Section 5.5 Solving indefinite Integral through substitution

Qucik guide to u substitution

1. Decide what u is 2. Find a substitute for dx 2a. Take the the derivative of u . . du = 3x^2 * dx Note dx is not defined here b.c we are doing dy/du 2b solve for dx: dx = du/3x^2 3. Substitute into the integral and cancel any terms- Does not always cancel 4. Then take the integral of (u)du which is always jsut the antiderivative of u or the height. . 5. multiply by any constants outside the integral 6. Finally sub u in terms back into the equation for u . . . Do not perform any more operations past this point.

Steps to u substitution

1. Pick 'u' so that the integral is easier 1tip: this is usually the 'inside' of someting 2. The derivative of u must be in the integrand 3. transform integral of f(x)dx to integral of (u)(du) 4. solve the integral 5 translate back to x or whatever variable you start with

Selecting the appropriate u

1: the derivative of u must be in the integrand somewhere. If you cant match it in the original integrand then it cant be u. Check your work by selecting u, taking the derivative and see if it matches, or else! 2. Usually it is inside of some operation 3. pick u so that the integral is easier

Recoginzing Aplitude in this section

Amplitude as i know it applied to trigonometric functions y = 2/5sin(2pit/5) aplitude = 2/5

Deciding on u for substitution

Another + for u is that the derivative of u be in the integral somewhere. ex u = tan du = sec^2x . . . Sec^2x is in the integral exactly and cancels out. The only exception so far would be if there had of been a constant we could have pulled out like 5 tanx, that would have also worked.

Step 3 of u substitution

How to write the integral in terms of u 1. Need to find u = x^2+1 2. du = derivative of u = (2x)(dx) Note on this step we do not simplify dx to one. . This is just normal derivative except for that.

integral all f(one variable) or d(one variable) Observation for this section

I have noticed that through all the substitution we put whats in the integral all in terms of u or one variable and solve the integral . . all in terms of one variable. . .

Solving for x if solved for dx in terms of u does not cancel out Rule

If du does not cancel out all the x terms in the integral then solve for x in the original substitute u equation. Sub that into where the uncancelled x term is so that the integral is all in terms of u

Finding Area from Derivative Review

If given Derivative Cannot just plug in time and know area of function. thats only the rate. Need to find the integral or ]x,x antiderivative. which is the F(b) - F(a)

Time function integral endpoints

If it is time, the start point will likely be 0 . . . The endpoint is when they ask the area up to that time.

Review antiderivative being ln(u) rule

If the antidirivative is ln(u) then it is ln(|u|) Which you may have to create a negative -(f(x)) unless the entire domain is always positive, then no action is requried for example u = e^x + 19. . 1/u does not need aboslute values b.c that function is always greater than 0

Constant rule for other variables

If there is a another variable in the integral besides the main main: if its dx on the end then x is main variable. Other variables are then considered constants and obey all the rules of constants excpet they stay a, the whole time

Common term F(b)-F(a) Integral Rule

If there is a common term within this operation, you can factor outside with the other constant

Exponent distribute review

If there is an exponent attached to (2a^2)^xx then the exponent distributes to the 2 and the a.

New Finding Rule for U d/dx U constant factor our Counts rule for u proper substitute

If very difficult look for only the inside of the operation. Ex choose whats inside of the square root. Also if you find the derivative of u and you can factor out a constant and that factored form of dy/du except for the constant so 9(what matches) then that can still be u. . dx = du/9(what matches) You do not have to take the 9 out of further operations. Just know that it works

Canceling the solved for dx from du

If you see that dx = du/5x^4 . . If you see x^4 cancels you can cancel that out strait away. Then just sub in 1/5du for dx. . Im not sure but maybe it has to be all in terms of u before takin the integral so maybe that x term always cancels out.

