CHAPTER 11
INTEGRAL CALCULUS

Integration is the inverse of differentiation.  The problem here is to find f(x), when f '(x) is given.  We call f(x) an antiderivative of f '(x), and the method of finding it is called integral calculus.

Integral calculus will be used for finding solutions of differential equations in chapter 12.  As well, integral calculus has many applications in business and economic problems.  One such problem is to find the original function when the marginal function, or the rate of change of one variable with respect to other variable is known.  For example, If marginal cost function is known, the total variable cost function can be found by finding the antiderivative of the marginal cost function.  

While the antiderivative of a function is a function, the integral of a function at a closed interval is, generally, a number.  For example, given the demand or marginal revenue functions, the increase or decrease in total revenue resulting from a price change can be found by integrating the marginal revenue function over the given price range. 
 
In the next section, indefinite integrals are discussed and some simple rules of integration are presented.  It is followed by the definite integrals and application of integral calculus to economic problems.

Tool Box
Mathematica Commands		Function
Integrate[f(x),  x]			Finds the indefinite integral of f(x) with respect to x, ∫ f(x)dx.
Integrate[f(x), {x, a, b}]		Finds the definite integral of f(x) in the interval [a, b], (∫_a^b) f (x) dx.

I.  Indefinite Integrals:  The Integrate[f(x), x] Command
Indefinite integration (antiderivative) is the process of reverse differentiation.  That is, given the derivative of a function with respect to some variable, often called derived function, the indefinite integral (denoted by ∫ ) is used to find the primitive function, or the function that gave rise to that derivative.  For example, ∫(x^2) dx = ((x^3)/3) + c because (d/dx)(((x^3)/3) + c) = (x^2).

If the derived function F'(x) = f(x), F(x) is an integral of f(x).  However, the general form of the integral of f(x) is F(x) + c, where c is an arbitrary constant.  Such a general form of an integral is called indefinite integral and is denoted by ∫ f(x)dx = F(x)+c where, f(x)dx is the differential of F(x), f(x) is the integrand function, and c is the constant of integration.

Example: Cost Function
A firms's marginal cost (MC) as a function of output (Q) is given as MC(Q) = 30 -2Q + .1(Q^2).  Find the total cost function (TC).
Solution:  What we know is the rate of change of total cost with respect to output, or marginal cost (the derived function), and we wish to uncover the primitive function TC.  This is the inverse of differentiation,  TC(Q) = ∫MC dQ = 30Q - (Q^2) + (1/30)(Q^3) + C, where C is the constant of integration, here interpreted as the fixed cost of the firm.
TC(Q) is the primitive function, which shows more than one function because C can be any number.  Therefore, additional information is needed to identify the function.  This additional information is called initial condition or boundary condition.  If we had the information that the fixed cost is 100, then TC(0) = C = 100, and TC(Q) = 30Q - (Q^2) + .033(Q^3) + 100.
To find the indefinite integral of a function using Mathematica, use the command Integrate as follows.
 Clear[Q] 
Integrate[30 - 2Q + .1Q^2, Q]
30.` Q - 1.` (Q^2) + 0.03333333333333333` (Q^3)

Rules of Integration:The following are some useful rules of integration.  It should be noted that there are integrands for which exact integrals cannot be found.  In those cases, only approximations will be available.

Rule 1.  Integration of a constant k is kx + c:   ∫k dx = kx + c.

Example: Find  ∫ 5dx       
Solution:  Here, k = 5.  Therefore, ∫ 5dx = 5x + c.  
In Mathematica, use the command "Integrate" as follows,
Clear[x]
Integrate[5, x]
5 x

Rule 2.  The Power Rule: Integral of (x^n) is (1/(n + 1))(x^(n + 1)) + c,  ∫ (x^n)dx = (1/(n + 1))(x^(n + 1)) + c,   n ≠ -1
          
Example:  Find  ∫ (x^3)dx
Solution:  Here, n = 3.  Therefore,  ∫ (x^3)dx = (1/4)(4^x)+ c.  
In Mathematica, 
Clear[x]
Integrate[x^3, x]
((x^4)/4)

Example: Find  ∫(((x^2))^(1/3)) dx.
Solution:  Since (((x^2))^(1/3)) = (x^((2/3))),  ∫(((x^2))^(1/3)) dx = (3/5)(x^((5/3)))+ c.  
In Mathematica, 
Clear[x]
Integrate[x^(2/3), x]
((3 (x^(5/3)))/5)

Example:  Find  ∫ (1/(x^3))dx .
Solution:  Since (1/(x^3)) = (x^(-3)) , we have n = -3.  Therefore,   ∫ (1/(x^3))dx = ((-1)/2)(x^(-2 )) + c = ((-1)/(2  (x^2)))  +  c.  
Using Mathematica,
Clear[x]
Integrate[1/x^3, x]
-(1/(2 (x^2)))

Example:  Variable Cost Function
The marginal cost function (MC) of a single product firm t is given as MC(Q) = (Q^2) - 7Q + 26,  where Q is the quantity of output.  Find the total variable cost function (TVC). 
Solution:  Total variable cost function represents the total variable cost of producing Q units of output.  Since the integral of a rate of change (marginal) is the total change, given the marginal cost MC(Q) the total variable cost TVC(Q) = ∫ MCdQ = ∫ ((Q^2) - 7Q + 26)dx. =  26 Q - ((7 (Q^2))/2) + ((Q^3)/3).
Using Mathematica,
Clear[Q]
Integrate[Q^2 - 7Q + 26, Q]
26 Q - ((7 (Q^2))/2) + ((Q^3)/3)

