Integral Calculus II Exercise 3.2 Text Book Solutions

Dear Samacheer Kalvi Students, here are the Integral Calculus II – Exercise 3.2 text book solutions in Business Maths Chapter 3 Integral Calculus II. If you have any doubts, please reach out to us in the comments section.

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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 3.3

Text Book Solutions for Integral Calculus I Exercise 3.4

Exercise 3.2

1. The cost of over haul of an engine is ₹10,000 The operating cost per hour is at the rate of 2x − 240 where the engine has run x km. Find out the total cost if the engine run for 300 hours after overhaul.

Answer:

$Total\; cost= Rs.10,000 +\displaystyle \int_{0}^{r} f(x)\; dx\\$
$= Rs.10,000 +\displaystyle \int_{0}^{300} (2x-240)\; dx\\$
$= Rs.10,000 + \left[ \dfrac{2x^{2}}{2}-240x \right]_{0}^{300}\\$
$= Rs.10,000 + \bigg[ x^{2}-240x \bigg]_{0}^{300}\\$
$= Rs.10,000 + (90,000-72,000)\\$
$= Rs.10,000 + Rs.18,000\\$
$TC =Rs.28,000$

$2. \; Elasticity \; of \;a \;function \;\dfrac{E_{y}}{E_{x}} \; is \;given \;by \; \dfrac{E_{y}}{E_{x}} = \dfrac{-7x}{(1-2x)(2+3x)}.\\$ $Find \; the \; function\; when \; x=2 \; and \; y=\dfrac{3}{8}$

Answer:

$Let\; \dfrac{E_{y}}{E_{x}} = \dfrac{-7x}{(1-2x)(2+3x)}\\$
$\dfrac{x}{y}\dfrac{dy}{dx}= \dfrac{-7x}{(1-2x)(2+3x)}\\$
$\dfrac{dy}{y}= \dfrac{-7x}{(1-2x)(2+3x)} \left( \dfrac{dx}{x} \right)\\$
$\dfrac{dy}{y}=-7\displaystyle \int \dfrac{dx}{(1-2x)(2+3x)}\\$
$Let \; \dfrac{1}{(1-2x)(2+3x)}= \dfrac{A}{(1-2x)}+\dfrac{B}{(2+3x)} \cdots (1)\\$
$\dfrac{1}{(1-2x)(2+3x)}= \dfrac{A (2+3x)+B (1-2x)}{(1-2x)(2+3x)}\\$
$1=A (2+3x)+B (1-2x)\cdots (2)\\$
$Put\; x=\dfrac{-2}{3} in \;(2)\\$
$1= A (0)+B (1-2\times\dfrac{-2}{3})\\$
$1= B\;(1+\dfrac{4}{3})\\$
$1= B\;(\dfrac{7}{3})\\$
$B=\dfrac{3}{7}\\$
$Put\; x=\dfrac{2}{7} in \;(2)\\$
$1= A (2+3\times\dfrac{1}{2})+0\\$
$1= A\;(2+\dfrac{3}{2})\\$
$1= A\;(\dfrac{7}{2})\\$
$A=\dfrac{2}{7}\\$
$Let \; \dfrac{1}{(1-2x)(2+3x)}= \dfrac{2}{7(1-2x)}+\dfrac{3}{7(2+3x)}\\$
$Substituting \; in \; Eq.(1)\\$
$\displaystyle \int \dfrac{dy}{y}=-7\displaystyle \int \dfrac{2}{7(1-2x)}+\dfrac{3}{7(2+3x)}\\$
$\displaystyle \int \dfrac{dy}{y}=-\dfrac{7}{7}\displaystyle \int \dfrac{2}{(1-2x)}+\dfrac{3}{(2+3x)}\\$
$log \left| y \right|= - \left[ \dfrac{2 \;log\left| (1-2x) \right|}{-2}+\dfrac{3 \;log\left| (2+3x) \right|}{3} +log \left| c \right|\right]\\$
$log \left| y \right|= log \left| (1-2x) \right|-log \left| (2+3x) \right|+ log \left| c \right|\\$
$log \left| y \right|= log \left| \dfrac{(1-2x)}{(2+3x)} \right|+ log \left| c \right|\\$
$log \left| y \right|= log \left| \dfrac{(1-2x)}{(2+3x)}(c) \right| \\$
$y=\dfrac{(1-2x)}{(2+3x)}(c)\cdots (3) \\$
$Substitute \; x=2 \; and\; y=\dfrac{3}{8} in Eq.\; (3)\\$
$= \dfrac{3}{8}= \left( \dfrac{1-4}{2+6} \right)k\\$
$= \dfrac{3}{8}= \left( \dfrac{-3}{8} \right)k\\$
$k = -1$
$y= \dfrac{1-2x}{2+3x} (-1) \\$
$y= \dfrac{2x-1}{2+3x}$

