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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of AP Calculus AB
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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Thu Nov 30, 2017 4:43 AM

Post by Magic Fu on November 23, 2017

Hello Professor Hovasapian,
I had a B on the first semester of AP Calc BC, is it good? Should I drop out to AB?

1 answer

Last reply by: Professor Hovasapian
Thu Aug 25, 2016 5:38 PM

Post by Isaac Martinez on August 25, 2016

Hello Professor Hovasapian,

I was wondering how you got 13.432 as an answer for  the second derivative of your example II, Function 3.

Thank you,



Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Antiderivatives 0:23
    • Definition of an Antiderivative
    • Antiderivative Theorem
  • Function & Antiderivative 12:10
    • x^n
    • 1/x
    • e^x
    • cos x
    • sin x
    • sec² x
    • secxtanx
    • 1/√(1-x²)
    • 1/(1+x²)
    • -1/√(1-x²)
  • Example I: Find the Most General Antiderivative for the Following Functions 15:07
    • Function 1: f(x) = x³ -6x² + 11x - 9
    • Function 2: f(x) = 14√(x) - 27 4√x
    • Function 3: (fx) = cos x - 14 sinx
    • Function 4: f(x) = (x⁵+2√x )/( x^(4/3) )
    • Function 5: f(x) = (3e^x) - 2/(1+x²)
  • Example II: Given the Following, Find the Original Function f(x) 26:37
    • Function 1: f'(x) = 5x³ - 14x + 24, f(2) = 40
    • Function 2: f'(x) 3 sinx + sec²x, f(π/6) = 5
    • Function 3: f''(x) = 8x - cos x, f(1.5) = 12.7, f'(1.5) = 4.2
    • Function 4: f''(x) = 5/(√x), f(2) 15, f'(2) = 7
  • Example III: Falling Object 41:58
    • Problem 1: Find an Equation for the Height of the Ball after t Seconds
    • Problem 2: How Long Will It Take for the Ball to Strike the Ground?
    • Problem 3: What is the Velocity of the Ball as it Hits the Ground?
    • Problem 4: Initial Velocity of 6 m/s, How Long Does It Take to Reach the Ground?

Transcription: Antiderivatives

Hello, and welcome back to www.educator.com, welcome back to AP Calculus.0000

Today, we are going to start our discussion of anti-derivatives.0005

In a couple of lessons, we are actually going to change the name of that and start calling them integrals.0009

This is the second half of calculus.0013

The first part is differential calculus, now it is integral calculus.0015

As you will see in a minute, it is actually the inverse process.0018

Let us jump right on in.0022

Up to now, we started with functions and we took derivatives of them.0026

Let us go ahead and write this.0032

Up to now, we started with some function f(x) and we found its derivative.0035

We have found its derivative, in other words, we differentiate it.0053

For example, if we had a sin x, we take the derivative of it and we ended up with cos x.0062

If we had, let us say, an x³, we take the derivative of it and we end up with something like 3x².0072

What if we begin with a function then ask is this the derivative of some function?0081

In other words, instead of starting with the function and going to the derivative,0089

let us say that this is a function and this the derivative of some previous function.0093

That is the question we are going to look at here.0099

What if we begin with a function f(x), then ask is f the derivative of some f(x).0100

Can I find this F, can I find this other f?0139

In other words, we are going to be giving you the cos x for the 3x².0147

And we are going to say, how do you recover the sin x?0151

How do you recover the x³, if that original function actually exists?0153

The answer is yes, fortunately.0158

Let us go ahead and start with a definition.0167

You know what, I think I will go ahead and put my definition.0170

The different color, I will go ahead and put in red.0175

The definition F(x) is called an anti-derivative, exactly what it sounds like,0180

it is just the reverse of the derivative, an anti-derivative of f(x).0194

If this f(x) happens to be the derivative of the F(x), we will just mark that with an f’, on a specified interval.0202

Let us go back to blue here, not necessary but what the heck.0225

All we are doing is going backward, that is it.0232

That is all this is.0243

Now the most difficult part is going to be remembering, am I going forward to differentiating0245

or am I going backward and taking the anti-derivative?0250

It is a different set of rules, it is a different set of formulas that you use to find that.0253

That is going to be probably the most difficult thing that you have to do is remember which direction you are going in.0258

