For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### Example Problems for the Fundamental Theorem

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
- Example I: Find the Derivative of the Following Function
- Example II: Find the Derivative of the Following Function
- Example III: Find the Derivative of the Following Function
- Example IV: Find the Derivative of the Following Function
- Example V: Evaluate the Following Integral
- Example VI: Evaluate the Following Integral
- Example VII: Evaluate the Following Integral
- Example VIII: Evaluate the Following Integral
- Example IX: Evaluate the Following Graph

- Intro 0:00
- Example I: Find the Derivative of the Following Function 0:17
- Example II: Find the Derivative of the Following Function 1:40
- Example III: Find the Derivative of the Following Function 2:32
- Example IV: Find the Derivative of the Following Function 5:55
- Example V: Evaluate the Following Integral 7:13
- Example VI: Evaluate the Following Integral 9:46
- Example VII: Evaluate the Following Integral 12:49
- Example VIII: Evaluate the Following Integral 13:53
- Example IX: Evaluate the Following Graph 15:24
- Local Maxs and Mins for g(x)
- Where Does g(x) Achieve Its Absolute Max on [0,8]
- On What Intervals is g(x) Concave Up/Down?
- Sketch a Graph of g(x)

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for the Fundamental Theorem

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

*Today, we are going to be doing some example problems for the fundamental theorem of calculus that we just covered.*0004

*These are going to be extraordinarily straightforward and simple, almost surprisingly.*0010

*Our first example is find the derivative of the following function.*0017

*The function is g(x) and it is defined as the integral from 1 to x, whatever x might be.*0021

*5/ t³ + 9 dt, nice and simple.*0027

*Since the upper limit is x, we would literally just put x where we have the t.*0032

*Our final answer is just 5/ x³ + 9, and we are done.*0039

*That is the beauty of the fundamental theorem.*0046

*When a function is defined as an integral or the upper integral is a variable,*0048

*this particular variable right here xx, we just literally put that in there.*0054

*Let me correct this, a little bit of a notational issue here.*0064

*g’ because we are looking for the derivative.*0071

*g’(x), that is a little more clear, I think.*0075

*We have 5/ x³ + 9.*0079

*Literally, when we take the derivative of this, we are just getting rid of this integral sign*0084

*because differentiation and integration or anti-differentiation are inverse processes.*0090

*That is the whole idea.*0095

*That is it, nice, straightforward, and simple.*0097

*Find the derivative of the following function.*0102

*g(x), upper variable is x, we do the same thing.*0105

*Therefore, the derivative of this g’(x), all we have to do is we put an x where we see a t.*0110

*We get x² sin³ x, that is it, very nice.*0117

*As long as this upper limit of integration is the variable that is inside the function defining it,*0128

*we can just put it straight into the function itself.*0144

*Let us try this one, find the derivative of the following function.*0150

*g(x) = the integral from 1 to x⁴ sin(t) dt.*0155

*Now this is x, this time, we have not x but we have a function of x.*0161

*The only thing we do know is we do the same thing.*0168

*We are just going to be putting in the x here.*0171

*g’(x) is equal to sin(x).*0173

*But because this is not x but it is a function of x, by the chain rule,*0177

*we are just going to multiply by the derivative of this function.*0181

*Let us write that out.*0186

*When the upper limit is not just x but a function of x, some more complicated function of x, we do the following.*0188

*This is the long process, the easy part is just multiply by the derivative of the upper limit, whatever the function happens to be.*0220

*Essentially, what you are doing is you are letting x⁴, the upper limit integration equals some letter u.*0227

*Essentially, you are just doing a substitution.*0239

*And then, du dx is going to equal 4x³.*0240

*And then, you rewrite this g(x) = the integral from 1 to u of sin(t) dt.*0250

*Then, when you do this, when you put this into here, in other words, when you do g’(x) is equal to the sin(u),*0263

*because u is a function of x, by the chain rule, the derivative is sin(u) du dx.*0277

*You are multiplying by du dx.*0285

*What you end up getting is g’(x) is equal to sin(x)⁴ × 4x³.*0292

*This part is exactly the same, you are just putting the upper limit into this variable to get the sin x⁴.*0303