Example:  Capital Formation
The net investment flow of a firm ( I ) at time t is given as I( t ) = 3\(t\^ .8\), and the initial capital stock (K) at t = 0 is K(0) = 1500.  Find the time path of the capital formation.  
Solution:  If the capital formation is continuous over time, the rate of capital formation at time t is (dk/dt) = I( t ) or dk = I( t )dt.  Therefore, K( t ) = ∫ I( t )dt.  Substituting for I( t ) yields, K( t ) = ∫ 3\(t\^ .8\)dt = 1.66667(t^1.8) + c.  Since K(0) = 1500, solve for c by setting t = 0 and K( t ) = 1500.  That is c = 1500 and  K( t ) = 1.66667(t^1.8) + 1500.
Using Mathematica, integrate the function and set the integral equal to sol.  Solve sol+c = 1500 for t = 0.
Clear[t, sol, c]
sol=Integrate[3t^.8, t]
1.6666666666666667` (t^1.8)`
Solve[sol+c==1500, c]/.t->0
{{c->1500.}}
Therefore, K( t ) = 1.66667(t^1.8) + 1500.  You can plot the time path of capital growth by plotting the K( t ) function  as,
Plot[1.66667t^1.8 + 1500, {t, 0, 20}]

⁃Graphics⁃

 Rule 2a.  Special Case of Power Rule:  If n = -1, then ∫ (x^n)dx = ∫(1/x) dx = lnx + c

Example:  Find   ∫ (5/x)dx 
Solution:   ∫ (5/x)dx = 5lnx + c.  
Using Mathematica,
Clear[x]
Integrate[5/x, x]
5 Log[x]

Rule 3.   Constant Multiple of a Function:  ∫ kf(x)dx = k∫ f(x)dx.

Example:  Find  ∫5(x^4)dx.
Solution:   ∫5(x^4)dx  = 5 ∫(x^4)dx = 5((1/5)(x^5) + c) = (x^5) + C where, C = 5c.
Using Mathematica,
Clear[x]
Integrate[5x^4, x]
(x^5)

Rule 4.  Integral of a sum of Functions:   ∫(u + v + w) dx  = ∫ udx + ∫vdx + ∫ wdx, where u, v, and w are a function of  x.  

Example:  Find  ∫(5(x^3)+ 3(x^2) + x)dx.
Solution:  ∫(5(x^3)+ 3(x^2) + x)dx = ∫5(x^3)dx  + ∫ 3(x^2)dx  + ∫ xdx = (5/4)(x^4) + c1+ (x^3) + c2 + (1/2)(x^2) + c3 = (5/4)(x^4) + (x^3) + (1/2)(x^2) + C,  where C = c1 + c2 +c3.
 Using Mathematica,
Clear[x]
Integrate[5x^3 + 3x^2 +x, x]
((x^2)/2) + (x^3) + ((5 (x^4))/4)

Example:  Find ∫(((x^3) +  2  (x^2)  -  1)/(10  x)) dx.               
Solution:  ∫(((x^3) +  2  (x^2)  -  1)/(10  x)) dx. = ∫((x^3)/(10  x)) dx + ∫(( 2  (x^(2 )))/(10  x)) dx. + ∫((- 1)/(10  x)) dx = ((x^3)/30) + ((x^2)/10) - (1/10)lnx + C.
Using Mathematica,
Clear[x]
Integrate[(x^3+2x^2-1)/(10x), x]
(1/10) ((x^2) + ((x^3)/3) - Log[x])

Rule 5.  Integration by Substitution or Integration by Change of Variable:  ∫ (u^n)du = (1/(n + 1))(u^(n + 1))du + c, where u = f(x) and n ≠ 1.
 
Example:  Find ∫ 2x((((x^(2 )) +  3 ))^5)dx.
Solution:  Set (x^2) + 3 = u so that, ((((x^(2 )) +  3 ))^5)= (u^5) , and (du/dx) = 2x or du = 2xdx.  Therefore,  ∫ 2x((((x^(2 )) +  3 ))^5)dx = ∫ ((((x^(2 )) +  3 ))^5) (2  x)dx = ∫ (u^5)du = (1/6)(u^6) + c = (1/6)((((x^(2  )) +  3))^6) + c.   Integration by substitution is applicable whenever the original integrand function, f(x), can be transformed as, ∫ f(x)dx = ∫ f(g(x))g' (x)dx so that we can set  f(g(x)) = f(u) where, u = g(x) and g'(x) =(du/dx).  The integral is then transformed to ∫ f(x)dx = ∫ udu.
Using Mathematica, the solution to the example is,
Clear[x]
Integrate[2x  (x^2+3)^5, x]          
2 (((243 (x^2))/2) + ((405 (x^4))/4) + 
            45 (x^6) + ((45 (x^8))/4) + ((3 (x^10))/2) + ((x^12)/12))
Factor[%]
(1/6) (x^2) (6 + (x^2)) (9 + 3 (x^2) + (x^4)) (27 + 
            9 (x^2) + (x^4))