If we begin with some f(x), that is the function that is given, we can go ahead and take the derivative which gives us f’.0265

Or we can go this way and take the anti-derivative which gives us our F(x).0276

Some function is going to be the function that is given to you.0284

There are two ways that you can go, depending on what problem that you are trying to solve.0285

That is what makes calculus incredibly beautiful, you start here.0291

As you will see in a few lessons, there is a relationship between derivative and anti-derivative.0296

They call it the fundamental theorem of calculus, it actually connects the two,0300

which as we said are inverse processes.0304

Let us note the following.0308

If we had some f(x) which is equal to sin x + 6 and we had some g(x),0320

let us say this is sin x + 4, these are not the same function.0329

If you put an x value in, you are going to get two different y values.0335

These are not the same function, very important to know that.0338

These are not the same function but you can probably see where this is going.0343

But f’(x), f’(sin x) + 6 is equal to cos(x).0357

Not f, I’m talking about g and g‘(x).0367

When I take the derivative of the sin x + 4, it is also going to equal cos(x).0370

These are the same derivative.0378

You get two different functions that end up going to the same derivative.0382

That could be a bit of an issue, but it is not, fortunately.0387

If we begin with h(x) = cos x, if we begin with the derivative and ask for its anti-derivative, which one do I choose?0393

Do is say that sin x + 6 is the anti-derivative?0426

Do I say sin x + 4 is the anti-derivative?0428

Is it sin x + 6?0436

Is it sin x + 4?0444

Or how about just sin x + c, where c can be any constant?0447

Because we know that the derivative of a constant = 0.0465

I does not matter what that number is, it can be sin x + π, because the derivative of π is 0.0470

Here is our theorem, let us go ahead and mark this in red.0477

Our theorem says, if f(x) is an anti-derivative of f(x) on an interval that we will just call I,0483

then f(x) + c, any other constant is the general solution.0514

It is the general solution to this anti-differentiation problem.0531

In other words, if I'm given cos(x) and if I take the anti-derivative of that,0551

I know that the anti-derivative is going to be sin x + something.0555

What is it that I’m going to choose for that something?0560

In general, if we are just speaking about general solutions and taking anti-derivatives, you are just going to write cos x + c.0562

There is going to be other information in the problem that allows you to find what c is.0569

sin x + 6 or sin x +, or sin x + 48, those are specific solutions, particular solutions to a particular problem where certain data is given to you.0576

When we speak of the general situation, we put the anti-derivative and you just stick a c right after it,0589

to make sure that it is formally correct.0594

Are you going to forget the c, yes you are going to forget the c.0597

I still forget the c, after all these years.0601

Do not worry about it but this is the general solution.0603

anti-derivative, without any other information, just add the constant to it.0605

I will say, technically, you must always include the c.0616

Try your best to remember it.0628

Once again, extra information will allow you to find what c is.0631

We will see that when we start doing some of the example problems.0648

To find what c is, extra information will allow you to find what c is, in a particular case.0652

Thus, giving you what we call the specific solution or often called the particular solution.0677

Let us do some examples.0690

Before we do the examples, I’m going to give you a list of the anti-derivatives that we actually already know.0703

Some anti-derivative formulas we already know.0711

I think I will go ahead and do this in red.0730

Here we have the function and here I'm going to put the anti-derivative.0732

Once again, we have to make sure that we know what direction we are going in.0745

If I have x ⁺n, the way to find the anti-derivative of x ⁺n is you take x ⁺n + 1/ n + 1.0749

In other words, if this were the function that we are given,0761

you know that what you do is to take the exponent, you bring it down here.0764

You subtract one from the exponent.0769

Differentiation is this way.0770

What we are saying is, if you are starting with a function anti-differentiation, this is the formula that you use.0773

Again, you will see in just a minute what we mean.0779

Right now, I’m just going to write down some formulas here.0782

1/x, the anti-derivative of that is the natlog of the absolute value of x e ⁺x.0786

The anti-derivative is e ⁺x.0795

If I have cos(x), I know the anti-derivative of that is sin(x).0798

If I’m given sin(x), the derivative is cos(x).0803

If I’m given cos(x), the anti-derivative is sin(x).0806

Direction is very important here.0810

Do not worry, you will make a thousand of mistakes, as far as direction is concerned.0812

