Informative line

Application Of Integration In Economics And Physics

Learn center of mass of centroid of complicated region & demand function. Practice to find the centroid of the region bounded by y & consumer surplus calculus.

Economic Applications

Consumer Surplus:

  • The demand function \(P(x)\) is the price that a company has to charge in order to sell \(x\) units of a commodity. so selling larger quantities will require lowering price.

  • The demand function is a decreasing function as \(x=\) number of units sold is increasing what P = price is decreasing.
  • We define the consumer surplus = amount of money  a consumer saves in purchasing the commodity at price P, corresponding amount demanded of X. 

Consumer surplus = \(\int\limits_0^X(P(X)-P)\;dx\)

when, P = current selling price, where demand is X.

Illustration Questions

The demand function of a certain item is given by \(P=50-0.2x\) , find the consumer surplus when the sales level is 100.

A 1000

B 500

C 5000

D 50

×

Consumer surplus = \(\int\limits_0^X(P(X)-P)\;dx\)

where, P = price when \(X\) item are sold.

\(P(X)\) = demand function.

In this case, 

\(P(X)=50-0.2X\)

\(X=100\)

\(\Rightarrow P(100)=50-0.2×100=50-20=30\)

\(\therefore\) Consumer surplus = \(\int\limits_0^{100}(50-.2X-30)\;dx\)

\(=\int\limits_0^{100}(20-.2X)\;dx=\Bigg[ 20x-\dfrac {0.2x^2}{2} \Bigg]_0^{100}\)

\(=\left ( 20×100 – 0.2×\dfrac {100^2}{2} \right)-0\)

\(=2000-1000=1000\)

The demand function of a certain item is given by \(P=50-0.2x\) , find the consumer surplus when the sales level is 100.

A

1000

.

B

500

C

5000

D

50

Option A is Correct

Producer Surplus

  • The supply function \(P_s(x)\) of a commodity gives the relation between the selling price and number of units that manufacturer will produce at that price.
  • \(P_s(x)\) is the increasing function of x, as manufacturer will produce more units if price is more.
  • Some producer will sell the commodity for lower selling price and receive more than the minimal price. The excess is called producer surplus.

Producer surplus = \(\int\limits_0^X(P-P_s(X))\; dx\)

where P is the price when, X items are sold.

Illustration Questions

Find the producer surplus for the supply function. \(P_s(X)=4+0.1X^2\). When the sales level is X = 15.

A 11.25

B 22.5

C 500

D 1.25

×

Producer surplus = \(\int\limits_0^X(P-P_s(X))\; dx\)

where,\(P_s(X)\)is the supply function and P is the price where X units are sold.

In this case 

\(P_s(X)=4+0.1X^2\),\(X=15\)

\(P(15)=4+.01×15^2=6.25\)

\(\therefore\) Producer surplus = \(\int\limits_0^{15}6.25-(4+0.1X^2)\; dx\)

\(=\int\limits_0^{15}(2.25-0.01X^2)\; dx=2.25x-\dfrac {0.01x^3}{3}\Bigg]_0^{15}\)

\(=2.25×15-.01×\dfrac {15×15×15}{3}=33.75-11.25=22.5\)

Find the producer surplus for the supply function. \(P_s(X)=4+0.1X^2\). When the sales level is X = 15.

A

11.25

.

B

22.5

C

500

D

1.25

Option B is Correct

Applications in Physics (Moments and Center of Mass)

  • The center of mass of any plate is the point P on which the plate balance horizontally.
  • If we have system of n particles with masses \(m_1, m_2, m_3,.....m_n\) located on the points \(x_1, x_2, x_3,.....x_n\) on X-axis then the center of mass is located at \(\overline x=\dfrac {m_1x_1+m_2x_2+....+m_nx_n}{m_1+m_2+....+m_n}=\dfrac {\sum\limits_{i=1} ^nm_ix_i}{\sum\limits_{i=1}^{n}m_i}\)

for n = 2

  • The value \(m_ix_i\) is called the moment of mass \(m_i\) about origin.
  • \(M=\sum\limits_{i=1}^{n}m_ix_i\) is called moment of the system about the origin.
  • If we consider the system of n particles with masses \(m_1,m_2....,m_n\) located at point \((x_1, y_1), (x_2, y_2),....,(x_n,y_n)\) in the X-Y plane. We define the moment of system about Y-axis as \(M_y=\sum\limits_{i=1}^nm_ix_i\) and moment of system about X axis to be \(M_x=\sum\limits_{i=1}^nm_iy_i\)

where, \(M_y=\) tendency of system to rotate about Y-axis

\(M_x=\) tendency of system to rotate about X-axis.

