Integral Calculus – II

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Questions:

Generic 5

4 x 5 = 20

Sketch the graph of y = |x - 5|. Evaluate \(\int _{ 0 }^{ 1 }{ |4x-5|dx } \)

The Marginal revenue for a commodity is MR=\(\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\), find the revenue function.

Find the area of the region bounded by the line y = x - 5, the x-axis and between the ordinates x = 3and x = 7

Calculate the area bounded by the parabola y² = 4ax and its latus rectum.

Generic 3

5 x 3 = 15

The marginal revenue function is given by \(R'(x)=\frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \). Find the revenue function and demand function if R(1) = 6

Find the producer’s surplus defined by the supply curve g(x) = 4x + 8 when x_o= 5.

The demand function of a commodity is y = 36 − x². Find the consumer’s surplus for y₀ = 11

The price of a machine is 6,40,000 if the rate of cost saving is represented by the function f(t) = 20,000 t. Find out the number of years required to recoup the cost of the function.

Find the area of the region lying in the first quadrant bounded by the region y = 4x², x = 0, y = 0 and y = 4

Multiple Choice Question

10 x 1 = 10

10.

The area bounded by the curve y = 4ax and the lines y² = 2a and Y-axis is _______ sq. units.

(a)

\(\frac{2a}{3}\)

(b)

2a²

(c)

\(\frac{a^2}{3}\)

(d)

\(\frac{2a^2}{3}\)

11.

The area of the region bounded by the curve y² = 2y - x and the y-axis _____ sq. units

(a)

\(\frac{4}{3}\)

(b)

\(\frac{2}{3}\)

(c)

(d)

\(\frac{16}{3}\)

12.

The area of the region bounded by the line y = 3x + 2, the X-axis and the ordinates x = - 1 and x = 1is_________ sq. units.

(a)

\(\frac{13}{3}\)

(b)

(c)

\(\frac{26}{3}\)

(d)

\(\frac{3}{13}\)

13.

The area enclosed by the curve y = cos²x in [0,\(\pi\)] the lines x=0, x = \(\pi\) and the X-axis is ________sq.units.

(a)

2\(\pi\)

(b)

2\(\pi\)

(c)

\(\frac{2}{\pi}\)

(d)

\(\frac{\pi}{2}\)

14.

The area of the region bounded by the line 2y = -x + 8, X - axis and the lines x = 2 and x = 4 is ________ sq.units.

(a)

\(\frac{1}{5}\)

(b)

\(\frac{2}{5}\)

(c)

(d)

\(\frac{5}{2}\)

15.

The area unded by the curves y = 2^x, x = 0 and x = 2 is________sq.units.

(a)

log_e2

(b)

3log_e2

(c)

\(\frac{3}{log_e2}\)

(d)

2log_e3

16.

The given demand and supply function are given by D(x) = 20 − 5x and S(x) = 4x + 8 if they are under perfect competition then the equilibrium demand is ________.

(a)

(b)

\(\frac{41}{2}\)

(c)

\(\frac{40}{3}\)

(d)

\(\frac{41}{5}\)

17.

The marginal revenue and marginal cost functions of a company are MR = 30 − 6x and MC = −24 + 3x where x is the product, then the profit function is ________.

(a)

9x² + 54x

(b)

9x² − 54x

(c)

54x - \(\frac { { 9x }^{ 2 } }{ 2 } \)

(d)

54x - \(\frac { { 9x }^{ 2 } }{ 2 } \) + k

18.

The demand and supply functions are given by D(x)= 16 − x² and S(x) = 2x² + 4 are under perfect competition, then the equilibrium price x is ________.

(a)

(b)

(c)

(d)

19.

Area bounded by the curve y = x (4 − x) between the limits 0 and 4 with x − axis is ________.

(a)

\(\frac{30}{3}\) sq.units

(b)

\(\frac{31}{2}\)sq.units

(c)

\(\frac{32}{3}\) sq.units

(d)

\(\frac{15}{2}\) sq.units

Generic 2

3 x 2 = 6

20.

Find the producer's surplus for the supply function p = x² + x + 3 when x_o = 4

21.

Find the consumer's surplus for the demand function p = 25 - x -x² when P_o= 19

22.

Find the demand function for which the elasticity of demand is 1

Match the following

5 x 1 = 5

23.

\(\frac{dp}{dx}\)

24.

