axis, $\quad\quad y_{2}=0$ $\therefore\ 27-3 x=0$ $\hspace{1.7cm} x=3 $ $\therefore\ A=(9,0)$ At the point of intersection of Line 1 and line 2, $\quad\quad y_{1}=y_{2}$ $\therefore\ 2(3 x)=27-3 x$ $\quad\quad 9 x=27$ $\quad\quad x=3$ $\therefore\ y=3(3)=9$ $\therefore\ B=(3,9)$ $\quad\quad$ Area of shaded region $\begin{aligned} &= \text { area of } \triangle O A B - \text { area under curve } \\\\ &= \dfrac{1}{2}(9)(9)-\displaystyle\int_{0}^{9}\left(3 x-x^{\frac{3}{2}}\right)\ d x \\\\ &= \dfrac{81}{2}-\left[\dfrac{3 x^{2}}{2}-\dfrac{2 x^{5 / 2}}{5}\right]_{0}^{9} \\\\ &= \dfrac{81}{2}-\dfrac{3(81)}{2}+\frac{2(243)}{5} \\\\ &= \dfrac{81}{5} \end{aligned}$




6. The diagram shows part of the graph of $y=2 \cos \left(x-\dfrac{\pi}{6}\right)$. The graph intersects the $y$ -axis at the point $A$, has a maximum point at $B$ and intersects the $x$ -axis at the point $C$.
   • Find the coordinates of $A$.
   • Find the coordinates of $B$.
   • Find the coordinates of $C$.
   • Find $\displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right) \mathrm{d} x$.
   • Hence find the area of the shaded region.





$y=2 \cos \left(x-\dfrac{\pi}{6}\right)$ When $x=0$, $y=2 \cos \left(-\dfrac{\pi}{6}\right)$ $=\sqrt{3}$ $\therefore\ A=(0, \sqrt{3})$ The maximum value of $y=2 \cos \left(x-\dfrac{\pi}{6}\right)$ is $2$ when $\cos \left(x-\dfrac{\pi}{6}\right)=1$ $x-\dfrac{\pi}{6}=0$ $x=\dfrac{\pi}{6}$ $B=\left(\dfrac{\pi}{6}, 0\right)$ When $y=0,$ $2 \cos \left(x-\dfrac{\pi}{6}\right)=0$ $\cos \left(x-\dfrac{\pi}{6}\right)=0$ $x-\dfrac{\pi}{6}=\dfrac{\pi}{2}$ $x=\dfrac{\pi}{2}+\dfrac{\pi}{6}$ $\therefore x=\dfrac{2 \pi}{3}$ $\therefore C=\left(\dfrac{2 \pi}{3}, 0\right)$ $\displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x$ let $u=x-\dfrac{\pi}{6}$ $\therefore\ d u=d x$ $\begin{aligned} \displaystyle\int 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x =&\displaystyle\int 2 \cos u\ d u \\\\ =&2 \sin u+C, c=\text { constant } \\\\ =&2 \sin \left(x-\dfrac{\pi}{6}\right)+C \end{aligned}$ $\begin{aligned} & \text { Area of shaded region } \\\\ =& \displaystyle\int_{0}^{\dfrac{2 \pi}{3}} 2 \cos \left(x-\dfrac{\pi}{6}\right)\ d x \\\\ =&\left[2 \sin \left(x-\dfrac{\pi}{6}\right)\right]_{0}^{\frac{2 \pi}{3}} \\\\ =& 2\left[\sin \dfrac{\pi}{2}-\sin \left(-\dfrac{\pi}{6}\right)\right]\\\\ =& 2\left[\sin \dfrac{\pi}{2}+\sin \dfrac{\pi}{6}\right] \\\\ =& 2\left[1+\dfrac{1}{2}\right]= 3 \\\\ \end{aligned}$




7. The diagram shows part of the curve $y=5+4 \tan ^{2}\left(\dfrac{x}{3}\right)$. Hence, find the exact area of the shaded region enclosed by the curve, the $x$ -axis and the lines $x=\dfrac{\pi}{2}$ and $x=\pi$





