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.

1000

500

5000

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.

11.25

22.5

500

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,

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

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

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

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

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

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

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

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

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

$\therefore$ $\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 $y=\sqrt x$, $y = 0$, $x = 8$

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

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

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

$\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 $(\overline x, \overline y)$

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 $y = x$.

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

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

$(6,8)$

$(-1, 1)$

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.

320$\pi^2m^3$

160$\pi^2m^3$

2$m^3$

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.

5000 $

125000 $

1255000 $

50 $

Option B is Correct

Application Of Integration In Economics And Physics

Economic Applications

Economic Applications

Consumer Surplus:

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.

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

Producer Surplus

Producer Surplus

Illustration Questions

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

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

Applications in Physics (Moments and Center of Mass)

Applications in Physics (Moments and Center of Mass)

Center of Mass or Centroid of Region

Illustration Questions

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

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

Illustration Questions

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

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

Theorem of Pappus

Theorem of Pappus

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 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.

Capital Formation Problem

Capital Formation Problem

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.

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.