*And then, you are just multiplying by the derivative of that because now it is a function of x, it is not just x itself.*0311

*I hope that makes sense.*0321

*Essentially, what you are doing is you are doing the same thing as you did before.*0322

*If this were just an x, you would put it in and it would be sin(x).*0325

*The derivative of x is just 1, you were technically doing du dx.*0331

*The derivative of x is just 1, which is why it just ends up staying that way.*0336

*But if it is a function, just multiply by its derivative.*0342

*That is your final answer, right there, very straightforward.*0345

*Find the derivative of the following functions, same thing.*0356

*Now π/3, the upper limit is cos(x).*0359

*We are just going to put this into the t's and then multiply everything by the derivative of cos(x).*0363

*g’(x), we would literally just read it off.*0372

*It is going to be 3 × cos(x) + 7 × cos² (x) × the derivative of that function cos(x) which is –sin(x).*0375

*We are done, that is all.*0391

*Just do what the first fundamental theorem of calculus says which is just put the upper limit into those things.*0399

*Then, multiply by ddx of the upper limit.*0415

*That is all, moving along nicely.*0431

*Evaluate the following integral.*0435

*We want to evaluate a definite integral.*0437

*We have a lower limit, an upper limit, they are both numbers.*0440

*We are going to find the antiderivative, the integral.*0443

*And then, we are going to put in 5, we are going to put in 1,*0446

*and we are going to subtract the value of what we get from 1, from the value of the upper limit 5.*0450

*Let us go ahead and do that.*0456

*The integral of this, let us go ahead and actually multiply this out first because we have x³ - 4x².*0457

*This is going to be the integral from 1 to 5 of x⁶ – 4x × that is going to be 4x⁴, then another 4x⁴.*0468

*It is -8x⁴ + 16x², I just multiply this out, dx.*0484

*And that is going to equal x⁷/7 - 8x⁵/5 + 16x³/3.*0495

*I'm going to evaluate this from 1 to 5.*0512

*In other words, I’m going to put 5 in for all of these and get a number, then I'm going to subtract by 1.*0514

*Putting them into all the x here.*0520

*When I put 5 into this, I get 78,000, 5⁷ is 78,125/7.*0522

*When I put 5 into here, multiply by 8 and divide by 5.*0535

*I get - 2500/5, I hope you are going to be confirming my arithmetic.*0539

*I’m notorious for arithmetic mistakes.*0544

*+ 2000/3 that takes care of the 5.*0546

*I’m going to subtract by putting 1 in.*0553

*This is going to be 1/7 - 8/5 + 16/3.*0556

*Find the antiderivative of the integrand, and then evaluate.*0565

*Put the upper number into this, you are going to get a value.*0570

*Put the lower number into the x values, you are going to get this and then you subtract.*0573

*The rest is just arithmetic.*0577

*I'm just going to go ahead and leave it like this.*0579

*You can go ahead and do the arithmetic, if you want, to get a single answer.*0581

*Evaluate the following integral, once again, we are using the second fundamental theorem of calculus.*0588

*The integrals of f from 2 to 8 is going to be the f(8) - f(2), F(8) – F(2), that is what the second fundamental theorem says.*0594

*Remember, it is this one, if from a to b of f(x) dx is just equal to f(b) – f(a), where F is the antiderivative of f.*0611

*This is going to be the integral from 2 to 8, I’m going to separate these out.*0629

*I’m going to write this as x²/3, it is going to be x/ x²/3 - 3/ x²/3.*0638

*x/ x²/3 dx - the integral from 2 to 8 of 3/ x²/3 dx.*0646

*That is going to equal, the integral from 2 to 8 of x¹/3 dx - the integral from 2 to 8 of 3 × x⁻²/3 dx.*0662

*I can go ahead and integrate these, not a problem.*0678

*1/3 + 1 is 4/3, I get x⁴/3/ 4/3 - 3 × x – 2/3 + 1 is 1/3/ 1/3.*0681

*That is going to equal ¾ x⁴/3 - 9x¹/3.*0700

*I'm going to evaluate that from 2 to 8.*0712

*When I put 8 in for here, I'm going to get 12.*0716

*When I put 8 into here, it is going to be -18, that is the first one.*0721

*-, when I put 2 into here, I end up with 1.89.*0727

*And then, - when I put 2 into here, I end up with 11.34.*0734

*There is your answer.*0740

*Once you have the antiderivative, you put the upper limit into the x value to evaluate, that gives you the first term.*0742