You will be asked to get an anti-derivative and you end up differentiating.0817

That is just the process that we go through, do not worry about it.0819

One more page here, let us write our function and let us write our anti-derivative.0825

If I have a sin x and I want the anti-derivative, it is actually going to be a -cos x.0841

Because if I’m given –cos x, the derivative of that is sin x.0847

Sec² x, the anti-derivative is the tan(x).0854

If I'm given sec x tan x, the anti-derivative sec x.0859

If I'm given 1/ 1 - x², all under the radical,0867

the anti-derivative of that is the inv sin(x) 1/ 1 + x², that anti-derivative is the inv tan(x).0872

One last one, -1/ 1 - x², all under the radical and that is going to be the inv cos(x), that is the anti-derivative.0886

If I were given the inv cos, the derivative of that would be -1/ √1 - x².0896

Now let us go ahead and jump right into the examples.0904

Find the most general anti-derivative for the following functions.0909

General means just add c to your answer, that is all that means.0912

1, 2, 3, 4, 5, let us go ahead and jump right on in.0917

Let me go ahead and make sure that I have everything here.0924

I have copied the functions, 14 x⁵, 4/3, and 3 ⁺x.0928

Let us start with number 1, I think I will go back to blue here, I hope you do not mind.0939

Number 1, our f(x) was x³ – 6x² + 11x – 9.0947

We said that the formula for the anti-derivative, when you are given some x ⁺n,0962

the anti-derivative of that is x ⁺n + 1/ n + 1.0968

Add 1 to the exponent , divide by the number that you get which is now the new exponent, very simple.0972

Therefore, our f(x), our anti-derivative is going to be x⁴/ 4.0979

How much easier can this possibly be?0987

-6, the constant, x³/ 3, we will simplify it, just a minute.0992

+ 11 x² because this is 1, add 1 to it and divide by that same number, -9x.1001

This is x⁰, x⁰, add 1 to the exponent, it becomes 1, divide by 1, it becomes 9x.1009

Now we can simplify, divide where we need to.1017

This is perfectly valid, you do not have to take the 6 and the 3, and divide it.1020

You can stop there, if you want to.1023

It just depends on what you teacher is going to be asking for.1025

We have x⁴/ 4, 6/3 is 2.1028

It is going to be -2x³, it is going to be +11 x²/ 2 – 9x + c.1035

I will go ahead and put that c here.1045

Again, we are going to add that c because it is the most general solution.1046

There you go, that is your anti-derivative.1051

You can always double check by differentiating your F.1055

The anti-derivative that you got, just differentiate and see if you get the original function.1066

That corroborates the fact that you have done it right, by differentiating f(x).1070

We have f(x) right here, therefore, f’, let us see what happens when I take the derivative of that.1085

It is going to be 4x³/ 4 - 6x² + 22x/ 2 – 9.1091

Sure enough, f’(x) is equal to x³ - 6x² + 11x – 9, which is exactly what the original = f(x).1111

F’(x) = f(x), that is what our theorem said, that is all we are doing.1128

We just have to remember which direction we are going in.1133

If we are taking the derivative of x³, it is going to be 3x², the original function.1137

If we are taking anti-derivative, it is going to be x⁴/ 4.1144

Direction is all that matters.1148

Let us go to function number 2, we had f(x) is equal to 14 × √x - 27 × 4√x.1151

We are going to write this with rational exponents.1167

This is going to be equal to 14 × x ^½ - 27 × x¹/4.1170

When we have an x ⁺n, when we take our anti-derivative, our formula is x ⁺n + 1/ n + 1.1180

That is it, you just subjected to the same thing.1188

It does not matter whether the exponent is rational or not.1191

This is going to be 14 × x, ½ + 1 is 3/2 divided by that number 3/2 – 27 × x.1194

¼ + 1 is 5/4 divided by 5/4.1206

We have to have our + c.1213

Therefore, we end up with, our final anti-derivative is going to be 14/ 3/2, that is going to be 28/3 × x³/2 - 4 × 27.1217

That is going to be 108 divided by 5 × x 5⁴ + c.1234

That is your most general anti-derivative.1244

If you took the derivative of this, you would get the original back.1248

Example number 3, we have f(x) = cos(x) - 14 × sin(x).1254

We are doing anti-derivative.1267

We go back to that list where we have the function and its anti derivative, which is also in your book or anywhere on the web.1270

