 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
print barcode image c# The Integral in Software
CHAPTER 4 The Integral QR Creation In None Using Barcode generator for Software Control to generate, create QR Code image in Software applications. Recognize QRCode In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. For our rst example, we calculate the area under a parabola EXAMPLE 44
Making Quick Response Code In Visual C#.NET Using Barcode creator for VS .NET Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications. Creating Denso QR Bar Code In .NET Framework Using Barcode printer for ASP.NET Control to generate, create QR Code image in ASP.NET applications. Calculate the area under the curve y = x 2 , above the xaxis, and between x = 0 and x = 2
Paint QR Code 2d Barcode In Visual Studio .NET Using Barcode creation for .NET framework Control to generate, create QRCode image in VS .NET applications. QR Code JIS X 0510 Printer In VB.NET Using Barcode generation for .NET Control to generate, create QR Code image in VS .NET applications. SOLUTION Refer to Fig 48 as we reason along Let f (x) = x 2
Create GTIN  13 In None Using Barcode encoder for Software Control to generate, create GS1  13 image in Software applications. Encode Barcode In None Using Barcode creator for Software Control to generate, create barcode image in Software applications. Fig 48
Bar Code Maker In None Using Barcode generator for Software Control to generate, create barcode image in Software applications. Generate Data Matrix ECC200 In None Using Barcode creation for Software Control to generate, create Data Matrix image in Software applications. Consider the partition P of the interval [0, 2] consisting of k + 1 points x0 , x1 , , xk The corresponding Riemann sum is Draw UPCA Supplement 5 In None Using Barcode creation for Software Control to generate, create UPCA Supplement 5 image in Software applications. Generate Code39 In None Using Barcode generator for Software Control to generate, create Code 39 Full ASCII image in Software applications. R(f, P ) = Generate Code 2/5 In None Using Barcode generator for Software Control to generate, create 2 of 5 Standard image in Software applications. Bar Code Encoder In Visual C#.NET Using Barcode generation for .NET Control to generate, create bar code image in .NET applications. j =1 Code 39 Full ASCII Recognizer In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. Read Barcode In VS .NET Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications. f (xj ) DataMatrix Creation In .NET Using Barcode creation for Reporting Service Control to generate, create DataMatrix image in Reporting Service applications. EAN128 Reader In C#.NET Using Barcode decoder for .NET Control to read, scan read, scan image in .NET applications. Of course x= and 2 xj = j k In addition, f (xj ) = j 2 k
GTIN  128 Scanner In Visual Basic .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. GS1 128 Drawer In ObjectiveC Using Barcode creation for iPad Control to generate, create GS1128 image in iPad applications. 2 2 0 = k k
4j 2 k2
As a result, the Riemann sum for the partition P is
The Integral
R(f, P ) = j =1 k
4j 2 2 k2 k 8 8j 2 = 3 3 k k
j =1 j 2 j =1 Now formula II above enables us to calculate the last sum explicitly The result is that
R(f, P ) = 8 2k 3 + 3k 2 + k 6 k3 4 8 4 = + + 2 3 k 3k 4 8 8 4 + + 2 = 3 k 3k 3
x 2 dx = lim R(f, P ) = lim
We conclude that the desired area is 8/3 The most important idea in all of calculus is that it is possible to calculate an integral without calculating Riemann sums and passing to the limit This is the Fundamental Theorem of Calculus, due to Leibniz and Newton We now state the theorem, illustrate it with examples, and then brie y discuss why it is true Theorem 41 (Fundamental Theorem of Calculus) Let f be a continuous function on the interval [a, b] If F is any antiderivative of f then You Try It: Use the method presented in the last example to calculate the area under the graph of y = 2x and above the xaxis, between x = 1 and x = 2 You should obtain the answer 3, which of course can also be determined by elementary considerations without taking limits EXAMPLE 45
Calculate
AM FL Y
In sum, f (x) dx = F (b) F (a) x 2 dx
CHAPTER 4 The Integral
SOLUTION We use the Fundamental Theorem In this example, f (x) = x 2 We need to nd an antiderivative F From our experience in Section 41, we can determine that F (x) = x 3 /3 will do Then, by the Fundamental Theorem of Calculus, x 2 dx = F (2) F (0) = 23 03 8 = 3 3 3 Notice that this is the same answer that we obtained using Riemann sums in Example 44 EXAMPLE 46
Calculate
sin x dx
SOLUTION In this example, f (x) = sin x An antiderivative for f is F (x) = cos x Then
sin x dx = F ( ) F (0) = ( cos ) ( cos 0) = 1 + 1 = 2
EXAMPLE 47
Calculate
ex cos 2x + x 3 4x dx
SOLUTION In this example, f (x) = ex cos 2x + x 3 4x An antiderivative for f is F (x) = ex (1/2) sin 2x + x 4 /4 2x 2 Therefore ex cos 2x + x 3 4x dx = F (2) F (1) = e2 24 1 sin(2 2) + 2 22 2 4
14 1 sin(2 1) + 2 12 2 4 9 1 = (e2 e) [sin 4 sin 2] 4 2 e1
You Try It: Calculate the integral
1 3 The Integral
x 3 cos x + x dx
Math Note: Observe in this last example, in fact in all of our examples, you can use any antiderivative of the integrand when you apply the Fundamental Theorem of Calculus In the last example, we could have taken F (x) = ex (1/2) sin 2x + x 4 /4 2x 2 + 5 and the same answer would have resulted We invite you to provide the details of this assertion Justi cation for the Fundamental Theorem Let f be a continuous function on the interval [a, b] De ne the area function F by F (x) = area under f , above the xaxis, and between 0 and x Fig 49
Let us use a pictorial method to calculate the derivative of F Refer to Fig 49 as you read on Now F (x + h) F (x) [area between x and x + h, below f ] = h h f (x) h h = f (x) As h 0, the approximation in the last display becomes nearer and nearer to equality So we nd that F (x + h) F (x) = f (x) h
But this just says that F (x) = f (x) What is the practical signi cance of this calculation Suppose that we wish to calculate the area under the curve f , above the xaxis, and between x = a and x = b Obviously this area is F (b) F (a) See Fig 410 But we also know that CHAPTER 4 The Integral
that area is
b a f (x) dx
We conclude therefore that
f (x) dx = F (b) F (a) y = f (x) F (b) _ F (a) Fig 410
Finally, if G is any other antiderivative for f then G(x) = F (x) + C Hence G(b) G(a) = [F (b) + C] [F (a) + C] = F (b) F (a) = f (x) dx
That is the content of the Fundamental Theorem of Calculus You Try It: Calculate the area below the curve y = x 2 + 2x + 4 and above the xaxis

