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Step by Step Calculus Solver

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Step by Step Calculus Solver

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Hello, Tom from everystepcalculus.com, everystepphysics.com. Don’t forget physics, either, in your schooling. I‚m going to do two problems in calculus: a definite integral, and a log problem. I’ll show you the diversity of my programs. And my programs turn the titanium into a calculus calculator with steps. A calculus calculator with steps‚ that’s exactly what my programs do. So let’s do it. Index 8 to get to my main menu. We’re going to scroll down to definite integral in the D’s. Definite integral in X because you only see X in the problem, right? We’re going to enter the function. You have to press alpha before you enter anything into these entry lines. You’re going to press alpha, and we‚re going to enter the function. 3 minus X to the cubed plus 4 times X. Now we show you what you’ve entered; you can change it if you want. I say it’s okay. I’m going to enter the range. Lower range is alpha minus 2. Upper range is alpha 2. I say that’s okay also. And we integrate it, which is this right here. We’ve integrated each one of those terms, separated by plus or minus signs. And at the upper range, X equals 2. You add these into the‚ you’re going to use, instead of quotation marks, you’re going to use parentheses around your additions into the main function. But it equals 10. And if X equals minus 2, the answer is minus 2. Upper minus lower is equal to 12 square units. Pretty neat, huh? All right, we’re going to go back to the main menu. Number two: And we’re going to scroll down. Now this is in the L section, logs. So I’m going to do this quick. Behind the simulator here I can only use, I only have one essential finger to do this with. On your calculator, the titanium, you can hold the 2nd down with your thumb or finger and press this. It’ll go screen by screen and really go quick down to logs. We have natural logs and all kinds diversity in my menus. I’ve done all the calculus problems, or most of the tests, of course. Nobody can do all of the calculus problems; there’s millions of derivations of that. Log problems, okay. We’re going to evaluate this log problem, number 3. We have to press 2nd alpha to get to the letter register to put log in. We’re going to enter the problem. We want to make, we have 2nd down here, but we want to turn it to black like that. Then we can put the logs in there, so that’s‚Äî The letters appear over the numbers, you can see them. And then we’re going to go back to numbers, which erases that black mark there, indication. And we’re going to put 3 parentheses 1 divided by 27. Close off the parenthesis. And I show you what you’ve entered. It looks pretty good to me. We’re going to press 1, and here’s the answer step-by-step. In other words, if you on your calculator, if you put 3 to the exponent minus 3, you’re going to come up with 127. All right. Pretty neat, huh? Everystepcalculus.com. Go to my site. Buy my program if you want to pass calculus or physics. Or subscribe so you can see more videos. Have a good one.

Implicit Differentiation Calculator With Steps

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Implicit Differentiation Calculator With Steps

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Hello everyone; Tom from everystepcalculus.com and everystepphysics.com. Don’t forget physics. I have turned through the use of my programs, or for programming, your titanium calculator into an implicit differentiation calculator with steps. And that’s exactly what my programs do. So let’s do it. Index8() to get to my menu. I’m already at implicit differentiation. I’ve scrolled there. On your titanium, if you want to go to my menu and go down quickly, you hold the 2nd button down here and then use the down cursor here. We’re already at that, so. And then enter the function. You have to press alpha before you enter anything in these entry lines here. And the problem is alpha Y squared plus 3 times X minus 8 times Y plus 3 equals 0. Now it will show you what you’ve entered, and you change it if you want. I say it’s okay. You’re differentiating all of the terms on both sides of the equals sign. You must have a 0 or a constant on the right of the equals sign. You have to use the algebra to make that happen if they give you a more difficult problem, or a different look at a problem. And so we differentiate each one of the terms. I do that for you; you write this on your paper as we go through it, exactly as you see it here. And we combine the DY DX terms, separate it from the other terms, and here’s the answer right here. Now we want to evaluate it at a point, so we’re going to press the X value is alpha 4 and alpha 3. I show you that also; it looks good to me. So we’re substituting 4 and 3 for the X and the Y values and then find the derivative. Cut them to one half and the slope of the line is 333 degrees. That means that the slope is going to be here’s 360 going this way. Here’s 270 going down. So the slope is going to be like this. Pretty neat, huh? Everystepcalculus.com. Go to my site. Buy my programs if you want help passing calculus. Or subscribe to enjoy more videos that I might make. Have a good one.