You can just look at table of anti-derivatives also called table of integrals.1278

The anti-derivative of cos x was sin x - 14 which is the constant.1284

The anti-derivative of sin x was -cos x + c.1296

When we simplify, we get sin x + 14 cos(x) + c.1305

Once again, if you want to go ahead and check, the derivative of sin x is cos x.1318

The derivative of 14 cos x is -14 sin x.1323

Let us see what we have got, number 4.1336

We have x⁵, let me write down f(x).1341

F(x) = x⁵ + 2 × √x/ x⁴/3.1349

Let us write with rational exponents here.1364

We have x⁵ + 2 × x ^½/ x⁴/3.1366

I'm going to go ahead and separate this out.1378

It is going to be x⁵/ x⁴/3 + 2x ^½ / x⁴/3.1380

This is going to be x⁵/ x⁴/3 + 2x ^½/ x⁴/3.1386

I do not like writing my fractions that way, sorry about that.1398

I’m going to write it as it is supposed to be written, x⁴/3.1402

We get this is equal to x ⁺15/3 - 4/3 + 2 × x³/6 – 8/6.1406

Our f(x) is actually equal to x ⁺11/3 + 2 × x⁻⁵/6.1430

We can go ahead and take the anti-derivative.1445

I have just simplified that and made it such that there was an x to some exponent, so that I can use my formula.1447

I hope that make sense.1457

I cannot do anything with this, I have to convert it to something where I have x ⁺n and x ⁺n.1458

Now I can apply the formula.1463

The anti-derivative of f(x), now it is equal to, it is going to be x ⁺11/3 + 1 / 11/3 + 1 + 2 × x⁻⁵/6 + 1 all divided by -5/6 + 1 + c.1465

We have f(x) = x ⁺14/3/ 14/3 + 2x⁻⁵/6 + 6/6 is x¹/6 divided by 1/6 + c.1490

Our final answer is going to be 3/14 x ⁺14/3 + 12x¹/6 + c.1511

Our final answer, slightly longer not a problem.1533

It was only because of the simplification that we have to do.1536

Our number 5, we have our f(x) is equal to 3e ⁺x - 2/1 + x².1542

Really simple, this we can just read off.1555

The anti-derivative of e ⁺x is e ⁺x.1558

This stays 3e ⁺x.1561

Hopefully, we recognize that 1/1 + x², the anti-derivative of that is the inv tan.1564

Sorry about that, it is -2 × inv tan(x).1574

Of course, we add our c to give us our most general anti-derivative.1580

There you go, that takes care of that.1585

Hopefully, those examples help.1589

Again, it is all based on the basic formulas.1590

Let us do example number 2, given the following, find the original function f(x).1595

This time, they have given us extra information.1600

They have not only given us the f’, we are going to find the anti-derivative which is the f.1603

They have given it to us as f’, we just need to find f.1614

They also gave us other information, they said that the original f at 2 is equal to 40.1618

This extra information now is going to allow us to find what c is, in a particular case.1623

We are going to find the general solution.1629

And then, we are going to use this extra information to find the particular constant.1630

Let us get started here.1637

I think I have the wrong number here.1646

F(2) = 40, let me double check and make sure that my numbers are correct here.1652

I think I ended up actually using a different number when I solve this.1661

I had f(2) = 47, sorry about that, slight little correction.1665

Number 1, we have that f’(x) is equal to 5x³ - 14x + 24.1671

They tell us that f(2) is equal to 47.1684

F(x), notice that if I’m using prime notation, f is the anti-derivative of f’.1691

This just becomes 5x⁴/4 - 14x²/ 2 + 24x + c.1701

x ⁺n, just add 1 to the exponent, put that new exponent also in the denominator.1716

Let us go ahead and simplify a little bit.1722

f(x) = 5/4 x⁴ - 7x² + 24x + c.1724

This is our f, they tell me f(2) is equal to 47.1740

I put 2 wherever I have an x, I set it equal to 47, and I solve for c.1744

They tell me that f(2) which is 5/4⁴ - 7 × 2² + 24 × 2 + c.1752

They are telling you that all of that actually = 47.1765

I hope that I have done my arithmetic correctly.1775

We have got 20 - 28 + 48 + c = 47 and that gives me a final c = 7.1777

I’m hoping that you will confirm.1790

Now that I have c which is equal to 7, I can go ahead and put it back in to my equation that I have got.1793