$\begin{aligned} y&=5+4 \tan ^{2}\left(\dfrac{x}{3}\right) \\\\ &=5+4\left(\sec ^{2}\left(\dfrac{x}{3}\right)-1\right) \\\\ &=4 \sec ^{2}\left(\dfrac{x}{3}\right)+1 \end{aligned}$ $\begin{aligned} &\text { Area of shaded region } \\\\ = & \displaystyle\int_{\dfrac{\pi}{2}}^{\pi}\left[5+4 \tan ^{2}\left(\dfrac{x}{3}\right)\right]\ d x \\\\ = & \displaystyle\int_{\dfrac{\pi}{2}}^{\pi}\left(4 \sec ^{2}\left(\dfrac{x}{3}\right)+1\right)\ d x\\\\ = & 4 \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \sec ^{2}\left(\dfrac{x}{3}\right)\ d x+\displaystyle\int_{\frac{\pi}{2}}^{\pi}1\ d x \\\\ = & \left[\dfrac{4}{3} \tan \left(\dfrac{x}{3}\right)\right]_{\frac{\pi}{2}}^{\pi}+\left[x\right]_{\frac{\pi}{2}}^{\pi} \\\\ = & \frac{4}{3}\left[\tan \dfrac{\pi}{3}-\tan \dfrac{\pi}{6}\right]+\left[\pi-\dfrac{\pi}{2}\right] \\\\ = & 4\left[\sqrt{3}-\dfrac{\sqrt{3}}{3}\right]+\dfrac{\pi}{2} \\\\ = & 8 \sqrt{3}+\dfrac{\pi}{2} \end{aligned}$




8. The diagram shows part of the curve $y=x^{3}+4 x^{2}-5 x+5$ and the line $y=5$. The curve and the line intersect at the points $A, B$ and $C .$ The points $D$ and $E$ are on the $x$ -axis and the lines $A E$ and $C D$ are parallel to the $y$ -axis.Calculate the total area of the shaded regions enclosed between the line and the curve. You must show all your working.





$y=x^{3}+4 x^{2}-5 x+5, y_{2}=5$ When $y=y_{2}$ $x^{3}+4 x^{2}-5 x+5=5$ $x^{3}+4 x^{2}-5 x=0$ $x\left(x^{2}+4 x-5\right)=0$ $\therefore x(x+5)(x-1)=0$ $\therefore x=-5$ or $x=0$ or $x=7$ $\quad\quad$ area of shaded region $\begin{aligned} &=\displaystyle\int_{-5}^{0}\left(y_{1}-y_{2}\right)\ d x+\displaystyle\int_{0}^{1}\left(y_{2}-y_{1}\right)\ d x\\\\ &=\displaystyle\int_{-5}^{0}\left(x^{3}+4 x^{2}-5 x\right)\ d x+\int_{0}^{1}\left(-x^{3}-4 x^{2}+5 x\right)\ d x \\\\\\ &=\left[\dfrac{1}{4} x^{4}+\dfrac{4}{3} x^{3}-\dfrac{5}{2} x^{2}\right]_{-5}^{0}+\left[-\dfrac{1}{4} x^{3}-\dfrac{4}{3} x+\dfrac{5}{2} x^{2}\right]_{0}^{1} \\\\ &=-\left[\dfrac{1}{4}(625)+\dfrac{4}{3}(-125)-\dfrac{5}{2}(25)\right]+\left[-\dfrac{1}{4}-\dfrac{4}{3}+\dfrac{5}{2}\right] \\\\ &=\dfrac{443}{6} \end{aligned}$




9. The diagram shows the curve $y=3 x^{2}-2 x+1 $ and the straight line $y=2 x+5$ intersecting at the points $P$ and $Q$. Showing all your working, find the area of the shaded region.





$y_{1}=2 x+5$ $y_{2}=3 x^{2}-2 x+1$ When $y_{1}=y_{2}$, $2 x+5=3 x^{2}-2 x+1$ $\therefore\ 3 x^{2}-4 x-4=0$ $(3 x+2)(x-2)=0$ $x=-\dfrac{2}{3}$ or $x=2$ When $x=-\dfrac{2}{3}, y=-\dfrac{4}{3}+5=\dfrac{11}{3}$ When $x=2, y=4+5=9$ $\therefore P=\left(-\dfrac{2}{3}, \dfrac{11}{3}\right)$ and $Q=(2,9)$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{-2 / 3}^{2}\left(y_{1}-y_{2}\right)\ d x$ $=\displaystyle\int_{-2 / 3}^{2}\left(-3 x^{2}+4 x+4\right)\ d x$ $=\left[-x^{3}+2 x^{2}+4 x\right]_{-2 / 3}^{2}$ $=-8+8+8-\left(\dfrac{8}{27}+\dfrac{8}{9}-\dfrac{8}{3}\right)$ $=\dfrac{256}{27}$