Learn more about implicit differentiation on your calculator

Implicit Differentiation test question solved by TI-89

Implicit Differentiation calculator – Video Example #2

Implicit Differentiation at a Point – Video Example #3

Implicit Differentiation – Video Example #4

Calculus Calculator with Steps

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Calculus Calculator with Steps

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Hello, Tom from everystepcalculus.com, everystepphysics.com. Don’t forget physics, either, in your schooling. I’m going to do two problems in calculus: a definite integral, and a log problem. I’ll show you the diversity of my programs. And my programs turn the titanium into a calculus calculator with steps. A calculus calculator with steps, that’s exactly what my programs do. So let’s do it. Index 8 to get to my main menu. We’re going to scroll down to definite integral in the D’s. Definite integral in X because you only see X in the problem, right? We’re going to enter the function. You have to press alpha before you enter anything into these entry lines. You’re going to press alpha, and we’re going to enter the function. 3 minus X to the cubed plus 4 times X. Now we show you what you’ve entered; you can change it if you want. I say it’s okay. I’m going to enter the range. Lower range is alpha minus 2. Upper range is alpha 2.
I say that’s okay also. And we integrate it, which is this right here. We’ve integrated each one of those terms, separated by plus or minus signs. And at the upper range, X equals 2. You add these into the you’re going to use, instead of quotation marks, you’re going to use parentheses around your additions into the main function. But it equals 10. And if X equals minus 2, the answer is minus 2. Upper minus lower is equal to 12 square
units. Pretty neat, huh? All right, we’re going to go back to the main menu. Number two: And we’re going to scroll down. Now this is in the L section, logs. So I’m going to do this quick. Behind the simulator here I can only use, I only have one essential finger to do this with. On your calculator, the titanium, you can hold the 2nd down with your thumb or finger and press this. It’ll go screen by screen and really go quick down
to logs. We have natural logs and all kinds diversity in my menus. I’ve done all the calculus problems, or most of the tests, of course. Nobody can do all of the calculus problems; there’s millions of derivations of that. Log problems, okay. We’re going to evaluate this log problem, number 3. We have to press 2nd alpha to get to the letter register to put log in. We’re going to enter the problem. We want to make, we have 2nd down here, but
we want to turn it to black like that. Then we can put the logs in there, so that’s—The letters appear over the numbers, you can see them. And then we’re going to go back to numbers, which erases that black mark there, indication. And we’re going to put 3 parentheses 1 divided by 27. Close off the parenthesis. And I show you what you’ve entered. It looks pretty good to me. We’re going to press 1, and here’s the answer step-by-step. In other words, if you on your calculator, if you put 3 to the exponent minus 3, you’re going to come up with 127. All right. Pretty neat, huh? Everystepcalculus.com. Go to my site. Buy my program if you want to pass calculus or physics. Or subscribe so you can see more videos. Have a good one.

