Integral Calculus II Exercise 3.3

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Important Formulas in Integral Calculus II

Text Book Solutions for Integral Calculus I Exercise 3.1

Text Book Solutions for Integral Calculus I Exercise 3.2

Text Book Solutions for Integral Calculus I Exercise 3.4

Exercise 3.3

1. Calculate consumer’s surplus if the demand function p = 50 − 2x and x = 20.

Answer:

$Let\; P=50-2x\\$
$when \; x=20 p=50-2(20)\\$
$p=50-40\\$
$p=10\\$
$p_{0}=10; x_{0}=20\\$
$C.S=\int_{0}^{x_{0}} f(x)\; dx-x_{0}p_{0} \\$
$=\int_{0}^{20} (50-2x)\; dx-20(10) \\$
$=\bigg[ 50x-\dfrac{2x^{2}}{2} \bigg]_{0}^{20}-200\\$
$=\bigg[ 50x-x^{2} \bigg]_{0}^{20}-200\\$
$=\left[ 50(20)-(20)^{2}- \left( 0-0 \right) \right]-200\\$
$=1000-400-200\\$
$=400\; units\\$

2. Calculate consumer’s surplus if the demand function p = 122 − 5x − 2x² and x = 6.

Answer:

$p= 122-5x-2x^{2}\\$
$when \; x= 6; p= 122-5(6)-2 (6)^{2}\\$
$=122-30-72\\$
$=20\\$
$p_{0}=20; x_{0}=6;\\$
$C.S= \displaystyle \int_{0}^{ x_{0}}f(x)\; dx- x_{0} p_{0}\\$
$=\displaystyle \int_{0}^{ 6}(122 − 5x − 2x^{2})\; dx- 6 (20)\\$
$=\bigg[ 122x-\dfrac{5x^{2}}{2}-\dfrac{2x^{3}}{3} \bigg]_{0}^{ 6}-120\\$
$=122(6)-\dfrac{5(6)^{2}}{2}-\dfrac{2(6)^{3}}{3} \bigg]_{0}^{ 6}-120\\$
$=732-90-144-120\\$
$=378\; units$

3. The demand function p = 85 − 5x and supply function p = 3x − 35 . Calculate the equilibrium price and quantity demanded. Also calculate consumer’s surplus.

Answer: At equilibrium price, quantity demanded = quantity supplied

Q_{d =}Q_s

$85-5x=3x-35\\$
$-5x-3x=-35-85\\$
$=-8x=-120\\$
$x=\dfrac{120}{8}\\$
$x=15\\$
$Let p=3x-35\\$
$=3(15)-35\\$
$=45-35\\$
$p=10$
$p_{0}=10 \; (equilibrium\; price)\\$
$x_{0}=15 \; (quantity\; demanded)\\$
$CS=\displaystyle \int_{0}^{x_{0}} f(x)\;dx-x_{0}p_{0}\\$
$=\displaystyle \int_{0}^{15} (85-5x)\;dx-15(10)\\$
$=\bigg[ 85x-\dfrac{5x^{2}}{2} \bigg]_{0}^{15}-150\\$
$=\bigg[ 85(15)-\dfrac{5(15)^{2}}{2} \bigg]_{0}^{15}-150\\$
$=1275-\dfrac{1125}{2}-150\\$
$=1125-\dfrac{1125}{2}\\$
$=\dfrac{2250-1125}{2}\\$
$=\dfrac{1125}{2}\\$
$C.S = 562.50 \;units$

4. The demand function for a commodity is p = e^-x .Find the consumer’s surplus when p = 0.5.

Answer:

$p=e^{-x}\\$
$when \; p= 0.5\\$
$0.5=e^{-x}\\$
$\dfrac{1}{2}=\dfrac{1}{e^{x}}\\$
$e^{x}=2\\$
$x=log 2\\$
$p_{0}=0.5\; x_{0}=log 2\\$
$CS=\displaystyle \int_{0}^{x_{0}} f(x)\;dx-x_{0}p_{0}\\$
$=\displaystyle \int_{0}^{log 2} (e^{-x})\;dx-log\;2(0.5)\\$
$=\bigg[ \dfrac{e^{-x}}{-1} \bigg]_{0}^{log 2}- \left( \dfrac{1}{2} \right)log\;2 \\$
$=-\bigg[ \dfrac{1}{e^{x}} \bigg]_{0}^{log 2}- \left( \dfrac{1}{2} \right)log\;2 \\$
$=- \left[ \dfrac{1}{e^{log\; 2}}-\dfrac{1}{e^{0}} \right]-\left( \dfrac{1}{2} \right)log\;2\\$
$= -\left( \dfrac{1}{2}-1 \right)- \left( \dfrac{1}{2} \right) log\;2\\$
$=\dfrac{1}{2}-\dfrac{1}{2} log\;2\\$
$C.S = \dfrac{1}{2}\left( 1-log 2 \right)\; units\\$

5. Calculate the producer’s surplus at x = 5 for the supply function p = 7 + x.

Answer:

$p=7+x\\$
$when \;x=5 \; p=7+5\\$
$p=12\\$
$p_{0}=12\; x_{0}=5\\$
$PS= x_{0}\; p_{0} \displaystyle \int_{0}^{x_{0}} g(x)\; dx\\$
$= 5(12)-\displaystyle \int_{0}^{5} (7+x)\; dx\\$
$=60 \bigg[ 7x+\dfrac{x^{2}}{2} \bigg]_{0}^{5}\\$
$=60- \left[\left( 7(5)+\dfrac{5^{2}}{2} \right)-(0+0)\right]\\$
$=60-\left( 35+12.5 \right)\\$
$=60-47.5\\$
$=12.5 \;units$

6. If the supply function for a product is p = 3x + 5x² .Find the producer’s surplus when x = 4.

Answer:

$p=3x+5x^{2}\\$
$when \; x=4\; p=3(4)+5(4)^{2}\\$
$p=12+80\\$
$p=92\\$
$p_{0}=92; x_{0}=4\\$
$PS= x_{0}\; p_{0} \displaystyle \int_{0}^{x_{0}} g(x)\; dx\\$
$= 4(92)-\displaystyle \int_{0}^{4} (3x+5x^{2})\; dx\\$
$=368-\bigg[ \dfrac{3x^{2}}{2}+\dfrac{5x^{3}}{3} \bigg]_{0}^{4}\\$
$=368- \bigg[ \left( \dfrac{3(4)^{2}}{2}+\dfrac{5(4)^{3}}{3} \right) -\left( 0+0 \right)\bigg]\\$
$=368-\left( 24+\dfrac{320}{3} \right)\\$
$=368-24-\dfrac{320}{3}\\$
$=\dfrac{1104-72-320}{3}\\$
$=\dfrac{712}{3}\\$
$PS=237.33\; units$

7. The demand function for a commodity is p = 36/x+4. Find the consumer’s surplus when the prevailing market price is Rs.6.

Answer:

$p=\dfrac{36}{x+4}\\$
$when \: p=6\; \\$
$6=\dfrac{36}{x+4}\\$
$6x+24=36\\$
$6x=36-24\\$
$6x=12\\$
$x=2\\$
$p_{0}=6; x_{0}=2\\$
$CS= \displaystyle \int_{0}^{x_{0}} f(x)\;dx \;- x_{0}\; p_{0}\\$
$=\displaystyle \int_{0}^{2} \dfrac{36}{x+4}\; dx- 2 (6)\\$
$=36\displaystyle \int_{0}^{2} \dfrac{1}{x+4}\; dx- 12\\$
$=36 \; \bigg[ log (x+4) \bigg]_{0}^{2}-12\\$
$=36 \;\left[ log\;6-log\;4 \right]-12\\$
$=36 \;log \dfrac{6}{4}-12\\$
$PS=36\; log \dfrac{3}{2}-12\; units$

8. The demand and supply functions under perfect competition are p_d= 1600 – x²and p_s = 2x²+ 400 respectively. Find the producer’s surplus.