  • \(\overline x=\dfrac {M_y}{m}, \) \(\overline y=\dfrac {M_x}{m}\), where \(m=\sum\limits_{i=1}^{n}m_i\)
  • Center of mass = \((\overline x , \overline y)\)

Center of Mass or Centroid of Region

  • Consider a flat plate also called the lamina with uniform density \(\rho\)that occupies region R of the plane.

The center of mass or the centroid of this plate is given by \((\overline x, \overline y)\) when

\(\overline x =\dfrac {\int\limits_a^bxf(x) dx}{\int\limits_a^bf(x) dx}\),  \(\overline y =\dfrac {\int\limits_a^b\,\dfrac{1}{2} (f(x))^2 dx} {\int\limits_a^bf(x) dx}\)

  • If the region R lies between two curve \(y=f(x)\) and \(y=g(x)\) when \(f(x)\geq g(x)\)\(f\)is higher than \(g\)), then

\(\overline x =\dfrac {\int\limits_a^bx\Big(f(x)-g(x)\Big) dx} {\int\limits_a^b\Big(f(x)-g(x)\Big) dx}\)\(\overline y =\dfrac {\int\limits_a^b\,\dfrac{1}{2} \Big((f(x))^2-(g(x))^2\Big) dx} {\int\limits_a^b\Big(f(x)-g(x)\Big) dx}\) and centroid is \((\overline x, \overline y)\).

Illustration Questions

Find the centroid of the region bounded by \(y=\sqrt x\), \(y = 0\), \(x = 8\)

A \(\left ( \dfrac {11}{2}, \dfrac {5}{2\sqrt2} \right) \)

B \(\left ( \dfrac {24}{5}, \dfrac {3}{2\sqrt2} \right) \)

C \(\left ( 5, \dfrac {1}{\sqrt2} \right) \)

D \(\left ( -\dfrac {1}{2}, -\dfrac {1}{2} \right) \)

×

Centroid  \(\equiv(\overline x, \overline y)\)

where, 

\(\overline x =\dfrac {\int\limits_a^bxf(x) dx}{\int\limits_a^bf(x) dx}\),  \(\overline y =\dfrac {\int\limits_a^b\,\dfrac{1}{2} (f(x))^2 dx} {\int\limits_a^bf(x) dx}\)

 

image

\(\int\limits_a^bx f(x) dx=\int\limits_0^8x× \sqrt x dx =\int\limits_0^8x^{3/2} dx\)

\(=\Bigg[\dfrac {x^{5/2}}{5/2}\Bigg]_0^8=\dfrac {2}{5}×8^{5/2}=\dfrac {2×(2\sqrt2)^5}{5}=\dfrac {2×32×4\sqrt 2}{5}\)

\(=\dfrac {256\sqrt 2}{5}\)

\(\int\limits_a^b f(x) dx=\int\limits_0^8 \sqrt x\; dx=\Bigg[\dfrac {x^{3/2}}{3/2}\Bigg]_0^8\)

\(=\dfrac {2}{3}×8^{3/2}=\dfrac {2}{3}×(2\sqrt 2)^3\)

\(=\dfrac {2}{3}×8×2\sqrt 2=\dfrac {32\sqrt 2}{3}\)

 

\(\int\limits_a^b \dfrac {1}{2} \Big(f(x)\Big)^2 dx=\int\limits_0^8 \dfrac {1}{2}× x\; dx=\Bigg[\dfrac {x^2}{4}\Bigg]_0^8\)

\(=\dfrac {64}{4}=16\)

\(\therefore\) \(\overline x =\dfrac {\int\limits_a^bxf(x) dx}{\int\limits_a^bf(x) dx}\)\(=\dfrac {\dfrac {256\sqrt2}{5}} {\dfrac {32\sqrt2}{3}}=\dfrac {24}{5}\)