Example:  Find ∫(x^2) (((2  (x^(3  )) +  1))^4)dx.
Solution:  Set 2(x^3) + 1 = u so that, (((2  (x^(3  )) +  1))^4) = (u^4), and (du/dx)= 6(x^2) or du = 6(x^2)dx.   Therefore, ∫(x^2) (((2  (x^(3  )) +  1))^4)dx = ∫(1/6) (u^4) du = (1/30)(u^5) + c = (1/30)(((2  (x^(3  )) +  1))^4) + c.
Using Mathematica,
Clear[x]
Integrate[x^2  (2x^2+1)^4, x]
\(x\^3\/3 + \(8\ x\^5\)\/5 + \(24\ x\^7\)\/7 + \(32\ x\^9\)\/9 + \(16\ x\^11\)\/11\)
Factor[%]
\(\(x\^3\ \((1155 + 5544\ x\^2 + 11880\ x\^4 + 12320\ x\^6 + 5040\ x\^8)\)\)\/3465\)

Rule 5a.  Special Case of Integration by Substitution:  If n = -1, then ∫ (u^n)du = ∫ (1/u)du = lnu + c.

Example:  Find ∫(1/(z  +  5)) dx.
Solution:  Set x + 5 = u, and note that (du/dx) = 1 or du = dx.  Hence, ∫(1/(x  +  5)) dx = ∫ (1/u)du = lnu + c = ln(x + 5) + c.
Using Mathematica,
Clear[x]
Integrate[1/(x+5), x]     
Log[5+x]

Example:  Find ∫((3  x)/((x^2)  +  1)) dx.     
Solution:  Set (x^2) + 1 = u.  Since (du/dx) = 2x, du = 2xdx.  Therefore,  ∫((3  x)/((x^2)  +  1)) dx =  ∫ (3/(2  u))du = (3/2) lnu + c = (3/2)ln((x^2) + 1).
Using Mathematica,
Clear[x]
Integrate[3x/(x^2+1), x]     
(3/2) Log[1 + (x^2)]

Example:  Find ∫(Sqrt[1 + 
                x]) (x^3) dx.
Solution:  Set u = 1+ x.  This gives du = dx and (x^3) = (((u - 1))^3) = (u^3) - 3(u^2) + 3u - 1.  Therefore, ∫(Sqrt[1 + 
                x]) (x^3) dx = ∫(Sqrt[u]) ((u^3) - 3  (u^2) + 3  u - 
              1) du = ∫((u^((7/2))) - 3  (u^((5/2))) + 
              3  (u^((3/2)))  -  (u^((1/2)))) du = ((2  (u^((9/2))))/9) - ((6  (u^((7/2))))/7)+ ((6  (u^((5/2))))/5) - (2/3)(u^((3/2))).  Substituting for u = 1 + x gives∫(Sqrt[1 + 
                x]) (x^3) dx = \(\@\(1 + x\)\ \((\(-\(32\/315\)\) + \(16\ x\)\/315 - \(4\ x\^2\)\/105 + \(2\ x\^3\)\/63 + \(2\ x\^4\)\/9)\)\).
Using Mathematica,
Clear[x]
Integrate[x^3 Sqrt[x+1], x] 
\(\@\(1 + x\)\ \((\(-\(32\/315\)\) + \(16\ x\)\/315 - \(4\ x\^2\)\/105 + \(2\ x\^3\)\/63 + \(2\ x\^4\)\/9)\)\)

Rule 6.  Integration of Exponential Functions of Base e:  ∫ (e^nx)dx = (1/n)(e^nx) + c.

Example:   Find ∫2(e^x) dx.     
Solution:  ∫2(e^x) dx = 2(e^x) + c. 
Clear[x]
Integrate[2E^x, x]
2 (e^x)

Rule 6a.  ∫ (e^u)du = (e^u) + c where, u = f(x).
      
Example:   Find  ∫ (e^(5  x))dx.    
Solution:  Let 5x = u so that, du = 5dx, or dx = (1/5)du.  Therefore, ∫ (e^(5  x))dx = ∫ (1/5) (e^u)du = (1/5)(e^u) + c = (1/5)(e^(5  x)) + c.
Using Mathematica,
Clear[x]
Integrate[E^(5x), x]
((e^(5 x))/5)

Example:   Find  ∫ 3x(e^u)dx,  where u =  (x^2) + 5.    
Solution:  Since du = 2xdx, we can write ∫ 3x(e^u)dx = ∫ (3/2)(e^u)du = (3/2)(e^u) + c.  Substituting for u results in  ∫ 3x(e^u)dx = (3/2)(e^(((x^2) + 5))) + c.
Clear[x]
Integrate[3x E^(x^2 + 5), x]
((3 (e^(5 + (x^2))))/2)

Example:  Cost Function
The marginal cost function of a single-product firm (MC) is given as MC(Q) = 2(e^((2 +  .1  Q))).   Find the total variable cost function (TVC) of the firm.
Solution:  The integral of a rate of change (marginal) is the total change.  Therefore, TVC(Q) = ∫ MCdQ = ∫2(e^((2 +  .1  Q))) dx.  
Using Mathematica,
Clear[x]
Integrate[2E^(2+.1Q), Q]
147.781121978613` (e^(0.1` Q))

Rule 7.  Integration of Exponential Functions of Base b:  ∫ (b^nx)dx = (1/lnb)(b^nx) + c.