Green’s Theorem

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Hello again, Tom from everystepcalculus.com, everystepphysics.com, a calculus video on Green’s Theorem. Let’s do it. Index 8 to get to my menu. We’re gonna scroll down to the “G”‘s and choose Green’s theorem. Normally here’s the way problems are given, you know, something before the dy and something before the dy, and so I’m gonna ask you to enter those into this– in any problem, but you can see an example of what problem we’re gonna be doing right now. This is from Patrick JMT’s site, exactly the way he does it. Of course I have the program so that you can add other variables et cetera, so that’s what’s neat about programming and using the Titanium. So we’re going to enter alpha before we enter anything into these entry lines here, alpha X for the dx, and then alpha minus X squared times y squared. Green’s theorem turns a line integral into a double integral and I have found that line integrals are very difficult, I’d say it’s a waste of time, but to me as most calculus is. In fact, even, I was searching and searching line integrals and there’s not even SI units for the answer. You get some some sort of answer like 8192 over 5 or something. And there’s not even– so I don’t know what that computes and nobody else does. It’s called a line integral and it’s not a length, it’s not a width, it’s not an area… so I question that. But of course the Green’s Theorem makes it even a little bit more easy or difficult, depending on what you can do with integrals. But anyways, here’s my program on it. I say it’s okay once we enter that and of course they always give you a triangle. They give you a square, a curve, a unit circle, or two circles, or one circle. Calculus always makes something difficult out of something simple. If that’s simple. So anyways, this is a triangle, which we’re gonna choose. The object within the region is a right triangle. It has to be a right triangle because of the pythagorean theorem, which is x squared and y squared in there, which calculus can work with exponents like Newton designed it. And the first vertices given is (0,0), we’re gonna choose that. The next one is (0,y) and we have to enter something for the y, because even though these are all ones, you can’t enter something different. You can enter 5– and I have to make sure I understand that when I program. But this is alpha 1 and then that changes the one here and the last third vertices can’t be anything different than one. And we’re gonna enter the x-value for that. And that’s 1, 2, alpha 1. And so I show you what the vertices given are, if that’s correct then choose “okay”. And then something that other programs don’t give you, I draw it, because you’re supposed to draw it on your paper– adding the vertices and stuff like that, so here’s the way it’s drawn. And y equals x. Why does this equal x? Because this slope, if you graph it, is x. Okay, now what if this was 5 and 7? Well then, slope is rise over run so this would be 5– or, if the x was 5 and the y was 7, rise over run is 7/5. So it would be 7/5 x here. Right now its (1,1) x here. But you don’t see it because 1 divided by 1 is 1. So i exclude that part, but those are the kind of things that upset me– and this is the tough part to me, this was always the tough part– getting the vertices and how to put it in the integral. So here’s the formula for the Green’s theorem integral, here’s the region, and I show you this is ax over bx, you know, these are the limits. And so we add those automatically. 0 over 1 for the x, and x over 1 for the y. We’re gonna do the order of integration is gonna be dy first and then dx. So here’s the– we’re integrating this and we’re just gonna go through it quick and keep this video kind of short. At y equals 1 here’s the answer. At y equals x here’s the answer when you substitute that. I show you the substitutions in here. Here’s one substitute of that. Now, you’ll use parenthesis for this instead, I use quotes because that’s the way that the calculator operates, but you’re going to use parentheses. Two times parenthesis “one” parenthesis cubed. Times– and that’s substituting 1 for the y. And here we’re substituting x for the y so there would be parenthesis around this x here. And of course the upper limit minus the lower limit, look at the minus signs here and how it works out. I always screwed that up when I was in calculus. That’s the reason I programmed, so I wouldn’t screw up. And now we’re gonna do this integral here from 0 to 1. And… at x equals one it equals minus one fifth and zero. Subtract the upper limit from the lower. Or, lower from the upper limit. And the answer is -1/5. Pretty neat, huh? everystepcalculus.com, go to my site, subscribe and see other videos that I make. Have a good one.

TI 89 Partial Fractions

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Partial Fractions solved on the TI-89 calculator.

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Hello Tom from EveryStepCalculus.com EveryStepPhysics.com there’s been many searches for ti-89 partial fractions and I’m going to show you how that works on your titanium.

You can download these programs into your titanium and do this exactly like I do it here a miracle of programming from the inventors MIT the geniuses over there. So I’m going to show you index8() to get to my menu. I’m already at partial fractions which you would scroll to when you know the title of what you’re looking for in calculus.

And we’re gonna enter the integral and that is quite long but I’m going to do it yet to press alpha before you enter anything in these entry lines here. Alpha left parenthesis 2 times X cubed minus 4 times x squared minus 15 times X plus 5 close off the parentheses.

Divide sign more parentheses x squared minus 2 times X minus 8. Remember calculus is the sudoko math in my opinion a lot of nonsense and this is, this is certainly true of partial fractions. Now you try this without a program, you can do it if you want, not me. Press ENTER, when it’s busy you’re loading the program I always show you what you’ve entered so you can change it if you want to X cubed minus 4x squared minus 15 X plus 5.