\(\overline y =\dfrac {\int\limits_a^b\,\dfrac{1}{2} (f(x))^2 dx} {\int\limits_a^bf(x) dx}\)\(=\dfrac{16}{\dfrac {32\sqrt2}{3}}=\dfrac {3}{2\sqrt 2}\)

\(\therefore\) Centroid is \(\left ( \dfrac {24}{5}, \dfrac {3}{2\sqrt2} \right) \)

Find the centroid of the region bounded by \(y=\sqrt x\), \(y = 0\), \(x = 8\)

A

\(\left ( \dfrac {11}{2}, \dfrac {5}{2\sqrt2} \right) \)

.

B

\(\left ( \dfrac {24}{5}, \dfrac {3}{2\sqrt2} \right) \)

C

\(\left ( 5, \dfrac {1}{\sqrt2} \right) \)

D

\(\left ( -\dfrac {1}{2}, -\dfrac {1}{2} \right) \)

Option B is Correct

Illustration Questions

Find the centroid of the region bounded by the curves  \(y=6-x^2\) and \(y = x\).

A \(\left ( \dfrac {5}{4}, \dfrac {3}{2} \right) \)

B \(\left (- \dfrac {1}{2}, 2 \right) \)

C \((6,8)\)

D \((-1, 1)\)

×

Centroid is given by  \((\overline x, \overline y)\)

\(\overline x =\dfrac {\int\limits_a^bx\Big(f(x)-g(x)\Big) dx} {\int\limits_a^b\Big(f(x)-g(x)\Big) dx}\)\(\overline y =\dfrac {\int\limits_a^b\,\dfrac{1}{2} \Big((f(x))^2-(g(x))^2\Big) dx} {\int\limits_a^b\Big(f(x)-g(x)\Big) dx}\) 

where \(f(x)\geq g(x)\)

 

image

So, \(f(x)=6-x^2, g(x)=x\)

Solve for the points of intersection A and B.

\(6-x^2=x\Rightarrow x^2+x-6=0\)

\(\Rightarrow (x+3)(x-2)\Rightarrow x=-3, 2\)

\(\therefore\) \(A=(-3,3)\) and \(B(2, 2)\)

\(\int\limits_a^b \Big(f(x)-g(x)\Big) dx=\int\limits_{-3}^{2}x(6-x^2-x) dx \)

\(=\int\limits_{-3}^{2}(6x-x^3-x^2) dx =3x^2-\dfrac {x^4}{4}-\dfrac {x^3}{3}\Bigg]_{-3}^{2}\)

\(=\left ( 3×4- \dfrac {16}{4} - \dfrac {8}{3} \right)\)– \(\left (36-\dfrac {81}{4} \right)\)

\(=\dfrac {16}{3}-\dfrac {63}{4}=\dfrac {64-189}{12}=\dfrac {-125}{12}\)

\(\int\limits_a^b \Big(f(x)-g(x)\Big) dx=\int\limits_{-3}^2 (6-x^2-x)\; dx=6x-\dfrac {x^3}{3}-\dfrac {x^2}{2}\Bigg]_{-3}^2\)

\(=\left ( 6×2-\dfrac {8}{3}-\dfrac {4}{2} \right) -\left ( -18+9-\dfrac {9}{2} \right)\)

 

\(=\left ( 10-\dfrac {8}{3} \right) -\left ( -\dfrac {27}{2} \right)\)

\(=10-\dfrac {8}{3}+\dfrac {27}{2}=\dfrac {60-16+81}{6}=\dfrac {125}{6}\)

\(\int\limits_a^b\,\dfrac{1}{2} \Big((f(x))^2-(g(x))^2\Big) dx= \dfrac {1}{2}\int\limits_{-3}^{2}\Big(6-x^2)^2-x^2\;dx \)

\(=\dfrac {1}{2}\int\limits_{-3}^{2}\;(36+x^2-13x^2)\;dx \) = \(=\dfrac {1}{2}\left [36x+\dfrac {x^5}{5} -\dfrac {13x^3}{3}\right ]_{-3}^2\)

