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.

#### 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.

#### 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 $$(\overline x, \overline y)$$.

#### 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,

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

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

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

$$=\dfrac {256\sqrt 2}{5}$$

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

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

$$=\dfrac {2}{3}×8×2\sqrt 2=\dfrac {32\sqrt 2}{3}$$

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

$$=\dfrac {64}{4}=16$$

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

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

$$\therefore$$ 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

#### 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 $$f(x)\geq g(x)$$

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.

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

#### 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