*You put the lower limit into the x, that gives you the second term.*0753

*You subtract the second term from the first term, f(b) - f(a).*0758

*Nice and straightforward.*0765

*Another good example, the integral from π/6 to 2π/ 3 of cosθ dθ.*0772

*This is going to equal where the integral of cosθ dθ is going to be sin(θ).*0780

*We are going to evaluate sin(θ) from π/6 to π/3.*0786

*Not a problem, that is going to be the sin(π/3) - sin(π/6).*0796

*We are just putting in upper and lower limit into the actual variable of integration.*0809

*The sin(π/3) is √3/2 – sin(π/6) which is ½.*0815

*The integral from 1 to 3 of 5/ x² + 1.*0835

*You should recognize that the integral of 1/ x² + 1 is equal to the inv tan.*0841

*Remember, the integral or the antiderivative, whatever you want to call it,*0847

*of 1 + x² dx that was equal to the inv tan(x).*0852

*Therefore, this is, that is fine, I will go ahead and separate it out.*0860

*This is just a constant, it is 5 integral from 1 to 3, 1/ x² + 1 dx = 5 × inv tan(x), evaluated from 1 to 3.*0866

*That is it, that is all it is.*0885

*This is just 5 × inv tan(3) – inv tan(1).*0887

*That is all, you can go ahead and multiply the 5 through.*0901

*You can do 5 or you can evaluate this from here to here.*0904

*This - this and then multiply the constant through.*0911

*The order actually does not matter.*0914

*I’m just somebody who tends to like to see the constant on the outside, personal choice.*0916

*g(x) is equal to the integral from 0 to x of f(t) dt.*0928

*f being the function whose graph is given below.*0934

*This graph right here, this is a function of f.*0937

*Answer the following questions.*0943

*We take a look at this graph, it looks like some sort of a modified sin or cos graph, definitely periodic.*0945

*We are taking a look real quickly, we notice that we have about a 3. something, then we have a 6.2 something.*0952

*It turns out that these are actually just π/2, π.*0961

*I’m going to go ahead and label them that way.*0970

*I left the labels just as is, but it is nice to recognize the numbers for what it is that they are.*0972

*Especially, since you are talking about a periodic function, then more than likely the graphs that you are going to be dealing with*0980

*are going to pass through the major points, π/2, π, π/6, π/3, π/4, things like that.*0985

*In this case, let me mark these off.*0992

*3.14, this is π.*1003

*This over here looks like it is π/2.*1010

*Over here, this looks like 2π and this looks like 3π/ 2, 4π/ 2.*1017

*It looks like we have got 5π/ 2.*1028

*At least that way we have some numbers that we are accustomed to seeing,*1030

*instead of just referring them as 3.1, 1.5, things like that.*1033

*Give the values of x for which g(x) achieves its local maxes and mins.*1040

*You have to be very careful here, this is a graph of f.*1047

*g is the integral of f from 0 to x.*1052

*Therefore, g is actually the area under the graph.*1059

*The graph actually falls below the x axis, the area is going to become negative.*1066

*Give the values of x for which g(x) achieves its local maxes and mins.*1075

*As we start moving from 0 x, moving forward g(x), as x gets bigger and bigger,*1080

*this value is just the area under the graph itself.*1091

*It is going to hit a high point, when I hit π/2.*1095

*And then, at π/2, because the graph itself falls below the axis, as I keep going,*1101

*as x keeps getting bigger and bigger, the integral itself is going to be negative.*1108

*The area is going to hit a maximum and then the integral is going to start to go negative.*1115

*The area is going to diminish.*1120

*At this point, the area is going to go down.*1123

*At some point, it is going to hit 0.*1127

*The area is going to rise to a certain number, that is up to here.*1130

*We are going to start subtracting from the area, at some point it is going to hit 0.*1134