\(=\dfrac {1}{2}\left [ \left (72+\dfrac{32}{5} -13× \dfrac{8}{3}\right) - \left (-108-\dfrac {243}{5} +13×\dfrac {27}{3}\right) \right ]\)

\(=\dfrac {250}{6}\)

 

\(\therefore\) \(\overline x= \dfrac {\dfrac {-125}{12}} {\dfrac {125}{6}} =\dfrac {-1}{2}\)

\(\overline y= \dfrac {\dfrac {250}{6}} {\dfrac {125}{6}} = 2\)

\(\therefore\)Centroid is at \(\left (-\dfrac {1}{2}, 2\right)\)

Find the centroid of the region bounded by the curves  \(y=6-x^2\) and \(y = x\).

A

\(\left ( \dfrac {5}{4}, \dfrac {3}{2} \right) \)

.

B

\(\left (- \dfrac {1}{2}, 2 \right) \)

C

\((6,8)\)

D

\((-1, 1)\)

Option B is Correct

Theorem of Pappus

  • The theorem states that if R is a plane region that lies entirely on one side of line \(\ell\) and is rotated about \(\ell\), then volume of resulting solid is the product of area A of R and distance 'd' traveled by centroid of R.

Illustration Questions

A truss is formed by rotating a circle of radius 4 m about a line in the plane of circle that is at a distance of 5m from center of circle. What is the volume of truss.  

A 320\(\pi^2m^3\)

B 160\(\pi^2m^3\)

C 2\(m^3\)

D 100000\(\pi^2m^3\)

×

Pappus's theorem states that if R is a plane region that lies entirely on one side of line \(\ell\) and is rotated about \(\ell\), then volume of resulting solid is the product of area A of R and distance 'd' traveled by centroid of R.

In this case,

Area which is rotated = Area of circle of radius 4 m

\(\pi×4^2=16\;\pi m^2\)

Centroid of circle is its center

\(\therefore\)  distance traveled by the center 

\(=2\pi×5\)

\(=10\;\pi m\)

\(\therefore\) By Pappus's theorem

Volume of truss = \(16\;\pi×10\;\pi m^3\)

\(=160\;\pi^2 m^3\)

A truss is formed by rotating a circle of radius 4 m about a line in the plane of circle that is at a distance of 5m from center of circle. What is the volume of truss.  

A

320\(\pi^2m^3\)

.

B

160\(\pi^2m^3\)

C

2\(m^3\)

D

100000\(\pi^2m^3\)

Option B is Correct

Capital Formation Problem

  • If the amount of capital that a company has at a time t is \(f(t)\) then the derivative \(f' (t)\) is called the net investment flow.

\(\therefore\) Capital formation function \(t_1\) to \(t_2\) \(=\int\limits_{t_1}^{t_2}\) \(f' (t) dt\)

Illustration Questions

If revenue flows into a company at the rate of \(g(t)=3000\sqrt {1+2t}\) where, \('t'\) is in year and \(g(t)\) in Dollars per year, then find total revenue observed in first twelve years.  

A 5000 $

B 125000 $

C 1255000 $

D 50 $

×

Total revenue = Capital formation

\(=\int\limits_{t_1}^{t_2}\) \(f' (t) dt\)

In this case, 

\(f'(t) = g(t) = 3000\sqrt {1+2t}\;dt\)

\(t_2=12,\;t_1=0\)

\(\therefore\) Total revenue = \(\int\limits_0^{12}3000\sqrt {1+2t}\;dt\)

\(=3000×\int\limits_0^{12}\sqrt {1+2t}\;dt\) \(=3000× \dfrac {(1+2\,t)^{3/2}}{3/2×2} \Bigg]_0^{12}\)

\(=3000× \dfrac {(1+2\,t)^{3/2}}{3} \Bigg]_0^{12}=1000×(1+24)^{3/2}\)

= 1000 × 125 = 125000 $

If revenue flows into a company at the rate of \(g(t)=3000\sqrt {1+2t}\) where, \('t'\) is in year and \(g(t)\) in Dollars per year, then find total revenue observed in first twelve years.  

A

5000 $

.

B

125000 $

C

1255000 $

D

50 $

Option B is Correct

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