That’s crew x squared minus 2x minus 8 that’s good, let’s say ok, and so let’s do it. First thing you do is factor the denominator here, it’s loading programs again but once it’s loaded it’s quick on your calculator but these are long programs. Because the numerator is higher than or equal to then than the denominator you have to use short division.

So 2x cubed divided by x squared is 2x and that’ll be added to the answer at the end. This is a quotient that’ll be added to the answer at the end, calculator is computing and you write all this on your paper of course. A X minus 4 B X plus 2 times the factored denominator and you get this. Write this stuff down, I’m not going to explain it to you, and x equals -2 it equals this here x equals 4. We’re solving for all the a B’s and C’s equals 1/2 and x equals 4 I do that for you amazing program.

I think a equals 3 2 3 and 3 halves here’s the answer, 3 halves over X minus 4 plus minus 1/2 over X plus 2 plus. Here’s the quotient we found with short division plus C and then we even do the integrals for you remember most of these are logs because when you have something on a denominator without an exponent it has to be a log problem. It has to be a log differentiation.

So good luck in your calculus class and subscribe to me, maybe see more videos or you can actually buy the programs and enjoy passing calculus. Have a good one.

U Substitution Calculator

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Hello, everyone. Tomfrom everystepcalculus.com and everystepphysics.com. I’m already at U Substitution. You scroll with these cursors, here to get to whatever in my menu that you’re interested in. You have to press alpha before you enter anything into these entry lines, here. And the problem is Alpha x times the square root of x squared plus sixteen. close off the parentheses. X times Square root of x squared plus sixteen. That’s cool. I always show you what you’ve entered. Now, how are you going to go about solving this yourself? Okay, think about it. In U Substitution notice this is a exponent of two, this is an exponent of one. okay. If the x has one less exponent than the inside of the parentheses, this is a U substitution problem. For instance, you should always in your mind to do with the derivative of something in the parentheses in Calculus. That’s one of the big deals and big tricks. So this is 2 X then. And there’s of course no derivative 16 at zero. So it’s 2x
Well here’s an X on the outside. Well, we can do something with the 2 to make
that x dx. So generally what I do what and what you should do. I say it’s okay because we’ve entered it right. You should rewrite the problem okay. So you rewrite the problem.
Square root of x squared plus 16 and you put the x dx over here so you can kinda see that you’re trying to match that somehow, okay. So then U equals x squared plus 16. Well then du equals 2x dx, okay. We take that 2 and the du divided by 2 through Algebra. And we come up with x dx. Notice that this matches the previous match of the function. so that a course you is equal to this group experts 16 So we have the Integral of u and of course the square root, in calculus you always change it to 1/2. It was a cube, it could be 1/3. If it was the 15 root, you’d be 1/15, okay. That’s Algebra 2. I mean also. So and then you have the DU divided by two so that’s a constant, you have to bring outside the integral, here. I do that for you. Here’s the 1/2 outside the integral and then we do the U to the 1/2 inside and
then integrate that part, okay. And that turns out to be this this this and the answer is 1/3, etc. So, have a good one. everystepcalculus.com

