x2 1 ln x 2 2 in Visual Studio .NET

Drawing QR Code in Visual Studio .NET x2 1 ln x 2 2

x2 1 ln x 2 2
Decode QR Code 2d Barcode In Visual Studio .NET
Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in VS .NET applications.
Printing QR-Code In Visual Studio .NET
Using Barcode drawer for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications.
x2 1 x2 ln x +C 2 2 2 x2 (2 ln x 1) + C 4
Scan Denso QR Bar Code In Visual Studio .NET
Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications.
Generate Bar Code In .NET Framework
Using Barcode creation for Visual Studio .NET Control to generate, create barcode image in .NET framework applications.
CHAP. 38]
Bar Code Reader In Visual Studio .NET
Using Barcode decoder for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications.
Painting Quick Response Code In C#.NET
Using Barcode creation for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in .NET framework applications.
INTEGRATION BY PARTS
Create QR Code In .NET Framework
Using Barcode maker for ASP.NET Control to generate, create QR Code JIS X 0510 image in ASP.NET applications.
Denso QR Bar Code Encoder In VB.NET
Using Barcode creator for .NET Control to generate, create QR Code 2d barcode image in Visual Studio .NET applications.
Thus, u dv = uv ln x dx = x ln x = x ln x v du 1 x dx x dx
Painting EAN-13 In VS .NET
Using Barcode creator for VS .NET Control to generate, create European Article Number 13 image in .NET applications.
EAN128 Creation In .NET Framework
Using Barcode maker for VS .NET Control to generate, create UCC.EAN - 128 image in .NET framework applications.
= x ln x x + C = x(ln x 1) + C (d) Sometimes two integrations by parts are necessary. Consider u = ex du = ex dx Thus, Let us try to nd ex cos x dx. Let u = ex and dv = cos x dx: dv = cos x dx v = sin x ex sin x dx (1)
Barcode Printer In Visual Studio .NET
Using Barcode creation for .NET framework Control to generate, create barcode image in .NET applications.
Make USPS POSTNET Barcode In .NET Framework
Using Barcode generator for .NET framework Control to generate, create Delivery Point Barcode (DPBC) image in .NET applications.
ex cos x dx = ex sin x
Making Barcode In None
Using Barcode generation for Office Word Control to generate, create bar code image in Microsoft Word applications.
Encode EAN13 In Java
Using Barcode creator for Java Control to generate, create EAN 13 image in Java applications.
ex sin x dx by another integration by parts. Let u = ex and dv = sin x dx: u = ex du = ex dx dv = sin x dx v = cos x ( ex cos x) dx ex cos x dx
ECC200 Recognizer In VB.NET
Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET applications.
Code 39 Extended Decoder In Java
Using Barcode scanner for Java Control to read, scan read, scan image in Java applications.
Thus,
ECC200 Recognizer In Java
Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications.
Universal Product Code Version A Recognizer In VB.NET
Using Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET applications.
ex sin x dx = ex cos x = ex cos x +
Scan EAN / UCC - 13 In VB.NET
Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET framework applications.
EAN-13 Supplement 5 Generator In Java
Using Barcode maker for Android Control to generate, create EAN-13 image in Android applications.
Substitute this expression for
ex sin x dx in (1) and solve the resulting equation for the desired antiderivative, ex cos x dx = ex sin x ex cos x + ex cos x dx = ex sin x + ex cos x 2 ex cos x dx
ex cos x dx
ex cos x dx = ex sin x + ex cos x = ex (sin x + cos x) ex cos x dx = ex (sin x + cos x) +C 2
Solved Problems
38.1 Find
Let u=x du = dx dv = e x dx v= e x dx = e x
xe x dx.
INTEGRATION BY PARTS
[CHAP. 38
Integration by parts gives xe x dx = xe x ( e x ) dx = xe x + e x dx
= xe x e x + C = e x (x + 1) + C Another method would consist in making the change of variable x = t and using example (a) of this chapter.
38.2 (a) Establish the reduction formula x n ex dx = x n ex n for (b) Compute
(a) Let u = xn du = nx n 1 dx and integrate by parts, x n ex dx = x n ex (b) For n = 1, (2) gives xex dx = xex ex dx = xex ex = (x 1)ex ex (nx n 1 ) dx = x n ex n x n 1 ex dx dv = ex dx v = ex
x n 1 e dx
x n ex dx x 2 ex dx.