Example:  Find ∫ (5^x)dx.    
solution:  ∫ (5^x)dx = ((5^x)/lnx) + c.
Using Mathematica,
Clear[x]
Integrate[5^x, x]
((5^x)/Log)[5]

Rule 8.  Integration by Parts:  ∫ udv = uv  - ∫ vdu,  where u = f(x) and v = g(x).   The formula follows from the derivative of the product uv with respect to x.  We know that, d(uv) = udv + vdu  or  udv = d(uv) - vdu.  Integrating both sides yields,  ∫ udv = uv  -  ∫ vdu.  Basically, integration by parts is used when the integral of an expression involving products or logarithms cannot be evaluated directly.  The integrand should then be transformed to standard integrable parts.
     
The application of integration by parts depends on the appropriate choice of u and dv, so that ∫ vdu and ∫ dv can be evaluated, even though ∫ udv cannot.  Therefore, it is advisable to choose u and v in such a way that: 1) dv is integrable; 2) when the integrand is a product of two functions, choose the more complicated one as part of dv, in order to make ∫ vdu as easily integrable as possible.
     
Example:   Find ∫ x(e^x)dx.     
Solution:  Since this integral cannot be evaluated directly, set u = x and dv = (e^x)dx, so that, du = dx and v = ∫ (e^x)dx = (e^x) + c.  Hence, ∫ x(e^x)dx  = ∫udv = uv - ∫ vdu = x(e^x)  -  (e^x)dx  =  x(e^x) -  (e^x) + c = (e^x)(x - 1).
Using Mathematica,
Clear[x]
Integrate[x E^x, x]
(e^x) (-1 + x)

Example:    Find ∫ xlnxdx.     
Solution:  Set x = dv and lnx = u so that,  v = (1/2) (x^2) and du = (1/x).  Using Rule 8, ∫ xlnxdx = ∫ udv = uv - ∫ vdu = (1/2)(x^2)lnx - ∫ (x/2)dx = (1/2)(x^2)lnx - ((x^2)/4) + c.
Clear[x]
Integrate[x Log[x], x]
-((x^2)/4) + (1/2) (x^2) Log[x]

Problems:
1- Find integrals of the following functions:
a)  ∫ (1/2)(x^3)dx			b)  ∫ 3 (x^((1/2)))dx	
c)  ∫ 3 (x^2)((((x^(3  )) -  4))^2)dx		d)  ∫ (1/(x  -  2))dx	
e)  ∫(2(x^3) + (1/(x^4)) + 2x)dx		 f)  ∫2(e^(4  x)) dx
g)  ∫ (1/2)((((e^(2  )) +  1))^2) (e^(2  x))dx	 h) ∫ ln(5^((2 - 2  x)))dx
i)  ∫ x(e^x)dx			  j)  ∫ (x^3)(e^(3  x))dx
k)  ∫x(e^(-x)) dx    			  l)  ∫ lnxdx
m) ∫ 5ln(x^4)dx			 n)  ∫(x^2) lnxdx
 0) ∫(Sqrt[1 + 
                (x^2)]) (x^5) dx    		 p) ∫(Sqrt[1 + 
                (x^3)]) (x^2) dx
2.  The marginal cost function (MC) of a single product firm is given as MC = 5 + 16Q,  where Q is the quantity of output.  Find the total variable cost function (TVC).      
3.  Suppose that the net investment flow of a firm ( I ) at time t is given as I( t ) = 3\(t\^ .5\), and the initial capital stock (K) at t = 0 is K(0) = 500.  Find the time path of the capital (Note that, I = (dK/dt)).   
4. The marginal propensity to save (MPS) of a country is given as MPS = f(Y) = .3 - .1\(y\^ .5\).  Find the saving function, given that S = 0 when Y = 81.    
5.  The marginal cost function of a single-product firm (MC) is given as MC = 2(e^((2 +  .1  Q))), and the fixed cost is FC = 90.   Find the total cost function.                      
                 
II.  Definite Integrals:  The Integrate[f(x),{x,a,b}] Command
Contrary to the indefinite integrals, which do not possess definite numerical values, definite integrals have specific numerical values free of the variable x and the arbitrary constant c.

Definite integrals are the result of what is famous to the Fundamental Theorem of Integral Calculus.  This theorem connects integral calculus to differential calculus, and in turn, it simplifies the solution of many integral problems.  The Fundamental Theorem enables us to compute areas under the functions and find integrals of functions without having to compute them as limits of sums.   The Fundamental Theorem consists of the following two parts:

Part 1.  If f(x) is continuous on [a, b], then function g(x) defined by 
		g(x) = (∫_a^x) f(x) dx , a ≤ x ≤ b 	
is continuous on [a, b] and differentiable on (a, b), and g'(x) = f(x).

Part2.  If f(x) is continuous on [a, b], then 
		(∫_a^b) f(x) dx = F(a) - F(b)
where F(x) is any antiderivative of f(x).  That is, a function that F'(x) = f(x).