Whats the difference between definate integration and indefinate integration by substituion

Indefinate you f(x) = u back into u and solve leaving an x variable and adding + C as the arbitrary constant. Definate, you sub the integral endpoints into u and find the new endpoints. At the end, you sub in the new endpoints in for u and get an exact answer without + C arbitrary constant

In the trig example time was not the endpoint

It takes 5 seconds to occur, so you would think ]0,5. . The wording - this does not describe the endpoints, rather the cycle of one breath. If there is no defined endpoint, and the x axis is time the integral is ]0,t so thats F(t) - F(0) . . However in this format you do not add + C to the end. Thats if there are no defined points of start and end

Simplify integral With u terms combining. and constant pull out again

New: (u-7) is a term, not a constant so it cannot be pulled outside of the integral. However it can be simplified within the integral, then we can take the integral Keeping it in term from, so not separately: (u^2 - u^3) = [1/3u^3 - 1/4u^4] Then distribute any constants that may be outside the integral... then sub back in the f(x) for u

Part two of u substituon

Not only can you sub in u for the 'inside' but you can can also sub in "du" for "dx"(base, change in width, rectangle) 1. If du = (2x)(dx) 2. Solve for dx = du/2x 3. sub in du/2x for dx in the integral After this you can sub in u and du/2x into the integral Note: the denominator of du will likely cancel out before taking the integral

Antiderivative ln rule Catch

Remember the antiderivative of 1/u = lnu however 1/u^2 = u^-2 = u^-1/-1 = -1/u for the antiderivative

Alternative Method Definiate integral U substituion to Finding new endpoints

Same applies that all must be u and take the antiderivative. However instead of finding new endpoints and subbing only 1 number in for u we can revert to old method used in indefintate when we did not sub anyhting into x just left in terms of x and added + C. So after antiderivative you can sub in the the f(x) for u . . keeping operation found by taking antiderivative. Now heres the the dif for definate. Sub the ends points in to where x is. . Then you will come up with an exact number with the antiderivative operations around it. To Find the integral just do the normal F(b) - F(a)

The constant Pull out rule Relevancy in this section

So Anytime you have a constant being multiplied in the integral use the constant pull out rule. This is especially helpful in finding the antiderivative ex: dx/u i dont know the antiderivative but dx*1/u = dx*ln(u)

Implications of taking the integral in Applications

So If you want to know how much water escaped a tank over an amount of time/ how much air is inhaled in lungs over time You have to find the rate that describes it f(u), and the derivative of the rate du If you have no idea what start and endpoint is ] and add + C If you dont know endpoint buts its in time ]0,t and F(b) - F(a) If you know both st/end points it ]-1,4 and F(b) - F(a) It will tell you the area of the function = the total units of the function that occured over the course of the function may be, air lungs/water in a tank If the Antiderivative is hard to compute, the only method we have is the u substitution method so far to make a complicated antiderivative easier

Finding the integral of composite functions by U substitution

So far we have been doing algebra with the integrand to get the integrand into least factored form and broken up into individual terms. Then taking the integral of each individual term. Some composite functions are complicated and we need a more efficient method: That's why we now learn about substitution

Substitution of dx into the integral critiquie

So to solve for dx take the derivative of u . . which with be x dx. . then du/x = dx. . Then sub in to integral before taking antiderivative. Cancel any terms if possible Note Then pull out whatever is left outside of the integral since it is not a function of x by *^/ etc. Note if only du is pulled outside the integral that simplifies to 1 . . (1)integral

Why the FDTCII Works

The Antiderivative of any function in F(b) - F(a) Gives the area of the original integrand from start to end point

Derivative Review

The derivative is always the number*dx. . However is previous sections we have simplified that to 1 . . But in some sections and again in this one we are leaving dx or x' as a part of the equation not simplfied to 1. . I think maybe b/c we are finding dy/du??

Derivative Review of constant pull out rule

The derivative of a constant is 0, however if a constant is attached to a f(x) that is a coefficient constant so it pulls out ex: 2-e^x = -1 d/dx e^x = -e^x dx

Whats the Point of u substitution with integrals

The goal is to make finding a difficult integral, easier. 2. In order to do an integration it must fit within the integration, If not you cannot do it.

Finding integral Review for substituion

There is an added step of subing in the endpoints in after antiderivatie, and after reintroducing f(x) in for u. . If there are two terms remember sub the endpoint in to both terms, thats F(b) . . then sub x into both terms. . thats F(a). . The F(b) - F(a) . . Thats two terms each, 4 terms total.

Writing integral as a sum of two integral rule

This case In the integrand we have 2 terms. 1 of which separated by a +sign. Distribute the 1 solo term to the + terms. . Via the integral distributive property we can write the integral of one term + the integral of the second term. Each integral is defined by F(b) - F(a). . Then add these two terms together

Solving for dx

This is kind of sketchy but if its not a constant usualy the derivative of u solve for dx if it has an x term cancels out with the original u integral

Section 5.5 Test tips

Tip: If its in this section; but gets wordy; It is most likely an application of definate integral by u substitution

Changing Integral Explanation

To find the endpoints of u finds the endpoint for u, not x but we do that separately instead of subing in the whole u equation into u and then pluging in the f(x) for u then subing in the endpoints which are the same things essentially. It just saves space while doing the integral to solve for the endpoints before hand, that way when subbing into the antiderivative our work is neater and we only have to sub in one number.

