Integral Calculus II Exercise 3.4 Text Book Solutions

Dear Samacheer Kalvi Students, here are the Integral Calculus II – Exercise 3.4 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 3.2

Text Book Solutions for Integral Calculus I Exercise 3.3

Exercise 3.4

$1.\; Area \;bounded\; by \;the\: curve\; y = x (4 − x) \;between \;the\; limits\; 0 \;and \;4 \;with \;x − axis \;is\\$
$a. \;\dfrac{30}{3}\; sq.units\\$
$b. \;\dfrac{31}{2}\; sq.units \\$
$c. \;\dfrac{32}{3}\; sq.units \\$
$d. \;\dfrac{15}{2}\; sq.units$

Answer:

$c. \;\dfrac{32}{3}\; sq.units \\$

$2. \;Area \;bounded\; by \;the\; curve\; y = e^{-2x} \;between\; the\; limits\; 0\le x\le \infty\; is \\$
$a. \;1\; sq.unit \\$
$b. \;\dfrac{1}{2}\; sq.unit \\$
$c. \;5\; sq.units \\$
$d. \;2\; sq.units$

Answer:

$b. \;\dfrac{1}{2}\; sq.unit \\$

$3.\; Area\; bounded \; by\; the \; curve\; y=\dfrac{1}{x}\; between\; the\; limits\; 1\; and \; 2\; is \\$
$a. \;log\;2 \; sq.units \\$
$b. \;log\;5 \; sq.units \\$
$c. \;log\;3 sq.units \\$
$d. \;log\;4 \; sq.units$

Answer:

$a. \;log\;2 \; sq.units \\$

$4. \;If\; the\; marginal \; revenue\; function \;of\; a\; firm\; MR=e^{\frac{-x}{10}}\; then\; revenue\; is \\$
$a.- \;10\;e^{\dfrac{-x}{10}}\\$
$b. \;1-e^{\dfrac{-x}{10}} \\$
$c. \;10 \left( 1-e^{\dfrac{-x}{10}} \right)\\$
$d. \;e^{\dfrac{-x}{10}}-10$

Answer:

$c. \;10 \left( 1-e^{\dfrac{-x}{10}} \right)\\$

$5.\; If\; MR\; and\; MC\; denotes\; the \; marginal \; revenue\; \\$
$and\; marginal \; cost\; functions\; then\; the\; profit\; functions\; is \\$
$a. \; P=\displaystyle \int \left( MR-MC\ \right)\; dx+k \\$
$b. \;P=\displaystyle \int \left( MR+MC\ \right)\; dx+k \\$
$c. \;P=\displaystyle \int (MR) (MC)\; dx+k \\$
$d. \;P=\displaystyle \int \left( R-C\ \right)\; dx+k$

Answer:

$a. \;P=\displaystyle \int \left( MR-MC\ \right)\; dx+k$

$6.\; The \; demand\; and\; supply\; functions\;are\; given\; by \; D(x)=16-x^{2}\; and \; S(x)=2x^{2}+4\; are\; under\; perfect\; competition\; then\; the\equilibrium\; price\; x\; is \\$
$a. \;2 \\$
$b. \; 3\\$
$c. \; 4 \\$
$d. \;5$

Answer:

$a. \;2 \\$

$7. \;The\; marginal\; revenue\;and\; marginal\;cost\;functions\; are \; given\; by\\$
$\; MR = 30 − 6x\; and \; MC = −24 + 3x\; where\; x\; is\; the\; product\; then\; \\$
$the\; profit\; function\; is\\$
$a. \; 9x^{2} +54x\\$
$b. \;9x^{2} -54x \\$
$c. \;54x-\dfrac{9x^{2}}{2} \\$
$d. \; 54x-\dfrac{9x^{2}}{2}+k$

Answer:

$d. \; 54x-\dfrac{9x^{2}}{2}+k \\$

$8. \;The \; given\; demand\; and\; supply\; functions\; are \; given\; by\; \\$
$D(x)= 20 − 5x\; and \; S(x)= 4x + 8\; if\; they\; are\; under\; \\$ $perfect\; competition\; then\; the\; equilibrium\; demand\; is\\$
$a. \; 40\\$
$b. \; \dfrac{41}{2} \\$
$c. \; \dfrac{40}{3} \\$
$d. \; \dfrac{41}{5}$