Proof of the fundamental theorem of integral calculus can be found in any standard calculus text book.  From expressions above it is clear that a definite integral can be evaluated by first finding the indefinite integral of the integrand function, and then substituting the upper and lower limits of integration for x to get F(a) and F(b).

Example: Evaluate the definite integral (∫_1^6)(x^2) dx.
Solution:  The indefinite integral of the integrand function is F(x) = (1/3)(x^3) + c.  Substitute the upper and lower limits of integration for x to find the definite integral, F (Subscript[((x)), 1]^6) = F(6) - F(1) = 72 + c - ((1/3) + c) = (215/3).  
In Mathematica, use the command Integrate as follows,
Clear[x]
Integrate[x^2, {x, 1, 6}]
(215/3)

Example:  Evaluate (∫_0^5)(x/((x^(2 )) +  1)) dx.
Solution:  The indefinite integral of the integrand function is F(x) = (1/2)ln((x^2) + 1) + c. Therefore,  (∫_0^5)(x/((x^(2 )) +  1)) dx = F(5) - F(0) = (1/2)(ln26 - lnl) = (1/2)ln26.  
In Mathematica,
Clear[x]
Integrate[x/(x^2 + 1), {x, 0, 5}]  
Log[26]/2

Example:  Application to Business
Marginal cost of a product is estimated to be MC(Q) = \(TraditionalForm\`25\/\@\(10  Q\  + \ 1\)\%3\) + (1/3) (Q^2).  Find the total cost of increasing output from 20 to 25 units. Find average variable cost (AVC) function and graph AVC and MC functions on the same set of axes.
Solution:  Since MC(Q) = ((dTC(Q))/dQ), to find total cost find the antiderivative of MC(Q) as, TC(Q) = ∫MC(Q) ddQ = ∫(\(25\/\@\(10  Q\  + \ 1\)\%3\)+ .3  (Q^2))dQ = ((Q^3)/9) + (15/4) (((1 + 10 Q))^((2/3))) + C, where ((Q^3)/9) + (15/4) (((1 + 10 Q))^((2/3))) is the total variable cost (TVC) and C can be interpreted as the fixed cost.   Total cost of increasing output from 20 units to 25 units is evaluated as \(TraditionalForm\`\((Q\^3\/9 + 15\/4\ \((1 + 10\ Q)\)\^\(2\/3\) + C)\)\_20\%25\) = 867.76.  The average variable cost (AVC) is , AVC(Q) = (TVC/Q) = ((Q^2)/9) + (15/(4  Q)) (((1 + 10 Q))^((2/3))).  The graph of MC(Q) and AVC(Q) are shown below.



As any standard MC and AVC graphs, MC passes through the lowest point (minimum) of AVC and continues to rise.
Using Mathematica, set TVC equal to indefinite integral of MC
Clear[Q, MC, AVC, TVC]
MC = 25/((10Q+1)^(1/3))+ (1/3)Q^2;
TVC = Integrate[ 25/((10Q+1)^(1/3))+ (1/3)Q^2, Q]
Increase in total cost for increase in Q from 20 to 30 units can be evaluated as the definite integral of MC in the range Q = 20 to Q = 30.
N[Integrate[25/((10Q+1)^(1/3))+(1/3)Q^2, {Q, 20, 25}]] 
867.762
The average variable cost is AVC = (TVC/Q).  Use the "Plot" command to graph the functions.
AVC = TVC/Q;
Plot[{MC, AVC}, {Q, 1, 20}, AxesLabel->{"Q", "Cost"}]


⁃Graphics⁃

Definite Integral As the Area Under the Curve:  The numerical value of a definite integral may be interpreted as the area under, or enclosed by, a curve.  Such an area is considered to be the limit of the sum of the subareas which result from a certain partitioning of the area in question.  For example, the graph of the continuous function f(x) = (x^2), in the closed interval [0, 4] is represented below, 



the area in the interval [0, 4] can be divided into n equal parts, with each subinterval △x = ((4 - 0)/n).  Suppose now that we designate the middle points of the subintervals by (Subscript[m, 1]), . . . . . ., (Subscript[m, n]); then the following sum, (Subscript[S, n]), is an approximation to the area under the curve, over the interval [0, 4], (Subscript[S, n]) = f((Subscript[m, 1]))△(Subscript[x, 1]) + . . . . . . . . + f ((Subscript[m, n]))△(Subscript[x, n]) =  (∑_(i = 1)^n) f((Subscript[m, 1])) Δ(Subscript[x, i]).  The size of n determines the goodness of the measurement of (Subscript[S, n]).  The larger n, the better is the approximation.  Denoting the exact area by S, we can write, (Underscript[(Lim (Subscript[S, n])), n -> ∞]) = S.  If the (Underscript[(Lim (Subscript[S, n])), n -> ∞]) exits, then ({{Lim}, {n -> ∞}})(∑_(i = 1)^n) f ((Subscript[m, i])) Δ(Subscript[x, i]) = (∫_0^4)f(x) dx = S.  S is the definite integral of f(x) from 0 to 4.  

It is interesting to note the similarities between the discrete summation, (∑_(i = a)^b) f (mi) Δxi, and continuous summation, (∫_a^b)f(x) dx. f(x) and dx are continuous counterparts of f(mi) and Δxi, with continuous summation ( ∫  ) replacing the discrete summation ( ∑ ).