Answer: At equilibrium price, p_d=p_s

$1600-x^{2}=2x^{2}+400\\$
$=-x^{2}-2x^{2}=400-1600\\$
$=-3x^{2}=-1200\\$
$x^{2}=\dfrac{1200}{3}\\$
$x^{2}=400\\$
$x=\pm 20\\$
$x=-20 \;is \; not \; possible.$
$x=20$
$Let \; p_{d}=1600-x^{2}\\$
$=1600-(20)^{2}\\$
$=1600-400\\$
$p_{d}=1200\\$
$p_{0}=1200; x_{0}=20\\$
$PS= x_{0}\; p_{0} \displaystyle \int_{0}^{x_{0}} g(x)\; dx\\$
$PS= 20 (1200)\displaystyle \int_{0}^{20} (2x^{2}+400)\;dx \\$
$=24000-\bigg[ \dfrac{2x^{3}}{3}+400x \bigg]_{0}^{20}\\$
$=24000-\bigg[ \left( \dfrac{2(20)^{3}}{3}+400(20) \right) -\left( 0+0 \right)\bigg]\\$
$=24000-\left( \dfrac{16000}{3}+8000 \right)\\$
$=24000-\dfrac{16000}{3}-8000\\$
$=16000-\dfrac{16000}{3}\\$
$=\dfrac{32000}{3}\\$
$PS=\dfrac{32000}{3}\; units\\$

9. Under perfect competition for a commodity the demand and supply laws are p_d= 8\x+1and p_s = x+3/2 respectively. Find the consumer’s and producer’s surplus.

Answer: At equilibrium price, p_d=p_s

$\dfrac{8}{x+1}-2=\dfrac{x+3}{2}\\$
$\dfrac{8}{x+1}-2-\dfrac{x+3}{2}=0\\$
$\dfrac{16-2(2)(x+1)-(x+1)(x+3)}{2(x-1)}=0\\$
$\dfrac{16-4x-4-(x^{2}+x+3x+3)}{2(x-1)}=0\\\$
$\dfrac{16-4x-4-x^{2}-x-3x-3)}{2(x-1)}=0\\$
$\dfrac{-x^{2}-8x-9}{2(x-1)}=0\\$
$-x^{2}-8x-9=0$
$x^{2} +8x+9=0\\$
$(x-1)(x+9)=0\\$
$x=1,-9 \\$
$-9 \; is\; not\; possible\; so\; x=1\;$
$Let \; p_{s}=\dfrac{x+3}{2}\\$
$At\; x=1\\$
$p_{s}=\dfrac{1+3}{2}\\$
$p_{s}=2\\$
$p_{0}=2; x_{0}=1\\$
$CS= \displaystyle \int_{0}^{x_{0}} f(x)\; dx\;- x_{0}\; p_{0}\\$
$=\displaystyle \int_{0}^{1} \left( \dfrac{8}{x+1}-2 \right)-1(2)\\$
$=\bigg[ \left( 8\; log (x+1)-2x \right) \bigg]_{0}^{1}-2\\$
$=\left( 8 \; log\; 2 -2 \right)- \left( log\;1-0 \right)-2 \\$
$=\left( 8 \; log\; 2 -2 \right)- \left( 0-0 \right)-2 \\$
$=8 \; log\; 2 -2-2\\$
$=8 \; log\; 2 -4\\$
$CS =8 \; log\; 2 -4\; units\$
$PS= x_{0}\; p_{0}- \displaystyle \int_{0}^{x_{0}} g(x)\; dx \\$
$= 1(2)-\displaystyle \int_{0}^{1} \left( \dfrac{x+3}{2} \right)\; dx\\$
$=2-\dfrac{1}{2} \displaystyle \int_{0}^{1} \left( x+3 \right)\; dx\\$
$=2-\dfrac{1}{2} \bigg[ \dfrac{x^{2}}{2}+3x \bigg]_{0}^{1}\\$
$=2-\dfrac{1}{2} \left[ \left( \dfrac{1}{2}+3 \right)- \left( 0+0 \right) \right]\\$
$=2-\dfrac{1}{2}\left( \dfrac{7}{2} \right)\\$
$=2-\dfrac{7}{4} \\$
$PS =\dfrac{1}{4} \; units\$

10. The demand equation for a products is x = √100 − p and the supply equation is x = p/2-10. Determine the consumer’s surplus and producer’s surplus, under market equilibrium.