3. The elasticity of demand with respect to price for a commodity is given by (4-x)/x where p is the price when demand is x. Find the demand function when price is 4 and the demand is 2. Also find the revenue function.

Answer:

$Given \;\eta_{d}=\dfrac{4-x}{x}\\$
$=\dfrac{-p}{x}\; \dfrac{dx}{dp}=\dfrac{4-x}{x}\\$
$=\dfrac{-x}{4-x}\; \left( \dfrac{dx}{x} \right)=\; \dfrac{dp}{p}\\$
$=\dfrac{dx}{x-4}=\dfrac{dp}{p}\\$
$\displaystyle \int\dfrac{dx}{x-4}=\displaystyle \int\dfrac{dp}{p}\\$
$log\; p = log\; (x-4)+ log \; k\\$
$p=(x-4)k\cdots \; (1)\\$
$Substitute \; p=4 \; and \; x=2 \; in \; (1)\\$
$4=(2-4)k\\$
$4=-2k\\$
$k=-2\\$
$p=(x-4)(-2)\cdots \; (1)\\$
$p=8-2x\\$
$R=px \; (Revenue \; function)\\$
$R=(8-2x)x\\$
$R=8x-2x^{2}$

4. A company receives a shipment of 500 scooters every 30 days. From experience it is known that the inventory on hand is related to the number of days x. Since the shipment, I (x) = 500 − 0.03×2 , the daily holding cost per scooter is ₹ 0.3. Determine the total cost for maintaining inventory for 30 days.

Answer:

$Total\; cost= \;C_{1}\displaystyle \int_{0}^{T} I(x)\; dx\\$
$=0.3\displaystyle \int_{0}^{30} 500-0.03x^{2}\; dx\\$
$=0.3 \bigg[ 500x-\dfrac{0.03x^{3}}{3} \bigg]_{0}^{30}\\$
$=0.03 \bigg[ 500x-0.01x^{3} \bigg]_{0}^{30}\\$
$=0.03 \left[ 15000-0.01(27000) \right]-0\\$
$=0.03 \left( 15000-270 \right)\\$
$=0.03 (14730)\\$
$Total \;cost =Rs. 4419$

5. An account fetches interest at the rate of 5% per annum compounded continuously. An individual deposits ₹1,000 each year in his account. How much will be in the account after 5 years. (e^0.25 = 1.284) .

Answer:

$A=\displaystyle \int_{0}^{n} P\; e^{rt}\; dt\\$
$=\displaystyle \int_{0}^{5} 1000\; e^{0.05t}\; dt\\$
$=1000 \bigg[ \dfrac{e^{0.05t}}{e^{0.05}} \bigg]_{0}^{5}\\$
$=\dfrac{1000}{0.05} \left[ \left( e^{0.05} \right)^{5}-e^{0} \right]\\$
$=20,000 \left( e^{0.25}-1 \right)\\$
$=20,000 \left( 1.284-1 \right)\\$
$=20,000(.284)\\$
$A= Rs. 5680$

6. The marginal cost function of a product is given by dC/dx = 100 −10x + 0.1x² where x is the output. Obtain the total and the average cost function of the firm under the assumption, that its fixed cost is ₹ 500.