My anti-derivative, my specific, my particular solution is going to be 5/4 x⁴ - 7x² + 24x + 7.1801

I found my constant and I have a particular solution, a specific solution, that is all I'm doing.1818

Do the anti-derivative and then use the information that is given to you to find the rest.1825

Number 2, we have an f’(x) is equal to 3 × sin(x) + sec² (x).1834

They are telling me that the f(π/6) happens to equal 5.1847

If this is f’, I take the anti-derivative.1856

This is going to be my f without the prime symbol.1858

It is going to be -3 × cos(x) because the anti-derivative of sin x is -cos(x).1861

The anti-derivative of sec² is tan(x).1868

This is going to be + c.1874

Now I use my information, f(π/6) is equal to -3 × cos(π/6) + tan(π/6) + c.1875

They are telling me that all of that is equal to 5.1892

Here we have cos(π/6) is going to be √3/2, -3 √3/2.1896

Tan(π/6) is going to be 1/ √3 + c is equal to 5.1905

Therefore, my c is going to equal, I’m not going to solve for of them, I’m just going to write it out straight.1913

It is going to be 5 – 1/ √3 + 3 √3/2, that is my c.1919

Therefore, I stick my c there and I get f(x) is equal to -3 × cos(x).1928

I will make my o's a little closer here, × cos(x) + tan(x) + whatever c I got which is 5 – 1/ √3 + 3 √3/ 2.1941

There you go, nice and simple.1957

Your teacher can tell you about the extent to which they want this simplified, put together, however they want to see it.1961

Number 3, let us see what we have got here.1970

Number 3, this one involves taking the anti-derivative twice.1977

They are telling me that f”(x) is equal to 8x - cos x.1982

They gave me two bits of information.1990

They are giving me f(1.5) is equal to 12.7.1992

They are telling me that f’(1.5) is equal to 4.2.1998

I have to take two anti-derivatives.2006

Therefore, I'm going to have two initial conditions.2009

One is going to be for f, one is going to be for f’.2011

I’m going to take the anti-derivative once, find f’.2014

Use this information, the f’(1.5) = 4.2, to find that constant.2017

I’m going to take the anti-derivative again, we will call it integration later, it is not a problem.2023

We are going to take the anti-derivative again, of the f’ to get our original function f.2030

We are going to use this first bit of information to find that constant.2034

Each step has a constant in it.2039

From f”, we are going to take the anti-derivative which means find f’.2042

This is going to be 8x²/ 2 - the anti-derivative of cos x which is sin x.2049

I will call this constant 1.2057

This is f’(x) = simplify a little bit, we have got 4x² – sin x + the constant of 1.2061

Now I use this information right here, the f(1) f’.2072

F' of 1.5 is equal to 4 × 1.5² – sin(1.5) + c1, they are telling me that it = 4.2.2076

I will write it all, that is not a problem.2096

When I solve this, I get 9 - 0.997 + c1 = 4.2.2097

I get that my c1 is equal to -3.803.2106

I found my first c1, that is the one that I’m going to plug in to here.2113

Therefore, my f’(x) is going to equal 4x² – sin x - 3.803.2119

Now that I have my f’, I want my original function f.2134

I’m going to take the anti-derivative again.2136

F(x) = 4x³/ 3 + cos(x) because the anti-derivative of sin x is -cos(x).2140

It is going to be -3.803x.2154

This is x⁰, it becomes x¹/1, and then now, + c2, always add that constant.2158

I know, I always forget.2165

I think the only reason that I actually remember is because I'm doing a lesson now, I’m trying hard to remember this, to put that c there.2169

Now we use this bit of information.2177

They are telling me that f(1.5) which is equal to 4/3 × 1.5³ + cos(1.5) - 3.803 × 1.5 + c2.2180

They are telling me that it = 12.7.2200

When I do that, I have got f(1.5) is equal to, it is going to be 4.5 + 0.0707 - 5.303 + c2 = 12.7.2207

When I solve, I get 13.432, that is my c2.2228

I put it back to my original and I end up with f(x) = 4/3 x³ + cos x - 3.803x + 13.432.2236

This is my particular solution to this particular anti-differentiation problem, given those two initial conditions.2256