Exponent Tricks

a^3/2 is sqrt (a ^ 3 . . . so that is . . . sqrt(a * sqrt(a * sqrt(a Another example a ^ 2/3 is the 3rt(a ^ 2 . . which is 3rt(a * 3rt(a . .. Remember top to top is the exponent then just bottom goes out in front as the root.

Odd and Even Function

an Even Function is one whose y reflects across the y axis - gives the same y for both positive and negative domain Odd function is one whose y reflects across the y axis and then across the x axis making it 180 about the origin

What is a constant: unexpected examples

anything not associated with the variable ex: 2/pie

direct integral using substitution example

cos(pit/2) . . u = piet/2

U sub Good finds

e^(1/x^3)/x^4 . . the thing in most inside is actually the exponent (1/x^3) Let that be U. another e^(-.02)(t) . . let -.02t be U

Example of du in integral

if du = 8 and the number 8 is anywhere in the integral, no matter the operations surround it, that can be u

sub endpoints into u = to find new endpoints for evaluation rule

if the endpoints are 0,1 then sub 0 into u the f(u) and solve for u. . thats start point. Then sub 1 into f(u) and solve for u. . thats endpoint Rewrite the integral with the new endpoints. . Then take the atiderivative of the integral with u subbed in. . evalute that at the endpoints. . this means instead of subing u back in you do F(b) - F(a) and sub previously solve for endpoints in for u. Again do not sub u=f(x) into U for definate integral integration by substituion

absolute value rule for fractions and substitution

if u is in the denominator of a fraction taking the antiderivative will most like be ln(u) however in that case u is absolute value. . ln(|u|)

Selected u locked in solved for x for third substitution into integral Rule

if you have selected u and it is a function of x. . . and x, just x is in the integral. . You can solve u = x - 8 for x . . x = u + 8 . . . and sub (u+8) into the integral. .

ln Correct answer format

ln(e) = 1 however ln(e+19) . . They just leave in in this form instead of distributing the ln

Odd integral, Integrated over symmetric bound/endpoints . . Integral = 0 Rule

odd integral = x sqrt(4-x^2 Why? b.c it gives the same answer y for each positive and negative domain same x. Then that would be even. However the x coefficient makes it the same y with a different sign thats y. By symmetrics bounds/endpoints = integral evaled at ]-2,2

double angle identity rule review

sin(2x) = 2sinxcosx cos(2x) = cos^(x)-sin^2(x) tan(2x) = 2tan(x)/1-tan^(x)

Equation of a semi-circle

sqrt(16-x^2 . . If the integral is ]-4,4 then you know the radius is 4. Since integral is area. Instead of taking the antiderivative of the integral of a semi-circle we can just apply the semicircle formula = 1/2(pi * r^2) . . again the radius ..

New Natural log revised rule

the antiderivative of 1/x is not ln(x) but is actually ln(|x|) how ever is the f(x) inside is always positive, Do not write the absolute value bars. We do not need a negative function b.c nowhere in the functions domain is f(x) given a negative y

selecting u for substitution

u can aslo be a term, for example 33 + cos^2x Remember the chaing rule for find the derivative of u is 2(cosx) * d/dx cosx = -2cosxsinx. . . If that term, coefficient or no coefficient is in the integral, then thats your u

Recognizing The period of tangent functions with different variable

y = 2/5sin(2piet/5) We know period for sin cos is 2pie/B Notice here that it can be rewritten (2pi/5 * t) . . This make 2pi/5 B . . Period equalts (5/2pi)*2pi = 5

Reimans sum Review

y = ]0,t f(t)dt or f(u)du or f(x)dx . . big thing all the inputs are x axis Try to not Forget the derivative on the end However f(u) = the rate or velocity, b.c it is the height at that point du: is the derivative of the rate which has no t variable, it is just distance The antiderivative is F(u) the true function of the graph You sub all designated n or column heights by u into the rate f(u) then multiply by change in u , d(u) from start point to endpoint with designated (n) columns. (b-a)/n