10. The figure shows part of the curve $y=3 x(x-2) .$ Find the total area enclosed by the curve $y=3 x(x-2)$, the $x$ -axis and the line $x=3$





$y=3 x(x-2)=3 x^{2}-6 x$ When $y=0$, $3 x(x-2)=0$ $\therefore x=0$ or $x=2$ The curve $y=3 x(x-2)$ cuts $x$ -axis at $(0,0)$ and $(2,0)$ $\quad\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{0}^{2}-y d x+\int_{2}^{3} y d x \\\\ =& \displaystyle\int_{0}^{2}\left(-3 x^{2}+6 x\right) d x+\displaystyle\int_{2}^{3}\left(3 x^{2}-6 x\right) d x \\\\ =&\left[-x^{3}+3 x^{2}\right]_{0}^{2}+\left[x^{3}-3 x^{2}\right]_{2}^{3} \\\\ =&[-8+12]+[27-27-8+12] \\\\ =& 8 \end{aligned}$




11. The diagram shows the curve $y=4+3 x-x^{2}$ intersecting the positive $x$ -axis at the point $A$. The line $y=m x+8$ is a tangent to the curve at the point $B$. Find
    • the coordinates of $A$,
    • the value of $m$
    • the coordinates of $B$,
    • the area of the shaded region, showing all your working.





Curve: $y_{1}=4+3 x-x^{2}$ Line: $y_{2}=m x+8$ When the curve ents $x$ -axis, $y=0$ $\therefore\ 4+3 x-x^{2}=0$ $(1+x)(4-x)=0$ $\therefore\ x=-1$ or $x=4$ $\therefore\ A=(4,0)$ Gradient of tangent to the curve $=y_{1}^{\prime}=3-2 x$ $\therefore m=3-2 x$ At the point of contact, $y_{1}=y_{2}$ $\therefore 4+3 x-x^{2}=(3-2 x) x+8$ $x^{2}-4=0$ $x^{2}=4$ $x=\pm 2$ By the diagran, $\quad\quad x=2$ $\therefore\ m=3-2(2)=-1$ $y_{2}=8-x$ When $x=2$ $y=-1(2)+8$ $\quad =6$ $\therefore\ B=(2,6)$ When the tangent cuts $x$ -axis, $8-x=0 \Rightarrow x=8$. $\quad\quad$ Area of shaded region $\begin{aligned} =& \displaystyle\int_{2}^{4}\left(y_{2}-y_{1}\right)\ d x+\int_{4}^{8} y_{2}\ d x \\\\ =& \displaystyle\int_{2}^{4}\left(x^{2}-4 x+4\right)\ d x+\int_{4}^{8}(8-x)\ d x \\\\ =&\left[\dfrac{1}{3} x^{3}-2 x^{2}+4 x\right]_{2}^{4}+\left[8 x-\dfrac{1}{2} x^{2}\right]_{4}^{8} \\\\ =&\left[\dfrac{64}{3}-32+16-\dfrac{8}{3}+8-8\right]+[64-32-32+8] \\\\ =& \dfrac{32}{3} \end{aligned}$




12. The diagram shows part of the curve $y=\dfrac{1}{x+1}$. The shaded region $R$ is bounded by the curve and by the lines $x=1, y=0$ and $x=p$.
    • Find, in terms of $p$, the area of $R$.
    • Hence find, correct to 1 decimal place, the value of $p$ for which the area of $R$ is equal to 2 .





$\quad\quad$ Area of $R$ $= \displaystyle\int_{1}^{p} \dfrac{1}{x+1}\ d x$ $=[\ln |x+1|]_{1}^{p}$ $= \ln |p+1|-\ln 2$ By the problem, $\ln |p+1|-\ln 2=2$ $\ln |p+1|=2+\ln 2$ $p+1=e^{2+\ln 2}$ $\therefore\ p =2 e^{2}-1=13.8$




13. The diagram, which is not drawn to scale, shows part of the graph of $y=8-\mathrm{e}^{2 x}$, crossing the $y$ -axis at $A$. The tangent to the curve at $A$ crosses the $x$ -axis at $B$. Find the area of the shaded region bounded by the curve, the tangent and the $x$ -axis.