Let us try another one of those.2270

This time we have, sorry about that, this is a double prime.2274

f”(x) is equal to 5/ √x.2283

We have f(2) is equal to 15 and we also have f’(2) is equal to 7.2289

Two initial conditions.2299

Let us write this in a way that we can manipulate.2301

f”(x) is equal to 5 × x⁻¹/2.2307

I take the anti-derivative so this is now going to become f’(x) and2314

this is going to be 5 × x⁻¹/2 + 1/ -1/2 + 1 which = 5x ^½/ ½, which is equal to 10x ^½ + c1.2318

There you go, that is my f’.2342

They tell me that f’(2) which is going to be 10 ^½ + c1 is equal to 7.2348

Therefore, my c1 is going to equal -7.14, when I do the calculation.2364

Therefore, I put this 7.14 into there and I get my f’, my specific solution f’(x) is equal to 10 x ^½ - 7.14.2371

f(x), I will do my f(x), I take the anti-derivative of this.2392

This is going to be 10x ½ + 1 is 3/2/ 3/2 - 7.14 × x + c2.2398

Let me see what I have got here, let me go to the next page.2416

I have got, when I simplify this, I have got f(x) = 20/3 x³/2 - 7.14 × x + c2.2429

They are telling me that f(2) which is equal to 20/3 × 2³/2 - 7.14 × 2 + c2.2445

They are telling me that that = 15.2459

When I solve for this, I get c2 is equal to 10.42.2461

I have my final f(x), my original function is 20/3 x³/2 - 7.14 × x + 10.42.2468

I think I did that right, I hope I did that right.2491

That is it, just anti-derivative, anti-derivative.2501

With each anti-derivative that you take, you want to go ahead and make sure to put the c2504

and then use the other information for wherever you are to find that c, and then take the next step.2509

Let us do a practical problem here.2518

A steel ball was dropped from rest from a tower 500 ft high, answer the following questions.2521

Take the acceleration of gravity to be 9.8 m/s2.2527

The first thing we want to do is find an equation for the height of the ball, after t seconds.2533

After I have dropped it, how long will it take for the ball to hit the ground?2537

What is the velocity of the ball as it hits the ground?2543

Part 4, if the stone is not dropped from rest, but if the stone is actually thrown downward with an initial velocity of 6 m/s,2546

how long does it take to reach the ground?2555

Let us see what we have got.2561

We have this tower, let us go ahead and draw this out.2568

This is the ground level, I’m just going to make this tower like that.2571

They tell us that this is 500 ft high.2575

I'm going to go ahead and take that as ground 0.2580

This is the 500, right.2584

Number 1 wanted the equation for the height of the ball after t seconds.2590

After a certain number of seconds, the ball is going to be like right there.2610

The height is going to be 500 - the distance that it actually traveled.2614

What I'm going to do is I'm going to find an equation for the distance that it actually traveled, and then take 500 – that.2621

That will give us our equation.2627

We will let s(t) be the distance function, how long it travels?2630

Be the distance function also called the position function.2638

You remember we called the position before.2644

That is the actual function that I’m looking for.2647

I’m looking for s(t).2649

S’(t), I know that the derivative of the position function is my velocity function.2651

It is my velocity and I know that if I take the derivative again, in other words, s”(t),2661

that is my acceleration, that is my acceleration function.2670

What do we know, we know that the acceleration of gravity is 9.8.2682

It is just a constant, that is it, you are just dropping it from rest.2686

The only force that is acting on this is the acceleration of gravity.2689

Therefore, our acceleration function s”(t) is actually just equal to 9.8, it is a constant.2693

Therefore, s’(t), when I take the first anti-derivative, that is just going to equal 9.8 t + c1.2704

What do I know, I know that s’(t) which is the velocity,2720

let us try it again, I know that s’ is the velocity.2731

I know that the velocity at time 0 which is the s’ at time 0 was starting from rest.2736