ഽ(2 + 5e^x) dx

25.

\(\int _{ 0 }^{ t }{(100-90x)dx } \)

26.

\(\int _{ 0 }^{ t }{ f(t)dt } \)

27.

\(\int _{ 0 }^{ t }{ 20,000t\ dt } \)

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Answers Key:

Generic 5

4 x 5 = 20

= {-(x-5) if x<0
= {+(x-5) if x>0
Required area \(=\int _{ 0 }^{ 1 }{ |x-5|dx } \)
\(=\int _{ 0 }^{ 1 }{ -(x-5)dx } \)
\(=\int _{ 0 }^{ 1 }{ (5-x)dx } \)
\(={ \left[ 5x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=\left[ 5(1)-\frac { 1 }{ 2 } \right] -0\)
\(=5-\frac { 1 }{ 2 } =\frac { 10-1 }{ 2 } \)
\(=\frac { 9 }{ 2 } \) sq.units.

Given that MR \(=\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\)
\(\int { MR } =\int { \left( \frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 } \right) } dx\)
\(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +k\)
When x=0, R=0
\(\Rightarrow 0=\frac { { e }^{ 0 } }{ 100 } +0+0+k\)
\(k=-\frac { 1 }{ 100 } [\because { e }^{ 0 }=1]\)
∴ Revenue \(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { 1 }{ 100 } \)

y = x - 5

x	0	5
y	-5	0

The required area lies partially above X-axis and partially below X-axis
∴ Area \(=\int _{ 3 }^{ 5 }{ -ydx } +\int _{ 5 }^{ 7 }{ ydx } \)
\(=\int _{ 3 }^{ 5 }{ (5-x)dx+ } \int _{ 7 }^{ 5 }{ (x-5) } dx\)
\(=\left[ \because \quad y=x-5\Rightarrow -y=5-x \right] \)
\(={ \left[ 5x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 3 }^{ 5 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -5x \right] }_{ 5 }^{ 7 }\)
\(=\left( 5(5)-\frac { 25 }{ 2 } \right) -\left( 15-\frac { 9 }{ 2 } \right) +\left( \frac { 49 }{ 2 } -35 \right) -\left( \frac { 25 }{ 2 } -25 \right) \)
\(=25-\frac { 25 }{ 2 } -15+\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -35-\frac { 25 }{ 2 } +25\)
\(=(25-15-35+25)+\left( \frac { 25 }{ 2 } +\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -\frac { 25 }{ 2 } \right) \)
\(=0+\left( \frac { -25+9+49-25 }{ 2 } \right) \)
\(=\frac { 58-50 }{ 2 } =\frac { 8 }{ 2 } \)
A = 4 sq.units

y²= 4 ax is the right open
⇒ y = \(\sqrt { 4ax } \) parabola.
The limits are from x = 0 to x = a
∴ Area \(=2\int _{ 0 }^{ a }{ ydx } \)
\(=2\int _{ 0 }^{ a }{ \sqrt { 4ax } } dx\)
\(=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx } \)
\(=4\sqrt { a } \int _{ 0 }^{ a }{ \sqrt { x } dx } =4\sqrt { a } \int _{ 0 }^{ a }{ { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=4\sqrt { 3 } { \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }=4\sqrt { a } \times \frac { 2 }{ 3 } { \left( { x }^{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ a }\)
\(=8\frac { \sqrt { a } }{ 3 } \left( { a }^{ \frac { 2 }{ 3 } }-0 \right) \)
\(=\frac { 8\sqrt { a } }{ 3 } (a\sqrt { a } )\)
\(=\frac { 8 }{ 3 } { a }^{ 2 }\) sq.units

Generic 3

5 x 3 = 15

Given \(R'(x)=\frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \)
\(\Rightarrow \int { R'(x) } =\int { \left( \frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \right) } dx\)
\(\Rightarrow R(x)=\frac { -3 }{ x } -2log\quad x+k\)
Given R(1) = 6 ⇒ when x = 1, R = 6
\(\Rightarrow 6=\frac { -3 }{ 1 } -2log1+k\)
⇒ 6 + 3 = k [∵ log 1 = 0]
⇒ k = 9
\(\therefore R(x)=-\frac { 3 }{ x } -2log\quad x+9\)
Demand function \(P=\frac { R }{ x } \)
\(\\ =\frac { 3 }{ { x }^{ 2 } } -\frac { 2log\quad x }{ x } +\frac { 9 }{ x } \)

g(x) = 4x + 8 and x₀ = 5
p₀ = 4(5) + 8 = 28
PS = x₀ p₀ – \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \) dx
= (5 × 28) - \(\int _{ 0 }^{ 5 }{ (4x+8) } \) dx
= 140 – \({ \left[ 4\left( \frac { { x }^{ 2 } }{ 2 } \right) +8x \right] }_{ 0 }^{ 5 }\)
= 140 – (50 + 40)
= 50 units
Hence the producer’s surplus = 50 units.