Answer:

$C=\displaystyle \int MC \; dx\\$
$C= \displaystyle \int 100 −10x + 0.1x^{2}\; dx\\$
$=100x-\dfrac{10x^{2}}{2}+\dfrac{0.1x^{3}}{3}+k\; \cdots (1)\\$
$When\; x=0;\; C= 500\\$
$In \; Eq \; (1)\; 500=0-0+0+k\\$
$k=500\\$
$C=100x-\dfrac{10x^{2}}{2}+\dfrac{0.1x^{3}}{3}+500\\$
$=100x-5x^{2}+\dfrac{0.1x^{3}}{3}+500\\$
$AC=\dfrac{C}{x}\\$
$=\dfrac{100x}{x}-\dfrac{5x^{2}}{x}+\dfrac{0.1x^{3}}{3x}+\dfrac{500}{x}\\$
$AC=100-5x+\dfrac{0.1x^{2}}{3}+\dfrac{500}{x}\\$

7. The marginal cost function is MC = 300x^2/5and fixed cost is zero. Find out the total cost and average cost functions.

Answer:

$C=\displaystyle \int MC \; dx\\$
$=\displaystyle \int 300x^{\frac{2}{5}}\; dx\\$
$=300 \displaystyle \int x^{\frac{2}{5}}\; dx\\$
$=300\left( \dfrac{x^{\frac{2}{5}+1}}{\frac{2}{5}+1} \right)+k\\$
$=300 \dfrac{x^{\frac{7}{5}}}{\frac{7}{5}}+k\\$
$=300\times \dfrac{5}{7} \left( x^{\frac{7}{5}}\right)+k\\$
$C=\dfrac{1500}{7} x^{\frac{7}{5}}+k\; \cdots (1)\\$
$when\; x= 0 \;C= 0\\$
$In \; Eq. \;(1)\; 0=0+k\\$
$k=0\\$
$C=\dfrac{1500}{7} x^{\frac{7}{5}}+0\\$
$C=\dfrac{1500}{7} x^{\frac{7}{5}}\\$
$AC=\dfrac{C}{x}\\$
$=\dfrac{1500x^{\frac{7}{5}}}{7x}\\$
$=\dfrac{1500x^{\frac{7}{5}-1}}{7}\\$
$AC=\dfrac{1500}{7}x^{\frac{2}{5}}\\$

$8. \; If \; the \; marginal \; cost\; function \; of \; x \; units\; of \; output\; is\;$

$\dfrac{a}{\sqrt{ax + b}} \; and \; if \;the \;cost \;of \;output \; is \;zero. \;Find \;the \;total \;cost \;as \;a \;function \;of \; x.$

Answer:

$C=\displaystyle \int MC \; dx\\$
$=\displaystyle \int\dfrac{a}{\sqrt{ax + b}}\; dx\\$
$= a \displaystyle \int \dfrac{dx}{(ax + b)^{\frac{1}{2}}}\\$
$= a \displaystyle \int (ax+b)^\frac{-1}{2}\; dx\\$
$= a \left[ \dfrac{(ax+b)^{\frac{-1}{2}+1}}{a\times \dfrac{-1}{2}+1}\right]+k\\$
$C= 2\sqrt{ax+b}+k\\$
$when \; x=0\; C=0\\$
$In \; Eq. \; (1)\; 0=2\sqrt{b}+k\\$
$k=-2\sqrt{b}$
$C=2\sqrt{ax+b}-2\sqrt{b}$

9. Determine the cost of producing 200 air conditioners if the marginal cost (is per unit) is C’ (x) = x² /200+4 .

Answer:

$C=\displaystyle \int MC \; dx\\$
$=\displaystyle \int\dfrac{x^{2}}{200}+4 \; dx\\$
$=\dfrac{x^{3}}{200\times3}+4x+k\\$
$=\dfrac{x^{3}}{600}+4x+k\; \cdots\; (1)\\$
$when\; x-0 \;c=0\\$
$In \; Eq\; (1)\; 0+0+k=0\\$
$k=0\\$
$C=\dfrac{x^{3}}{600}+4x \\$
$when\; x-200\\$
$C=\dfrac{200^{3}}{600}+4(200)\\$
$=\dfrac{8000000}{600}+800\\$
$C=Rs. 14,133.33$