Answer:

$c. \; \dfrac{40}{3} \\$

$9.\; If\; the \; marginal \; revenue\; MR= 35+7x-3x^{2}\; then\; the\; average\; revenue\; AR \; is \\$
$a. \; 35x+\dfrac{7x^{2}}{2}-x^{3}\\$
$b. \; 35+\dfrac{7x}{2}-x^{2}\\$
$c. \; 35+\dfrac{7x}{2}+x^{2}\\$
$d. \; 35+7x+x^{2}$

Answer:

$b. \; 35+\dfrac{7x}{2}-x^{2}\\$

$10.\; The\; profit\; of\; a\; function \; p(x) \; is\; maximum\; when \\$
$a. \; MC-MR=0\\$
$b. \; MC=0\\$
$c. \; MR=0 \\$
$d. \; MC+MR=0$

Answer:

$a. \;MC-MR=0\\$

$11.\; For\; the \;demand\; function\; p(x),\; the\; elasticity\; of\; demand\; with \\$ $\;respect \;to \;price \;is \;unity \;then \\$
$a. \; revenue\; is \;constant\\$
$b. \;cost \;function\; is \;constant \\$
$c. \;profit\; is \;constant \\$
$d. \;none\; of\; these$

Answer:

$a. \; revenue\; is \;constant\\$

$12.\; The\; demand \;function \;for \;the \;marginal \;function \;MR = 100 − 9x^{2}\; is \\$
$a.\; 100 − 3x^{2}\\$
$b.\; 100x − 3x^{2} \\$
$c.\; 100x − 9x^{2} \\$
$d.\; 100 + 9x^{2} \\$

Answer:

$a. \; 100 − 3x^{2}\\$

$13.\; When\; x_{0}=5 \;and \;p_{0}= 3 \;the \;consumer's\; surplus \;for \;the \;demand \\$ $\;function\; p_{d} = 28 − x^{2} is \\$
$a.\; 250 \;units\\$
$b.\; \dfrac{250}{3} \;units \\$
$c.\; \dfrac{251}{2}\; units \\$
$d.\; \dfrac{251}{3}\; units \\$

Answer:

$b. \;\dfrac{250}{3} \;units \\$

$14.\; When\; x_{0}=2 \;and \;p_{0}= 12 \;the \;producer's\; surplus \;for \;the \\$ $\;supply\;function\; p_{d} = 2x^{2}+4 is \\$
$a.\; \dfrac{31}{5} \;units\\$
$b.\; \dfrac{31}{2} \;units \\$
$c.\; \dfrac{32}{3}\; units \\$
$d.\; \dfrac{30}{7}\; units \\$

Answer:

$c. \;\dfrac{32}{3}\; units \\$

$15.\; Area \; bounded\; by\; y = x \;between\; the \;lines\; y = 1,\; y = 2 \;with\; y = axis \; is \\$
$a.\; \dfrac{1}{2} \;sq.units \\$
$b.\; \dfrac{5}{2} \;sq.units\\$
$c.\; \dfrac{3}{2}\; sq.units\\$
$d.\; 1 sq.unit\\$

Answer:

$c. \;\dfrac{3}{2}\; sq.units\\$

$16. \; The \; producer's \; surplus\; when\; the \; supply\; function \; for \; a \; \\$ $commodity \; is \; P = 3 + x \; and \; x_{0} = 3\; is \\$
$a.\; \dfrac{5}{2} \\$
$b.\; \dfrac{9}{2} \\$
$c.\; \dfrac{3}{2} \\$
$d.\; \dfrac{7}{2}$

Answer:

$b.\; \dfrac{9}{2} \\$

$17.\; The\; marginal\; cost\; function\; is \;MC =100\sqrt{x} \;find \;AC\; given\; \\$ $\; that\; TC =0 \;when\; the \;output \; is \;zero \;is \\$
$a.\; \dfrac{200}{3} x^{\frac{1}{3}} \\$
$b.\; \dfrac{200}{3} x^{\frac{3}{2}} \\$
$c.\; \dfrac{200}{3x^{\frac{3}{2}}} \\$
$d.\; \dfrac{200}{3x^{\frac{1}{2}}}$

Answer:

$a.\; \dfrac{200}{3} x^{\frac{1}{3}} \\$

$18.\; The \; demand \; and \; supply \; function \; of\; a \; commodity\; are\\$ $\; P(x)= (x − 5)^{2} \; and \; S(x)= x^{2} + x + 3 \; then \; the \; equilibrium \; quantity \; x_{0} \; is \\$
$a.\; 5 \\$
$b.\; 2 \\$
$c.\; 3 \\$
$d.\; 1 \\$

Answer:

$b.\; 2 \\$

$19.\; The \;demand \;and \;supply \;function \;of\; a \;commodity \;are \\$ $\; D(x)= 25 - 2x \;and \;S(x)= \dfrac{10+x}{4}\;then \;the\; equilibrium \;price\; P_{0}\; is \\$
$a.\; 5 \\$
$b.\; 2 \\$
$c.\; 3 \\$
$d.\; 10$

Answer:

$a.\; 5 \\$

$20. \; If \;MR \;and \;MC \;denote \;the \;marginal\; revenue \;and \;marginal \;cost \\$ $\;and\; MR-MC = 36x - 3x^2 - 81,\; then\; the\; maximum \;profit \;at x \;is\; equal\; to \\$
$a.\; 3 \\$
$b.\; 6 \\$
$c.\; 9 \\$
$d.\; 5$

Answer:

$c.\; 9 \\$

$21.\; If\; the \;marginal \;revenue\; of \;a \;firm \;is \;constant, \;then \;the \\$ $\;demand \;function\; is \\$
$a.\; MR \\$
$b.\; MC \\$
$c.\; C(x) \\$
$d.\; AC$

Answer:

$a.\; MR \\$

$22.\; For \; a \; demand \; function\; p \; if\; \displaystyle \;\int \dfrac{dp} {p}=k\int\dfrac{dx}{x} \;then\; k \;is \;equal\; to \\$
$a.\; \eta_{d} \\$
$b.\;-\eta_{d} \\$
$c.\;\dfrac{-1}{\eta_{d} } \\$
$d.\;\dfrac{1}{\eta_{d} }$

Answer:

$c.\;\dfrac{-1}{\eta_{d} } \\$

$23.\; Area \; bounded\; by \; y = e^x\; between\; the \; limits \; 0 \; to\; 1 \; is$
$a. \; (e-1) \; sq.units \\$
$b.\; (e+1) \; sq.units \\$
$c.\; \left( 1-\dfrac{1}{e} \right) \; sq.units \\$
$d.\; \left( 1+\dfrac{1}{e} \right) \; sq.units$

Answer:

$a. \; (e-1) \; sq.units \\$

$24.\; The \;area \;bounded \;by \;the \;parabola \;y^2 = 4x\; bounded\; by\; its\; latus \;rectum \;is \\$
$a. \;\dfrac {16}{3} \; sq.units \\$
$b. \;\dfrac {8}{3} \; sq.units \\$
$c.\; \dfrac {72}{3} \; sq.units \\$
$d. \;\dfrac {1}{3} \; sq.units$

Answer:

$b. \;\dfrac {8}{3} \; sq.units \\$

$25.\; Area \;bounded \;by \;y = \left| x \right| \;between \;the \;limits \;0 \;and \;2 \; is \\$
$a.\; 1 \; sq.units \\$
$b.\; 3 \; sq.units \\$
$c. \;2 \; sq.units \\$
$d. \; 4 \; sq.units \\$

Answer:

$c. \;2 \; sq.units \\$

If you have any doubts in this exercise, please let us know in the comments section.

Integral Calculus II Exercise 3.4 – Text Book Solutions 12th Business Maths

Exercise 3.4

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