Given y = 36 − x² and y₀ = 11
11 = 36 – x²
x² = 25
x = 5
CS = \(\int _{ 0 }^{ x }{ \text{(demand }\ \text{ function)dx–(Price×quantity demanded)}}\)
= \(\int _{ 0 }^{ 5 }{ (36-{ x }^{ 2 })dx-5\times 11 } \)
= \({ \left[ 36x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 5 }-55\)
= \(\left[ 36(5)-\frac { { 5 }^{ 3 } }{ 3 } \right] -55\)
= \(180-\frac { 125 }{ 3 } -55=\frac { 250 }{ 3 } \)
Hence the consumer’s surplus is = \(\frac { 250 }{ 3 } \)

Saving Cost S(t) = \(\int _{ 0 }^{ t }{ 20000t } \ dt\)
= 10000 t²
To recoup the total price,
10000 t² = 640000
t² = 64
t = 8
When t = 8 years, one can recoup the price.

Given curve y = 4x² is an open upward parabola
\(\Rightarrow \frac { y }{ 4 } ={ x }^{ 2 }\)
The limits are from y = 0 to y = 4
Since the shaded region lies to the right of Y-axis,
required area \(=\int _{ 0 }^{ 4 }{ xdy } \)
\(=\int _{ 0 }^{ 4 }{ \sqrt { \frac { y }{ 4 } } dy } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ 4 }{ \sqrt { y } dy } =\frac { 1 }{ 2 } \int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)
\(=\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}=\frac{1}{\not2} \times \frac{\not2}{3}\left[y^{\frac{3}{2}}\right]_{0}^{4}\)
\(=\frac { 1 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-0 \right] =\frac { 1 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 1 }{ 3 } { 4 }^{ 1 }\sqrt { 4 } =\frac { 1 }{ 3 } .4(2)\)
A = \(\frac{8}{3}\) sq.units

Multiple Choice Question

10 x 1 = 10

10.

(d)
\(\frac{2a^2}{3}\)

11.

(a)
\(\frac{4}{3}\)

12.

(a)
\(\frac{13}{3}\)

13.

(d)
\(\frac{\pi}{2}\)

14.

15.

16.

17.

(d)
54x - \(\frac { { 9x }^{ 2 } }{ 2 } \) + k

18.

(a)
2

19.

Generic 2

3 x 2 = 6

20.

Given supply function is P = x + x + 3 and X_o = 4
∴ P_o = 4² + 4 + 3
= 16+ 4 + 3 = 23
∴ p₀x₀= 23(4) = 92
Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=92-\int _{ 0 }^{ 4 }{ \left( { x }^{ 2 }+x+3 \right) dx } \)
\(=92-{ \left( \frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +3x \right) }_{ 0 }^{ 4 }\)
\(=92-\left( \frac { { 4 }^{ 3 } }{ 3 } +\frac { { 4 }^{ 2 } }{ 2 } +3(4) \right) \)
\(=92-\left( \frac { 64 }{ 3 } +8+12 \right) \)
\(=92-\frac { 64 }{ 3 } -20=72-\frac { 64 }{ 3 } \)
\(=\frac { 216-64 }{ 3 } \)
\(PS=\frac { 152 }{ 3 } \) units

21.

Given demand function is p = 25 - x - X²
and p₀= 19
⇒ 19 = 25-x-x²
⇒ x²+ x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units

22.

Given ηd = 1
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =1\)
\(\Rightarrow \frac { dx }{ x } =\frac { -dp }{ p } \)
⇒ log x = log p + log k
⇒log x+ log p = log k
⇒ log px = log k
⇒ px = k
Demand function P = \(\frac{k}{x}\)

Match the following

5 x 1 = 5

23.