10. The marginal revenue (in thousands of Rupees) functions for a particular commodity is 5 + 3e^-0.03x where x denotes the number of units sold. Determine the total revenue from the sale of 100 units. (Given e^-3 = 0.05 approximately)

Answer:

$MR= \left( 5+3e^{-0.03x} \right)\; dx$
$R=\displaystyle \int_{0}^{100}\left( 5+3e^{-0.03x} \right)\; dx\\$
$=\bigg[ 5x+\dfrac{3e^{-0.03x}}{-0.03} \bigg]_{0}^{100}\\$
$=\bigg[5x-\dfrac{e^{-0.03x}}{-0.01} \bigg]_{0}^{100}\\$
$=\left[ 5(100)-\dfrac{e^{-0.03(100)}}{-0.01} \right]-0-\dfrac{e^{-0}}{.01}\\$
$=500-\dfrac{e^{-3}}{0.01}+ \dfrac{1}{0.01}\\$
$=500-\dfrac{0.05}{0.01}+ \dfrac{1}{0.01}\\$
$=500-5+100\\$
$=595\\$
$Total\; Revenue = Rs.595\times1000\\$
$=Rs. 595,000$

11. If the marginal revenue function for a commodity is MR = 9 − 4x² . Find the demand function.

Answer:

$R= \displaystyle \int MR \; dx\\$
$= \displaystyle \int 9-4x^{2} \; dx\\$
$=9x-\dfrac{4x^{3}}{3}+k\; \cdots (1)\\$
$when \;x=0\; R=0\\$
$0=0-0+k\\$
$k=0\\$
$R=9x-\dfrac{4x^{3}}{3}+0\\$
$R=9x-\dfrac{4x^{3}}{3}\\$
$Demand \; Function\; p = \dfrac{R}{x}\\$
$=\dfrac{9x}{x}-\dfrac{4x^{3}}{3x}\\$
$p=9-\dfrac{4x^{2}}{3}$

12. Given the marginal revenue function 4\(2x+3)² -1 , show that the average revenue function is P = 4\(6x+9) -1.

Answer:

$R=\displaystyle \int MR \; dx\\$
$=\displaystyle \int \dfrac{4}{(2x+3)^{2}}-1\; dx\\$
$=4 \displaystyle \int (2x+3)^{-2}-1\; dx\\$
$= 4\left( \dfrac{(2x+3)^{-1}}{-1 (2)} \right)-x+k\\$
$R=\dfrac{-2}{2x+3}-x+k\; \cdots (1)\\$
$when \;x=0\; R=0\\$
$0=\dfrac{-2}{3}-0+k\\$
$k=\dfrac{2}{3}\\$
$In\; Eq. (1)\; R=\dfrac{-2}{2x+3}-x+\dfrac{2}{3}\\$
$AR= p=\dfrac{R}{x}\\$
$p=\dfrac{-2}{(2x+3)x}-\dfrac{x}{x}+\dfrac{2}{3x}\\$
$=\left[ \dfrac{-2}{(2x+3)x}+\dfrac{2}{3x} \right]-1 \\$
$=\left[ \dfrac{-2 (3)+2(2+3x)}{3x(2x+3)} \right]-1\\$
$=\left[ \dfrac{-6+4x+6}{3x(2x+3)} \right]-1\\$
$= \dfrac{4x}{3x(2x+3)}-1\\$
$=\dfrac{4}{3(2x+3)}-1\\$
$=\dfrac{4}{6x+9}-1\\$
$Proved$

13. A firm’s marginal revenue function is MR =20e^-x/10 (1-x\10). Find the corresponding demand function.

Answer:

$R=\displaystyle \int MR \; dx\\$
$=\displaystyle \int 20e^{\frac{-x}{10} }\left( 1-\dfrac{x}{10} \right)\; dx\\$
$=20 \displaystyle \int e^{\frac{-x}{10} }\left( 1-\dfrac{x}{10} \right)\; dx\\$
$=20 \displaystyle \int e^{\frac{-x}{10} }\left(-\dfrac{1}{10}(x)+1 \right)\; dx\\$
$a=\dfrac{1}{10}\; f(x)=x\; f'(x)=1\\$
$We\; know\; that\; \displaystyle \int e^{ax} \left( a f(x)+f'(x) \right)\; dx = e^{ax} f(x)+c$
$R= 20\;e^{\frac{-x}{10}} (x)+k\; \cdots (1)\\$
$when \;x=0 \;R=0\\$
$0=20e^{0} (0)+k\\$
$k=0\\$
$R=20\;e^{\frac{-x}{10}} (x)+0\\$
$R=20\;e^{\frac{-x}{10}} (x)\\$
$Demand\; function\; p=\dfrac{R}{x}\\$
$p=\dfrac{20\;e^{\frac{-x}{10}}(x)}{x}\\$
$p=20\;e^\frac{-x}{10}$

14. The marginal cost of production of a firm is given by C'(x) = 5 + 0.13x , the marginal revenue is given by R'(x) = 18 and the fixed cost is ₹ 120. Find the profit function.

Answer:

$C=\displaystyle \int MC\; dx\\$
$=\displaystyle \int (5 + 0.13x )\; dx\\$
$=5x+\dfrac{0.13x^{2}}{2}+k_{1}\; \cdots (1)\\$
$when \; x=0\; C=120\\$
$120=0+0+k_{1}\\$
$k_{1}=120\\$
$C=5x+\dfrac{0.13x^{2}}{2}+120$
$R=\displaystyle \int MR\; dx\\$
$=\displaystyle \int 18\; dx\\$
$=18x+k_{2}\; \cdots (1)\\$
$when \; x=0\; R=0\\$
$0=0+k_{2}\\$
$k_{2}=0\\$
$R=18x\\$
$P=R-C\\$
$=18x-\left( 5x+\dfrac{0.13x^{2}}{2}+120 \right)\\$
$=18x-5x-\dfrac{0.13x^{2}}{2}-120\\$
$=13x-\dfrac{0.13x^{2}}{2}-120\\$
$p=13x-0.065x^{2}-120\\$

15. If the marginal revenue function is R'(x)= 1500 − 4x − 3x² . Find the revenue function and average revenue function.

Answer:

$R= \displaystyle \int MR \; dx\\$
$=\displaystyle \int (1500-4x-3x^{2})\; dx\\$
$=1500x-\dfrac{-4x^{2}}{2}-\dfrac{3x^{3}}{3}+k\\$
$R=1500x-2x^{2}-x^{3}+k\; \cdots (1)\\$
$when\; x=0\; R=0\\$
$0=0-0-0+k\\$
$k=0\\$
$R=1500x-2x^{2}-x^{3}+0\\$
$R=1500x-2x^{2}-x^{3}\\$
$AR=\dfrac{R}{x}\\$
$=\dfrac{1500x}{x}-\dfrac{2x^{2}}{x}-\dfrac{x^{3}}{x}\\$
$= 1500-2x-x^{2}$

16. Find the revenue function and the demand function if the marginal revenue for x units is MR= 10 + 3x − x² .

Answer:

$R= \displaystyle \int MR\; dx\\$
$=\displaystyle \int (10-3x-x^{2})\; dx\\$
$=10x-\dfrac{3x^{2}}{2}-\dfrac{x^{3}}{3}+k\; \cdots(1)\\$
$when\: x=0\; R=0\\$
$0=0+0-0+k\\$
$k=0\\$
$=10x-\dfrac{3x^{2}}{2}-\dfrac{x^{3}}{3}+0\\$
$=10x-\dfrac{3x^{2}}{2}-\dfrac{x^{3}}{3}\\$
$Demand \; function\; p= \dfrac{R}{x}\\$
$p=\dfrac{10x}{x}+\dfrac{3x^{2}}{2(x)}-\dfrac{x^{3}}{3x}\\$
$p=10+\dfrac{3x}{2}-\dfrac{x^{2}}{3}$

17. The marginal cost function of a commodity is given by MC= 14000/√7x+4 and the fixed cost is ₹18,000. Find the total cost and average cost.