Answer:

$Let \; x=\sqrt{100-p}\; (demand \;function)\\$
$=squaring\; on \; both \; sides\\$
$x^{2}=100-p\\$
$p=100-x^{2}\\$
$p_{d}=100-x^{2}\\$
$Let x=\dfrac{p}{2}-10\; (supply\; function)\\$
$x=\dfrac{p-20}{2}\\$
$2x=p-20\\$
$p=2x+20\\$

At market equilibrium, p_{d =}p_s

$100-x^{2}=2x+20\\$
$100-x^{2}-2x-20=0\\$
$-x^{2}-2x+80=0\\$
$x^{2}+2x-80=0\\$
$(x+10)(x-8)=0\\$
$x=-10, 8\\$
$x=-10 \; is \; not\; possible\\$
$so\; x=8\\$
$Let \; p_{s}=2x+20\\$
$=2(8)+20\\$
$=16+20\\$
$p=36\\$
$p_{0}=36\; x_{0}=8\\$
$CS=\displaystyle \int_{0}^{x_{0}} f(x)\; dx\; -x_{0}\;p_{0}\\$
$\displaystyle \int_{0}^{8} (100-x^{2})\; dx\; -8 (36)\\$
$=\bigg[ 100x-\dfrac{x^{3}}{3} \bigg]_{0}^{8}-288\\$
$=100(8)-\left( \dfrac{8^{3}}{3} \right) -\left( 0-0 \right)\\$
$=800-\dfrac{512}{3}-288\\$
$=\dfrac{2400-512-864}{3}\\$
$CS=\dfrac{1024}{3}\; units\\$
$PS=x_{0}\;p_{0}-\displaystyle \int_{0}^{x_{0}} g(x)\; dx\\$
$=8(36)-\displaystyle \int_{0}^{8} (2x+20)\; dx\\$
$=288-\bigg[ \dfrac{2x^{2}}{2}+20x \bigg]_{0}^{8}\\$
$=288-\bigg[ \dfrac{2(8)^{2}}{2}+20(8) \bigg]\\$
$=288-\left( 64+160 \right)\\$
$=288-224\\$
$PS=64\; units$

11. Find the consumer’s surplus and producer’s surplus for the demand function p_d = 25 − 3x and supply function p_s = 5 + 2x.

Answer: At market equilibrium, p_{d =}p_s

$25-3x=5+2x\\$
$-3x-2x=5-25\\$
$-5x=-20\\$
$x=\dfrac{20}{5}\\$
$x=4\\$
$Let \; p_{d}=25-3x\\$
$=25-3(4)\\$
$=25-12\\$
$p=13\\$
$p_{0}=13\; x_{0}=4\\$
$CS=\displaystyle \int_{0}^{x_{0}} f(x)\; dx\; -x_{0}\;p_{0}\\$
$\displaystyle \int_{0}^{4} (25-3x)\; dx\; -4 (13)\\$
$=\bigg[ 25x-\dfrac{3x^{2}}{2} \bigg]_{0}^{4}-52\\$
$=\left[ \left(25(4)- \dfrac{3\times4^{2}}{2} \right) -\left( 0-0 \right) \right]-52\\$
$=100-24-52\\$
$=100-76\\$
$CS=24\; units\\$
$PS=x_{0}\;p_{0}-\displaystyle \int_{0}^{x_{0}} g(x)\; dx\\$
$=4(13)-\displaystyle \int_{0}^{4} (5+2x)\; dx\\$
$=52-\bigg[ 5x +\dfrac{2x^{2}}{2}\bigg]_{0}^{4}\\$
$=52-\bigg[5x+x^{2} \bigg]_{0}^{4}\\$
$=52-\left( 5(4)+4^{2}\right)\\$
$=52-36\\$
$=16\\$
$PS=16\; units$