Integral Calculator with Steps

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∫ 6x^4*(3x^5+2)^6 dx

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Hello, everyone. This is Tom from everystepcalculus.com. There’s been many, there’s always requests for Integral Calculator with Steps and that’s exactly what my programs do. And Integration is one of the toughest things in Calculus. It was for me when I was in class. I hated
Calculus. You probably feel the same way and I’ve found nobody that likes Calculus. Except
maybe Professors. But anyways, I’m going to show you how my programs work on U Substitution. Index 8 to get to my menu. I’m already at U Substitution. I have scrolled there. We’re going to enter our function. You have to press Alpha before you enter anything into these entry lines, here. Alpha 6 times x to the fourth power times the quantity parentheses
3 times x to the fifth power plus 2, close off the parentheses to the sixth power. I always show you what you’ve entered. 6x 4 times 3 x quantive to the 5 plus 2 to 6. Looks good to me. I say it’s okay. And we’re going to work the problem. Busy means the program is loading. We’re going to evaluate this. First we rewrite it where all constants come out of the integral. And then we put the x to the 4. The way that you know that any problem is U Substitution is that you look immediately at what’s inside the parentheses. You take the derivative, right now you should be able to do, this is 15x to the 4. Right now in one second, you should be able to know the derivative of that. And you notice that x4 is on the outside, too. If it isn’t, it’s not a U Substitution problem. It has to be converted, okay. So I do all that for you, really. But that you have some understanding of how you do U Substitution. U is equal to this, du is equal to this and then we make this the other trick, this whole system here took me about a year to figure this out in a system that works, you know. And so, this always has to be x to the 4 dx over here so you have to take the 15 and divide the du by that on the other side using Algebra, of
course. And so it works the problem. 6, you take of course the du with the 6 here, you have a du with a 15 notice the 15 come out of the, here’s the constant again so that comes out of the integral and goes in front right here, see it right here. 6 times that, of course, with the 2 fifths. etc, etc. Here’s the answer to your problem right here. Have a good one. everystepcalculus.com

Evaluate Log Problem on TI-89-Video

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evalutate log problem

 

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Hello, everyone. Tom from everystepcalculus.com and everystepphysics.com. I’m going to integrate a log to a base problem, in this video.Let’s do it. Index 8 to get to my menu. We’re going to use second and the cursor here to go to screen by screen to get down to the L section where there’s log problems. We’re going to choose that one. And here’s log problems. And we’re gonna, we want to integrate. So we’re going to choose integrate, number 9 there. You could press a number or scroll to it, either one. And we’re going to enter our functions, here. Now this is a little tricky because we have to add l o g in here. So we’re going to have to press second, Alpha twice so it this becomes black. You see this black here? Then we can go, 4 minus 7 is l o g. (Excuse me). And then we press Alpha to go back to the number register. And we’re going to put three and then they close parentheses or left parentheses, three times X and then the right parentheses. And I always show you what you’ve entered, you can change it if you want. I say it’s okay. Now in a log problem, you have to change it and ln(x) before you can integrate it so that’s this system here. In other words this goes in the log and you divide it by the log of the base, three. So then we’re into integration by parts and you mark all this stuff exactly as you see in your paper. Here’s the formula. VU minus the integral of VDU. And we keep working a problem multiplying things and whatever we have to do. And here’s the answer. Pretty neat, huh? everystepcalculus.com. Go to my site. You can see, I’m going to go back for just for a second. You can see all the things you can do solve for x or exponential form, logarithmic form, etc.

Line Integral r(t) Points Showing Work on TI-89-Video

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line integral

 

 

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Hello, everyone.  Tom from everystepcalculus.com and everystepphysics.com.  A problem on line integral, today from my menu.  And this is off the internet from Patrick JMT. So, he’s my favorite. Um, let’s do it.  Index 8 to get to my menu. And we gotta press second and the cursor down so you can go screen by screen to get to the L section.  Which is Line Integral, here.  And since they give us rt in the problem, you can see the problem on your screen. We choose RT from the menu.  And that was the, um.  Let’s go back and look at that menu, quick. Notice we have RT which is the original function XT plus YT plus ZT and then we have the magnitude of this of the derivative of that.  But in other words, here’s exactly what it is.  XT, YT, ZT which is the function there times the magnitude which is x prime of t squared, y prime of t squared, z prime of t squared.  Write this all on your paper, of course.  We’re going to enter our function.  You have to press Alpha before you enter anything into these entry lines.  You’re going to press Alpha 2 times x times y plus 3 times z.  Again, I show you what you’ve entered.  You can change it if you’ve made a mistake.  Looks pretty good.  Now we’re going to enter our lower point from the problem.  So you’re going to press Alpha 0, 0, 0. I say it’s okay.  Upper limit or point.  Alpha 2, 5, 4. Now notice that the farthest point, really really we’re taking so it’s different from 0 0 0 in the beginning point, we would subtract the beginning from the end point here. And then when you integrate this with respect to T, that’s where the whole system starts.  Then we get XT which is 2t, 5t, and 4t.  Then we start putting it together here.  Here’s the original function XT, YT, ZT  and the magnitude square root of 45. And then we, in integration, we always take the constants out of the integral before you work the integral.  So, we factored the inside, took the 4 out and the square root of 45 out and this is what is left here for integration.  So we’re going to integrate that, you know add 1 to the 2 here to get 3 and divide by 3. And so then we have this here which we’re going to do over the limit of 1 to 0.  Write all this down, this is exactly the way it’s done.  T equals 1 then.  Substitute 1 for all the t’s in the integral.  And so here, I use quotation marks here because of the calculator but you’re going to use parentheses when you put this on your paper. So you look like you know what you’re doing.  These calculations are exactly right.  We come up with 38 times the square root of 5. And then at t = 0 and here’s the replacements of t with all the zeroes in it and that of course equals zero.  We’re going to take the upper minus the lower which is 38 square root of 5 minus 0.  So the answer is 38 square root of 5.