Answer:

$C=\displaystyle \int MC\; dx\\$
$=\displaystyle \int\dfrac{14000}{\sqrt{7x+4}}\; dx\\$
$=14000 \displaystyle \int\dfrac{dx}{(7x+4)^\frac{1}{2}}\\$
$=14000 (7x+4)^\frac{-1}{2}\; dx\\$
$=14000\dfrac{(7x+4)\frac{1}{2}}{\frac{1}{2}(7)}+k\\$
$=14000\times \dfrac{2}{7} (7x+4)^\frac{1}{2}+k \\$
$C= 4000 (7x+4)^\frac{1}{2}+k\; \cdots(1)\\$
$when \; x=0\; C=18000\\$
$18000= 4000\left( (7(0)+4)^\frac{1}{2}+k \right)\\$
$18000=4000(2)+k\\$
$18000=8000+k\\$
$10000=k\\$
$C=4000(7x+4)^\frac{1}{2}+10000\\$
$AC=\dfrac{C}{x} \\$
$=\dfrac{4000\sqrt{(7x+4)}}{x}+\dfrac{10000}{x}\\$

18. If the marginal cost (MC) of a production of the company is directly proportional to the number of units (x) produced, then find the total cost function, when the fixed cost is ₹ 5,000 and the cost of producing 50 units is ₹ 5,625.

Answer:

$MC\propto x\\$
$MC=k(x)\\$
$C=\displaystyle \int MC \; dx\\$
$C=\displaystyle \int k(x)\; dx\\$
$C=k\displaystyle \int (x)\; dx\\$
$C=k \left( \dfrac{x^{2}}{2}+k_{1} \right)\; \cdots (1)\\$
$when \; x=0\; C=5000\\$
$5000=0+k_{1}\\$
$k_{1}=5000\\$
$C=\dfrac{kx^{2}}{2}+5000\; \cdots (2)\\$
$when \; x=50\; C=5625\\$
$5625=\dfrac{k(50^{2})}{2}+5000\\$
$5625-5000=\dfrac{k(2500)}{2}\\$
$625=\dfrac{k(2500)}{2}\\$
$625\times 2=k(2500)\\$
$1250=2500k\\$
$k=\dfrac{1250}{2500}\\$
$k=\dfrac{1}{2}\\$
$C=\dfrac{\frac{1}{2} x^{2}}{2}+5000\\$
$C=\dfrac{x^{2}}{4}+5000$

19. If MR = 20 − 5x + 3x² , find total revenue function.

Answer:

$R=\displaystyle \int MR \; dx\\$
$=\displaystyle \int \left( 20-5x+3x^{2} \right)\; dx\\$
$R=20x-\dfrac{5x^{2}}{2}+\dfrac{3x^{3}}{3}+k\; \cdots(1)\\$
$when x=0\; R=0\\$
$0=0-0+0+k\\$
$k=0\\$
$R=20x-\dfrac{5x^{2}}{2}+\dfrac{3x^{3}}{3}+0\\$
$R=20x-\dfrac{5x^{2}}{2}+\dfrac{3x^{3}}{3}\\$
$R=20x-\dfrac{5x^{2}}{2}+x^{3}\\$

20. If MR = 14 − 6x + 9x² , find the demand function.

Answer:

$R=\displaystyle \int MR \; dx\\$
$R=\displaystyle \int \left( 14-6x+9x^{2} \right) \; dx\\$
$=14x-\dfrac{6x^{2}}{2}+\dfrac{9x^{3}}{3}+k\; dx\\$
$=14x-3x^{2}+3x^{3}+k\; \cdots(1)\\$
$when \; x=0\; R=0\\$
$0=0-0+0+k\\$
$k=0\\$
$R=14x-3x^{2}+3x^{3}+0\\$
$R=14x-3x^{2}+3x^{3}\\$
$Demand\; function\; p=\dfrac{R}{x}\\$
$=\dfrac{14x}{x}-\dfrac{3x^{2}}{x}+\dfrac{3x^{3}}{x}\\$
$=14-3x+3x^{2}$

If you have any doubts or questions, in this chapter, please reach out to us in the comments section.

Integral Calculus II Exercise 3.2 – Text Book Solutions 12th Business Maths

Exercise 3.2

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