Example: Find the area under the curve f(x) = (x^2) - 8x + 7 and the x-axis in the interval x = 0 to x = 1.
Solution:  The numerical value of the definite integral (∫_0^1)f(x) dx is the area under the curve and the x axis.  Therefore, (∫_0^1)f(x) dx = (∫_0^1)((x^2)  - 8  x  +  7) dx = (Subscript[(((1/3) (x^3)  - 4  (x^2)  + 7  x  +  c)), 0]^1)  = (10/3).  
In Mathematica, 
Clear[x]
Integrate[x^2-8x+7, {x, 0, 1}]
(10/3)

One important property of definite integrals is that they can be expressed as a sum of a finite number of integrals.  That is, if  a < b < c < d, then (∫_a^d)f(x) dx = (∫_a^b)f(x) dx + (∫_b^c)f(x) dx + (∫_c^d)f(x) dx.  This property of definite integrals is used in evaluating the area under a curve, when a portion of the curve under consideration lies below the x axis.

If a portion of the curve under consideration lies below the x axis, the total area should be broken into the sum of two or more integrals by appropriate changes of the limits of integration.  The total area for the interval will be the sum of the absolute values of the sub-areas calculated by definite integral for the sub-intervals.

Example:  Application to Production
A manufacturer of sods has estimated that the sales of sods follows a seasonal pattern given by Q = 12000(1 + Sin((2  π(t  -  60))/365)) where Q is measured in square feet per day, t is time in days, with t = 1 representing january 1.  The manufacturer wants to set up a schedule to produce a uniform amount of sods each day.  Graph the Q for a year and find the average daily production of sods.
Solution:  By substituting values for t, this sine function can be graphed as, 


the total sale is the area under the curve in the range t = 1 to t = 365.  Therefore, find the definite integral of the Q in this range then divide it by 365 to find the average daily production.
Annual Sales = (∫_1^365)12000(1+Sin((2  π(t  -  60))/365))dt = \(TraditionalForm\`12000\ \((t - Cos[2\ π\ \((\(-60\)\  + \ t)\)]\/\(2\ π\))\)\_1\%365\) = 4.3780901406828975`*^6
Average Daily Production = 4.3780901406828975`*^(6/365)≈ 12000 feet per day.
Note:  ∫Sin x dx = - Cos x + c and ∫Sin u du = ((-1)/u)'Cos u + c where, u is a function of x and u' = (du/dx).
           ∫Cos x dx = Sin x + c and ∫Cos u du = (1/u)'Sin u + c where, u is a function of x and u' = (du/dx).
Clear[q]
q=12000(1+Sin[2Pi (t-60)/365]);
Plot[q, {t, 1, 365}, AxesLabel->{"Days", "Daily Sales"}]
N[(1/365)Integrate[q, {t, 0, 365}]]
12000.

Example:  Find the area under the curve for f(x) = (x^2) - 8x + 7, for the interval x = 0 to x = 10. 
Solution:  A look at the graph of the function, below, shows that a part of the graph in the interval of the interest falls below the x-axis.  

Set f(x) = 0 to find the points where f(x) crosses the x-axis.  These points are x =1 and x =7.  The interval [0, 10] can be broken into three sub-intervals, [0, 1], [1,7], and [7, 10].  The areas in the sub-intervals [0, 1] and [7, 10] lie above the x-axis and the area in the interval [1, 7] lies below the x-axis.  Therefore, (∫_0^10)f(x) dx = (∫_0^1)f(x) dx + (∫_1^7)f(x) dx + (∫_7^10)f(x) dx. 
The three areas are calculated as, (Subscript[(((1/3) (x^3)  - 4  (x^2)  + 7  x  +  c)), 0]^1) ,  (Subscript[(((1/3) (x^3)  - 4  (x^2)  + 7  x  +  
              c)), 1 ]^7), and (Subscript[(((1/3) (x^3)  - 4  (x^2)  + 7  x  +  c)), 7]^10), the numerical values of which are (10/3), -36, and 36, respectively.  However, the area of interest is the sum of the absolute value of these three number, (10/3)+ 36+36 = (226/3) = 75.33.
To solve the problem using Mathematica, first plot the function and find zeroes of the function.  Then find the total area as the sum of the absolute values of the integrals.  
Clear[x, area1, area2, area3]
Plot[x^2 -8x+7, {x, 0, 10}, AxesLabel->{"X", "f(X)"}]
Solve[x^2 -8x+7==0, x]
{{x->1},{x->7}}
area1 = Integrate[x^2 -8x+7, {x, 0, 1}];
area2 = Integrate[x^2 -8x+7, {x, 1,7}];
area3 = Integrate[x^2 -8x+7, {x, 7, 10}];
area = N[Abs[area1]+ Abs[area2]+Abs[area3]]
75.3333

Example:  Revenue of the Firm
Marginal revenue of a firm is estimated to be MR = 100 - .2Q.  A price change in the market results in a quantity change of 8 units from the current Level of 34 units.  Find the change in total revenue of the firm.
Solution:  Total revenue generated from the price change is the area under the marginal revenue in the interval [34, 42], (the sum of all marginal revenues between 34 and 48 units, the graph below).  