(n = 1, 2, 3, . . .).
as in example (a). We omit the arbitrary constant C until the end of the calculation. Now let n = 2 in (2), x 2 ex dx = x 2 ex 2 xex dx = x 2 ex 2((x 1)ex )
= (x 2 2(x 1))ex = (x 2 2x + 2)ex + C
38.3 Find
tan 1 x dx.
u = tan 1 x 1 dx du = 1 + x2 Hence, tan 1 x dx = x tan 1 x = x tan 1 x = x tan 1 x = x tan 1 x 1 2 x dx 1 + x2 2x dx 1 + x2
dv = dx v=x
1 ln 1 + x 2 + C 2 1 ln (1 + x 2 ) + C 2
[by Quick Formula II, Problem 34.5] [since 1 + x 2 > 0]
CHAP. 38]
INTEGRATION BY PARTS
38.4 Find
cos2 x dx.
u = cos x du = sin x dx Then,
dv = cos x dx v = sin x (sin x)( sin x) dx sin2 x dx (1 cos2 x) dx 1 dx cos2 x dx
cos2 x dx = cos x sin x = cos x sin x + = cos x sin x + = cos x sin x +
Solving this equation for
cos2 x dx, 2 cos2 x dx = cos x sin x + cos2 x dx = 1 dx = cos x sin x + x
1 (cos x sin x + x) + C 2
This result is more easily obtained by use of Theorem 26.8, cos2 x dx = 1 2 1 = 2 sin 2x +x +C 2 2 sin x cos x 1 + x + C = (sin x cos x + x) + C 2 2 (cos 2x + 1) dx = 1 2
38.5 Find
x tan 1 x dx.
u = tan 1 x 1 du = dx 1 + x2 Then, But and so x2 dx = 1 + x2 Hence 1 dx x tan 1 x dx =
dv = x dx v= x2 2 x2 dx 1 + x2
x2 1 tan 1 x 2 2
x2 (1 + x 2 ) 1 1 + x2 1 1 = = =1 1 + x2 1 + x2 1 + x2 1 + x2 1 + x2 1 dx = x tan 1 x + C 1 + x2
x2 1 tan 1 x (x tan 1 x) + C1 x tan 1 x dx = 2 2 1 2 1 = (x tan x (x tan 1 x)) + C1 2 1 2 1 = (x tan x x + tan 1 x) + C1 2 1 = ((tan 1 x)(x 2 + 1) x) + C1 2
INTEGRATION BY PARTS
[CHAP. 38
Supplementary Problems
38.6 Compute: (a) (e) (i) (m) (q) (u) x 2 e x dx x cos x dx eax cos bx dx xe3x dx x sin 2x dx x 2 tan 1 x dx (b) (f ) (j) (n) (r) (v) ex sin x dx x 2 sin x dx sin2 x dx x sec2 x dx x sin x 2 dx ln (x 2 + 1) dx (c) (g) (k) (o) (s) (w) x 3 ex dx cos (ln x) dx cos3 x dx x cos2 x dx ln x dx x2 x3 1 + x2 (d) (h) (l) (p) (t) dx (x) sin 1 x dx x cos (5x 1) dx cos4 x dx (ln x)2 dx x 2 e3x dx x 2 ln x dx
[Hint: Integration by parts is not a good method for part (r).] 38.7 Let R be the region bounded by the curve y = ln x, the x-axis, and the line x = e. (a) Find the area of R. (b) Find the volume generated by revolving R about (i) the x-axis; (ii) the y-axis. 38.8 Let R be the region bounded by the curve y = x 1 ln x, the x-axis, and the line x = e. Find: (a) the area of R; (b) the volume of the solid generated by revolving R about the y-axis; (c) the volume of the solid generated by revolving R about the x-axis. [Hint: In part (c), the volume integral, let u = (ln x)2 , v = 1/x, and use Problem 38.6(s).] 38.9 Derive from Problem 38.8(c) the bounds: 2.5 e 2.823. [Hint: By Problem 30.3(c), 0
Copyright © OnBarcode.com . All rights reserved.