Techniques of Integration

Chapter 2

22 mai 2017 be seen from the second derivative (if it exists). 6. Page 7. DMM summer 2017 ... Calculate the gradient of fA
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15 nov. 2012 determinant derivative of inverse matrix
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b − Ax2. 2. = (b − Ax)T (b − Ax). = bT b − (Ax)T b − bT Ax + xT AT Ax. = bT b − 2bT Ax + xT AT Ax Because mixed second partial derivatives satisfy.
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determinant derivative of inverse matrix
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Techniques of Integration

apparent that the function you wish to integrate is a derivative in some EXAMPLE 8.1.2 Evaluate ∫ sin(ax + b) dx assuming that a and b are constants ...
calculus Techniques of Integration

Week 3 Quiz: Differential Calculus: The Derivative and Rules of

Question 2: Find limx→2f(x): f(x) = 1776. (A) +∞. (B) 1770. (C) −∞. (D) Does not exist! (E) None of the above. Answer: (E) The limit of any constant

Assignment 2 — Solutions

If a1b2 = a2b1 show that this equation reduces to the form y′ = g(ax + by). Solution Substituting a2 = λa1 and b2 = λb1 into the equation yields:.
Assignment Solutions

Introduction to Linear Algebra 5th Edition

To see that action I will write b1
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1 mar. 2016
ORF S Lec gh

Order and Degree and Formation of Partial Differential Equations

When a differential equation contains one or more partial derivatives of an (viii) z = ax e' +. 1. 2. Sol. (1) We are given z = (2x + a) (2 y + b).
partial differential equations unit

146123 8

Techniques of Integration

Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with? x 10dx we realize immediately that the derivative ofx11will supply anx10: (x11)?= 11x10. We don"t want the "11", but constants are easy to alter, becausedifferentiation "ignores" them in certain circumstances, so d dx111x11=11111x10=x10. From our knowledge of derivatives, we can immediately writedown a number of an- tiderivatives. Here is a list of those most often used: x ndx=xn+1 n+ 1+C,ifn?=-1 x -1dx= ln|x|+C e xdx=ex+C sinxdx=-cosx+C 163

164Chapter 8 Techniques of Integration

cosxdx= sinx+C sec

2xdx= tanx+C

secxtanxdx= secx+C ?1

1 +x2dx= arctanx+C

?1 ⎷1-x2dx= arcsinx+C Needless to say, most problems we encounter will not be so simple. Here"s a slightly more complicated example: find?

2xcos(x2)dx.

This is not a "simple" derivative, but a little thought reveals that it must have come from an application of the chain rule. Multiplied on the "outside" is 2x, which is the derivative of the "inside" functionx2. Checking: d dxsin(x2) = cos(x2)ddxx2= 2xcos(x2), so

2xcos(x2)dx= sin(x2) +C.

Even when the chain rule has "produced" a certain derivative, it is not always easy to see. Consider this problem:? x 3?

There are two factors in this expression,x3and?

1-x2, but it is not apparent that the

chain rule is involved. Some clever rearrangement reveals that it is: x 3?

1-x2dx=?

(-2x)? -12? (1-(1-x2))?1-x2dx. This looks messy, but we do now have something that looks likethe result of the chain rule: the function 1-x2has been substituted into-(1/2)(1-x)⎷ x, and the derivative

8.1 Substitution165

of 1-x2,-2x, multiplied on the outside. If we can find a functionF(x) whose derivative is-(1/2)(1-x)⎷ xwe"ll be done, since then d dxF(1-x2) =-2xF?(1-x2) = (-2x)? -12? (1-(1-x2))?1-x2 =x3? 1-x2

-1 8

Techniques of Integration

Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with? x 10dx we realize immediately that the derivative ofx11will supply anx10: (x11)?= 11x10. We don"t want the "11", but constants are easy to alter, becausedifferentiation "ignores" them in certain circumstances, so d dx111x11=11111x10=x10. From our knowledge of derivatives, we can immediately writedown a number of an- tiderivatives. Here is a list of those most often used: x ndx=xn+1 n+ 1+C,ifn?=-1 x -1dx= ln|x|+C e xdx=ex+C sinxdx=-cosx+C 163

164Chapter 8 Techniques of Integration

cosxdx= sinx+C sec

2xdx= tanx+C

secxtanxdx= secx+C ?1

1 +x2dx= arctanx+C

?1 ⎷1-x2dx= arcsinx+C Needless to say, most problems we encounter will not be so simple. Here"s a slightly more complicated example: find?

2xcos(x2)dx.

This is not a "simple" derivative, but a little thought reveals that it must have come from an application of the chain rule. Multiplied on the "outside" is 2x, which is the derivative of the "inside" functionx2. Checking: d dxsin(x2) = cos(x2)ddxx2= 2xcos(x2), so

2xcos(x2)dx= sin(x2) +C.

Even when the chain rule has "produced" a certain derivative, it is not always easy to see. Consider this problem:? x 3?

There are two factors in this expression,x3and?

1-x2, but it is not apparent that the

chain rule is involved. Some clever rearrangement reveals that it is: x 3?

1-x2dx=?

(-2x)? -12? (1-(1-x2))?1-x2dx. This looks messy, but we do now have something that looks likethe result of the chain rule: the function 1-x2has been substituted into-(1/2)(1-x)⎷ x, and the derivative

8.1 Substitution165

of 1-x2,-2x, multiplied on the outside. If we can find a functionF(x) whose derivative is-(1/2)(1-x)⎷ xwe"ll be done, since then d dxF(1-x2) =-2xF?(1-x2) = (-2x)? -12? (1-(1-x2))?1-x2 =x3? 1-x2

-1