Curve: $y=8-e^{2 x}$ When $x=0$ $y=8-e^{0}=7$ $\therefore\ A=(0,7)$ Gradient of tangent to the curve is $y^{\prime}=-2 e^{2 x}$ When $x=0, y^{\prime}=-2$ $\therefore$ Equation of targent at $A$ is $y=-2 x+7$ When the tangent cuts $x$ -axis, $\quad y=0$ $-2 x+7=0$ $x=\dfrac{7}{2}$ $\therefore B=\left(\dfrac{7}{2}, 0\right)$ $8-e^{2 x}=0$ $e^{2 x}=8$ $2 x=\ln 8$ $x=\dfrac{\ln 8}{2}$ $\quad\quad$ Area of shaded region $=\displaystyle\int_{0}^{\frac{\ln 8}{2}}\left(-2 x+7-8+e^{2 x}\right)\ d x+\displaystyle\int_{\frac{\ln 8}{2}}^{7 / 2}(-2 x+7)\ d x$ $=\left[-x^{2}-x+\dfrac{1}{2} e^{2 x}\right]_{0}^{\frac{\ln 8}{2}}+\left[-x^{2}+7 x\right]_{\frac{\ln 8}{2}}^{7 / 2}$ $=-\left(\dfrac{\ln 8}{2}\right)^{2}-\dfrac{\ln 8}{2}+4-\dfrac{1}{2}-\dfrac{49}{4}+\dfrac{49}{2}+\left(\dfrac{\ln 8}{2}\right)^{2}-\dfrac{7}{2} \ln 8$ $=7.4322$




14. The diagram shows the curve $y=x^{3}-3 x^{2}-9 x+k$, where $k$ is a constant. The curve has a minimum point on the $x$ -axis.
    • Find the value of $k$.
    • Find the area of the shaded region.





$y=x^{3}-3 x^{2}-9 x+k$ Let the mimiun porint be $(a, 0)$ $\therefore a^{3}-3 a^{2}-9 a+k=0$ $y^{\prime}=3 u^{2}-6 x-9$ At $(a, 0), y=0$ $\therefore 3 a^{2}-6 a-9=0$ $a^{2}-2 a-3=0$ $(a+1)(a-3)=0$ $\therefore a=-1$ or $a=3$ By the diagran, $a=3$ $\therefore$ minimun ponnt $=(3,0)$ When $a=3, k=27$. $\therefore y=x^{3}-3 x^{2}-9 x+27 .$ $\quad\quad$Area of shaded region $=\displaystyle\int_{0}^{3}\left(x^{3}-3 x^{2}-9 x+27\right)\ d x$ $=\left[\dfrac{x^{4}}{4}-x^{3}-\dfrac{9}{2} x^{2}+27 x\right]_{0}^{3}=\dfrac{135}{4}$




15. The diagram shows the curve $y=x(x-1)(x-2)$, which crosses the $x$ -axis at the points $O(0,0)$, $A(1,0)$ and $B(2,0)$.
    • The tangents to the curve at the points $A$ and $B$ meet at the point $C$. Find the $x$ -coordinate of $C$.
    • Show by integration that the area of the shaded region $R_{1}$ is the same as the area of the shaded region $R_{2}$





Curve: $y=x(x-1)(x-2)$ $\hspace{1cm}y=x^{3}-3 x^{2}+2 x$ $\therefore y^{\prime}=3 x^{2}-6 x+2$ At $x=1, \quad y^{\prime}=-1$ At $x=2, \quad y^{\prime}=2$ Equation of tangent $A$ is $y-0=-1(x-1)$ $y=1-x$ Equation of tangent at $B$ is $y-0=2(x-2)$ $y=2 x-4$ Whent the two tangent intersect each other, $1-x=2 x-4 $ $ x=\dfrac{5}{3}$ When $ x=\dfrac{5}{3}$, $y=1-\dfrac{5}{3}=-\dfrac{2}{3}$ $\therefore\ C=\left(\dfrac{5}{3},-\dfrac{2}{3}\right)$ $\begin{aligned} R_{1} &=\displaystyle\int_{0}^{1}\left(x^{3}-3 x^{2}+2 x\right)\ d x \\\\ &=\left[\dfrac{x^{4}}{4}-x^{3}+x^{2}\right]_{0}^{1} \\\\ &=\dfrac{1}{4}-1+1 \\\\ &=\dfrac{1}{4}\\\\ R_{2} &=\displaystyle\int_{7}^{2}\left(x^{3}-3 x^{2}+2 x\right)\ d x \\\\ &=\left[\dfrac{x^{4}}{4}-x^{3}+x^{2}\right]_{1}^{2} \\\\ &=4-8+4 \dfrac{1}{4}-1+1 \\\\ &=\dfrac{1}{4} \\\\ & \therefore\ R_{1}=R_{2} \end{aligned}$