Critical Points-Work Shown on TI-89-Video

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Transcripts

Hello, Tom from everystepcalculus.com,and everystepphysics.com. A problem in calculus dealing with critical numbers in critical points of a function. Let’s do it. Index 8 to get to my menu. Then your going to scroll down to critical point or numbers.I always have you start a graph on your paper. When you enter the function you have to press Alpha before you enter anything into these entry lines. Your going to press Alpha X cubed minus 3 times X. I always show you what you’ve entered. You can change it if you want. I say it’s okay. And we’re gonna choose number five, critical points.

Triple Integral: Work Shown on TI89-Video

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Screen shot 2015-02-04 at 12.32.17 PM

 

 

 

Transcripts

Hello again, everyone. This is Tom from everystepcalculus.com and everystepphysics.com. I’m going to do a Triple Integral for Calculus 3 right now. This is an example of a Patrick JMT, my favorite instructor on the internet, on YouTube. So I’m going to show you how it works on my program. I don’t know anybody can do that problem. He can do it because he’s a genius. But for us students,etc. How do we do it? Let’s get started. Index 8 to get to my menu. I’m going to scroll up because I can go to the bottom of the menu then instead of going down quicker to go to the T’s section. And we’re going to choose Triple Integral. And we’re going to enter our function. You have to press Alpha before you enter anything into these entry lines here in programs, okay. Alpha x times sin of Y. I always show you what you’ve entered. You can change it if you want. And we’re going to use the order of integration which is dx, dz, dy which is in the example. You have the other choices in case that’s given on test also. And we’re going to enter region q. Enter these limits. This is Alpha 0 for the x one Alpha square root of 4 minus z squared I made a mistake so I gotta go back. Choose number 2. Alpha 0, Alpha square root of 4 minus z squared. Close up the parentheses. That’s better. I say it’s okay. Next one for the y is Alpha 0. Alpha pi. That looks okay. and Alpha 0 for z. Alpha 2. That’s okay. So here’s what you write on your paper. The way you write it with triple integral with dx dz dy order of integration. Here’s the function in here. So you’re going to do the dx first and you put this over here with these lines. Showing you’r doing a range over this integration here. And here’s the integral of the first function okay. And if x equals the upper range. I show quotation marks here but you put you put parentheses in there. Because you’re substituting this amount for an X in the integral. And it equals this, minus sin, etc. And then we do the lower integral. X equals 0 and there’s 0 and you put parentheses around this instead of quotation marks, okay? And here’s the answer, you have the upper range minus the lower range equals this right here. So that becomes the new integration function. And I show you that here. dz dy is left, okay. So now we integrate that. Come up with this. Minus sin, etc. over this range here 0 2. Add z equals 2 Here’s the answer here. And z equals 0. Plug these in for all the Z’s in the problem. And the answer is this. 8, the upper range minus the lower range is 8 sin y divided by 3. Now we’re going to use that for the integration function. With the range of 0 and pi. At y equals pi minus 8 cosine is 8 thirds.