I’m just dropping it, it is 0.2742

S’(0) is 0, let us plug it in here.2746

That means that s’(0) is equal to 9.8 × 0 + c1.2749

I know that that = 0, that implies that c1 is actually equal to 0.2762

When I plug that in to my original equation, to my s’, I get s’(t) = 9.8 t.2768

I found an equation for the velocity at time t, it is 9.8 t.2777

Now I'm looking for s(t), now I’m going to take the anti-derivative of that.2786

Now I have got s’(t) = 9.8.2794

Therefore, s(t) is going to equal 9.8 t²/ 2 + c2.2797

What else do I know, now I need to find c2.2812

I know that s(t) or s(0), in other words the position at time 0 is 0.2817

I take that as my 0 position, that also = 0.2828

I’m going to put that in here, s(0) = 4.9 × 0² + c2 is equal to 0, that implies that c2 is equal to 0.2833

Therefore, my position function s(t) is equal to 4.9 t².2849

That means after a certain number of seconds, t seconds, I have actually traveled 4.9 × t² ft, m, whatever the length is.2857

Therefore, that means that is this distance.2869

After a certain number of seconds, I traveled this distance.2876

Therefore, my height above the ground is going to be 500 - this distance,2879

my height function is going to be 500 - 4.9 t².2884

Again, this is only based from the fact that I chose this as my 0.2892

You could have chosen this as your 0, just a different frame of reference.2896

I hope that make sense.2902

Let us go back to blue here.2910

How long before the ball strikes the ground, in other words, how many seconds go by before it?2915

How long before the ball strikes the ground?2930

Let us see, we came up with our s(t) which was going to be 4.9 t².2937

That was our position function.2948

We are falling 500 ft, our tower was this one.2953

We need to find out, we are going to set s(t) which is 4.9 t², we are going to set it to 500.2959

In other words, how many seconds does it take to go 500 ft?2969

When I solve this, I get t = 10.10 s.2973

Nice and simple, I have my position function.2982

How long does it take to go the 500 ft?2986

Number 3, velocity, as the ball hits the ground.2990

We said that our velocity function which was our first anti-derivative, s’(t), we said that that = 9.8 × t.3008

After 10.10 s which is when the ball is hitting the ground, I get the velocity at 10.10 s = 9.8 × 10.10 s.3019

It is going to be 99 m/s.3033

That is it, very straight forward.3038

Let us do the last one, number 4, if our initial velocity is 6 m/s downward, what is the velocity of the ball as it hits the ground?3044

Let us do this again, let us start from the beginning?3083

We have s”(t) which is our acceleration function, we know that that = 9.8.3085

When I take the anti-derivative of that, that is going to give my velocity function.3092

That is what I’m interested in.3095

Again, I have s’(t) which is my velocity function, that is going to equal 9.8 t + c1.3096

Standard anti-differentiation but now it is slightly different.3109

Now my initial velocity, in other words, my s’(0) which is my v(0) is now 6, it is not 0.3113

My velocity at time 0 which is 9.8 × 0 for t + c1 is equal to 6 m/s.3126

This implies that my c1 is actually equal to 6, that goes in here.3137

Therefore, my s’(t) function is actually equal to 9.8 t + 6.3143

When I find my s(t), I take my anti-derivative again,3156

I’m going to get my 9.8 t²/ 2 + 60 + c(2) which is equal to 4.9 t² + 60 + c/2.3160

I have a different function now and I also know that s(0) is still 0.3182

My 0 point is my starting point.3189

Therefore, I have got s(0) is equal to 4.9 × 0² + 6 × 0 + c2 = 0, which implies that our c2 is equal to 0.3191

Therefore, I get s(t) is now equal to 4.9 t² + 6t, that is my equation.3207

I need to find out how many seconds it takes, now that I have thrown it with an initial velocity which introduces the second term,3222

which was not there before, now I need to set this equal to, s(t) = 4.9 t² + 60.3227

I need to set that equal to 500, when I do that, I get t is equal to 9.51 s.3242

Exactly, what I expect, I threw it down with initial velocity instead of dropping from it rest.3251

It is going to take less time for it to get to the ground.3256

It took 10.1 seconds, now it is only taking 9.51 seconds.3258

This 9.51 is now, what I actually am going to put into my velocity function.3263

I knew velocity function which includes this extra term for the initial velocity.3275

S’(t) which is my velocity function is equal to 9.8 t + 6.3286

Therefore, the velocity of 9.51 = 9.8 × 9.51 + 6.3295

I get my velocity 1.51 = 99.2 m/s.3306

Not a lot faster but certainly faster.3314

There you go, that takes care of anti-derivatives.3320

Thank you so much for joining us here at www.educator.com.3323

We will see you next time, bye.3325