*Over here, it is going to hit a minimum because now this area under the x axis is a lot bigger than the area above the x axis.*1138

*The integral from 0 to π is going to be a negative number.*1147

*It is actually going to hit a maximum.*1154

*One of the maximums that it is going to hit is right there.*1157

*And then, it is going to hit a minimum right there.*1161

*After that, it is going to climb again, the area under the graph.*1164

*That is what we are doing, g(x) is the area of f, the area under the graph of f because this is the graph of f.*1169

*The area is going to get bigger again.*1178

*At this point, it is going to hit another maximum.*1180

*And then at this point, the area is going to be negative.*1185

*We are going to hit another minimum, 2π.*1187

*We are going to increase the area again.*1191

*It is going to pass this and it is going to hit another maximum.*1194

*That is what is going to happen.*1198

*Give the values of x for which g(x) achieves its local maxes and mins.*1199

*It is going to achieve its local maxes and mins where the area under the graph hits a maximum and a minimum.*1205

*It is a maximum at π/2, a minimum at π.*1215

*Maximum at 3π/ 2, a local minimum at 2π, local max and min at 5π/ 2.*1218

*We have π/2, π, 3π/ 2, 2π, and 5π/ 2.*1225

*The reason is because g(x) is the net area under the graph of f.*1241

*This red is the graph of f.*1248

*Where does g(x) achieves its absolute max, between 0 and 8, this point right here.*1255

*Let us go ahead and draw this out.*1266

*Basically, what we are going to have is this.*1267

*Let me do this in black.*1270

*g(x) is going to look like this, it is going to rise.*1272

*It is going to hit a maximum here and it is going to hit a minimum.*1277

*It is going to start rising again.*1282

*It is going to hit a maximum and it is going to hit a minimum, and it is going to hit a maximum.*1286

*It is going to start coming down to there.*1296

*The area keeps getting bigger and bigger.*1300

*The absolute value, the area keeps getting bigger and bigger.*1304

*Our highest point is going to be this point right here, which is going to be our 5π/ 2.*1307

*Again, because g(x), that is what we want, where does g(x) achieve its absolute max.*1316

*g(x) is the net area under the graph.*1322

*Positive, if it is above the x axis, negative if it is below.*1326

*This is the graph of g.*1330

*On what intervals is g(x) concave up or concave down?*1335

*For concave up or concave down, we want to see the sin(g”).*1341

*g’ is equal to f.*1356

*By the fundamental theorem of calculus, if I take the derivative of this, I get rid of this.*1360

*I just have f(x), this is g’ that mean g“ is equal to f’.*1364

*f’ is just the derivative of the red graph.*1373

*The derivative is the slope.*1379

*It is going to be concave up where the slope is positive.*1383

*It is going to be concave down where the slope is negative.*1387

*If the slope is positive, that is f’.*1392

*f’ happens to be g”.*1395

*Therefore, we have from here to here, is concave up.*1400

*From here to here, it is concave down.*1415

*From here to here, concave up.*1424

*From here to here, concave down.*1429

*From here to here, concave up.*1433

*And from here to 8, we got concave down which is exactly what we see.*1437

*Concave up graph, this is g that we are looking at.*1443

*Now it is concave down, it starts to be concave up again up to here.*1445

*From here to here, it is concave down, it matches.*1450

*I hope that makes sense.*1457

*You are given a graph f, if g is defined as the integral from 0 to x of f.*1460

*It is the net area under that particular graph.*1467

*I hope that made sense.*1472

*Let us take a look at what it actually looks like.*1475

*The original function f was this one right here.*1478

*This one, this is the actual g(x).*1484

*This was our f(x).*1488

*It is a maximum here, it is a minimum here.*1494

*Local, hits a local max here, it is right there.*1502

*It is a local min here, it is right there.*1507

*Hits a local max here, that is right there.*1511

*It is exactly what we said before.*1515

*Thank you so much for joining us here at www.educator.com.*1518

*We will see you next time, bye.*1520

1 answer

Last reply by: Professor Hovasapian

Mon Feb 8, 2016 1:56 AM

Post by nathan lau on January 31, 2016

at 13:30, isn't it 2pi/3, not just pi/3?