Therefore, find (∫_5^6)(1000 -  .2  Q) dQ = (Subscript[((1000  Q -  .1  (Q^2)  +  C)), 5]^6) = 739.20.
Using Mathematica,
Clear[Q]
Integrate[100-.2Q, {Q, 34, 42}]
739.2

Example:  Total Cost 
Marginal cost of  manufacturing Q units of goods is MC(Q) = 300 - Q + .06(Q^2).  Find the increase in total cost if the production level is increased from 20 units to 30 units.
Solution:  Total cost of increasing Q from (Subscript[Q, 1]) to (Subscript[Q, 2]) is the area under the MC(Q) function in the interval [(Subscript[Q, 1]), (Subscript[Q, 2])] (the graph below).  


Therefore,  total cost of producing ten more units is (∫_20^30)(300 - 
              Q +  .06  (Q^2)) dQ = \(TraditionalForm\`\((300  Q - \(1\/2\) Q\^2 +  .02  Q\^3 + C)\)\_20\%30\) = 3130.
Clear[Q]
Integrate[300-Q+.06Q^2,{Q, 20, 30}] 
3130.

Area of the Region Between a Curve and Y-axis:  To find the area of a region between a curve and the y-axis write x as a function of y, integrate the new function with respect to y, and evaluate it for the defined boundaries.  

Example:  Find the area of the region between y = (x^(1/4)), y =1, and the y-axis.
Solution:  The graph below shows the region between y = (x^(1/4)), y = 1, and the y-axis.


To find the region of the interest, write x in terms of y as x = (y^4) and find (∫_0^1) (y^4) dy = (Subscript[((((y^5)/5) +  c)), 0]^1) = (1/5).
Using Mathematica,
Clear[y]
Integrate[y^4,{y,0,1}]
(1/5)

Example:  Application to Market
A consumer's demand for a product, which is selling at the market price of (Subscript[P, e]) = 10, is given as P = (100/(Q  +  4)).  Find the consumer's surplus.
Solution:  Consumer surplus for every unit of a good is the difference between the price that consumer is willing to pay for that unit and the price that consumer actually pays.  Therefore, for a demand function P = f(Q), consumer surplus is the region between the demand curve and P-axis that lies above the market price (Subscript[P, e]).  The graph below shows that region.


At Q = 0, P = 25.  Since we are interested in the region between the vertical axis (P-axis) and the function, write Q in terms of P, as Q = ((100  -  4  P)/P), integrate it with respect to P, and evaluate it in the boundries P = 10 and P = 25.
Consumer surplus = (∫_10^25)(( 100 - 
                4  P)/P) dP = (Subscript[((100  lnP  -  4  P)), 10]^25) = 31.63
Clear[P]
N[Integrate[(100-4P)/P, {P, 10, 25}]]
31.6291

Example: Learning Curve
Learning curve shows the increase in productivity of labor as labor learns by doing the same job over time.  The number of units produced between (Subscript[t, 1]) and (Subscript[t, 2]) periods is the area under the learning curve and bounded by (Subscript[t, 1]) and (Subscript[t, 2]).   Suppose that the learning curve of new employees for producing a certain product is estimated to be (dQ/dt) = 50(1 - (100/(((t  +  10))^2))) where, Q is the number of units and t is the time.  Find the number of units produced by a new employee between day 1 and day 10.  Find the number of units produced by the same employee between day 11 and day 20.  What is the increase in average daily productivity between the two periods? 
Solution:  The following graph shows the learning curve (dQ/dt) = 50(1 - (100/(((t  +  10))^2))) .  
   
    
    
The number of units produced between day 1 and day 10 is Q1 = (∫_1^10)50(1-(100/(((t  +  10))^2)))dt = 245.5 and the number of units produced by the same employee between the days 11 and 20 is Q2 = (∫_11^20)50(1-(100/(((t  +  10))^2)))dt = 378.6.  The increase in production in the second period is 133.1 units, with an average daily productivity increase of 13.31 units.
Using Mathematica,
 Clear[t, Q1, Q2, AP]
 Q1 = N[Integrate[50(1-100/(t+10)^2), {t, 1, 10}]]
245.455 Null
 Q2 = N[Integrate[50(1-100/(t+10)^2), {t, 11, 20}]]
378.571
AP=(Q2-Q1)/10
13.3117

Area of a Region Between Two Functions:  The area (A) of the region bounded by the curves y = f(x), y = g(x), and the lines x = a, x = b, where f(x) and g(x) are continuous and f(x) ≥ g(x) for all x in [a, b], is A = (∫_a^b)[f(x)  -  
              g(x)] dx.  

Example:  Find the area of the region bounded between y = (e^(1.  2  x)), y = 10 + 18x - 2(x^2), and bounded on the sides by x = 1 and x = 2.
Solution:  The following graph shows the graph of the functions and the lines of boundary.  

Here, y = 10 + 18x -2(x^2) lies above y = (e^(1.  2  x)) in [1, 2].  Therefore, we write the area as:
 A = (∫_1^2)[(10 + 18  x - 2  (x^2)) - 
              (e^(1.  2  x))] dx = (Subscript[(10  x+9  (x^2)-((2  (x^3))/3)-((e^(1.  2  x))/(1.  2))), 1]^2)= 25.91
 Using Mathematica,
Clear[x]
Integrate[10+18x-2x^2-E^(1.2x),{x,1,2}]
25.9141

Example:  Income Distribution
Lorenz curve is a continuous, increasing, and concavely upward function that passes through points (0, 0) and (1, 1) and describes income distribution of households in a country.  The x-axis represents the percentage of households and the y-axis represents the percentage of income.  A point such as (.50, .25) on the curve means that the bottom 50% of the households earn 25% of total national income.  Absolute equality of income is represented by a straight line y = x.  The area between the Lorenz curve and the y = x line measures how much the income distribution differs from absolute equality.  The so called coefficient of inequality is the ratio of the area between the lorenz curve and the line y = x to the area under y = x line.
Suppose that the Lorenz curve for a country is given as L(x) = .4(x^(2 ))+ .6x.  What is the percentage of total income received by the bottom 50% of the households?  What is the coefficient of inequality?
Solution:  To find the percentage of income received by the bottom 50% of the population find L(.50) = .40.  That is, the bottom 50% of the population earn 40% of the income.  To find the coefficient of inequality, graph L(x) and y = x.


The coefficient of inequality is the ratio of the area between the curves to the area under y = x and bounded by [0, 0] and [1, 1].   Since y = x lies above the L(x) curve, to find the area between the curves, write the area as the integral of x - L(x) between 0 and 1.
A1 = (∫_0^1)[x  -  ( .4  (x^2) +  .6  
                    x) ] dx = (Subscript[(( .2  (x^2) - (( .4  (x^3))/3) +  c)), 0]^1) = .067
The total area under y = x and bounded by [0, 0] and [1, 1] is 
A2 = (∫_0^1) x dx = (Subscript[((((x^2)/2) + c)), 0]^1) = (1/2).
Therefore, the coefficient of inequality is \(TraditionalForm\` .067\/ .5\) = .133
Using Mathematica,
Clear[L,x]
L=.4x^2+.6x;
y=x;
A1=Integrate[y-L,{x,0,1}]
0.0666667
A2=Integrate[y,{x,0,1}]
(1/2)
coef=A1/A2
0.133333

Symmetry Rule:  The following symmetry rules simplify the calculation of integrals of functions that posses symmetry properties:

1)  If f(x) is even function [f(-x) = f(x)], then (∫_(-a)^a) f(x) dx = 2(∫_0^a) f(x) dx

2)  If f(x) is an odd function [f(-x) = -f(x)], then (∫_(-a)^a) f(x) dx = 0

Example:  Find \(TraditionalForm\`∫\_\(-3\)\%3\ \(x\^4\) dx\)
Solution:  Here, f(x) = (x^4) is an even function, f(-x) = f(x).  Therefore, \(TraditionalForm\`∫\_\(-3\)\%3\ \(x\^4\) dx\) = 2(∫_0^3) (x^4) dx = (486/5).
Using Mathematica,
Clear[x]
Integrate[x^4,{x,-3,3}]
(486/5)

Example:  Find (∫_(-3)^3) 
              (x/((x^(4 )) +   (x^2)  +  
                1)) dx.
Solution:  Here, f(x) = (x/((x^4) +  (x^2)  + 1)) is an odd function with f(x) = f(-x).  Therefore, (∫_(-3)^3) 
              (x/((x^(4 )) +   (x^2)  +  
                1)) dx = 0.
Using Mathematica,
Clear[x]
Integrate[x/(x^4+x^2+1),{x,-3,3}]
0

Problems:
1.  Find the following definite integrals:
a) (∫_(-1)^2)(2  (x^2)  - 5  x +3)dx

b) (∫_0^2)(((2 (( (x^2))^( - 1))))^2) dx

c) (∫_2^∞)3(x^(-2)) dx

d) (∫_1^3)(1/(1  +  x)) dx
2.  For the following demand functions find consumer surplus, given that market price of the product is $5.
	a) p = -3q + 10
	b) p = (e^((2  -  q)))
	c) p = (25/(1  +  q))
	d) What will the consumer surplus be if the product price changes to $4?
3.  The marginal cost function of a single-product firm is given as, MC = (Q^2) - 8Q + 18,
	a) Find the producer surplus, if the market price of the product is $6.
	b) What will the increase in producer surplus be if the market price changes to $7?	
4.   A monopolist has the following demand and total cost functions:
	P = 2Q + 23
	TC = (1/3)(Q^3) - 4(Q^2) + 18Q + 1000	
	a)  Find the profit-maximizing level of output and the price.
	b)  Find the social loss of monopoly. (Hint:  the social loss of monopoly is the difference between the 		     consumer surplus and the producer surplus at the price that monopolist is selling the product.)	
5.    Suppose that the capital formation of a business is continuous over time, with (dk/dt) = I(t) = 3\(t\^ .5\).  If the initial stock of the capital at t = 0 is k(0) = 100,
	a) What is the time path of the capital k?
	b) What is the capital formation during the time interval t1 and t2?	
6.   The profit of a business in thousands of dollars is estimated to be p = 8((Sqrt[t]) + 45).  Find the total profit for the next 10 years and the average profit of the business over the next ten years.
7.   A manufacturer depreciates the value of machines used in production by the following rule (dv/dt)= 200000(t - 6) where v is the value of the machines after t years.  What is the total loss of the value of the machines over the first 3 years?  What is the total loss of the value of the machines over the first five years?