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A). In order to write the acceleration equation, consider the relationship between acceleration and velocity. Then, think about what time the acceleration would be smallest.

B). How do upward and downward movement relate to rightward and leftward movement?

C). Average acceleration is the change in velocity over a time interval. What does this say about average velocity?

D). Recall the relationship between velocity and position to write your position equation?

E). Refer to 4.3 Slide 5 (f) … Remember to consider changes in velocity when writing integrals. Also keep in mind: Can “total distance” include negatives?

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Look at 4.7 page 1 to understand the relationship between g(x) and f(x)

Draw a number line of g’(x)

When looking at a number line of the first derivative what does a maximum look like?

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Remember that g’(x)= f(x)

Remember that the point of inflection of g(x) is when g’’(x)changes signs, so what does that mean for the slopes of g’(x)?

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Use “CCC” to find absolute minimum

1. Find Critical points

2. Compare values of v at critical points and end points

3. Write a Conclusion

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Using the given information, come up with a function rule using an integral.

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Think about which theorem would apply to this question and use TUNE!

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A): Refer to 4.2 Slide 2 #1

B). Consider the difference between “total distance” and the value you compute when using an integral. Can “total distance” include negatives? How can you change the integral to get “total distance” as opposed to “net change?”

C). Recall the factors that relate to increasing/decreasing speed, and how the relationship of those factors determines the effect on speed.

D). In order to determine position remember you have to remember two important factors:

1. Your starting point

2. How does position relate to velocity? How can you use that relationship to write an integral that answers the question?

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Think about the relationship between the sign of the velocity and acceleration on this interval

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Think about what does the graph of v(t) would have to look like in order for a(t)<0

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Remember your integral rule:

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Think about the sign (+ or -) of f' and what it means in relation to the graph of f. Drawing a number line could also help.

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Think about which trig functions have the derivatives of sin(2x) and cos(2x). You may have to add a coefficient to the antiderivative to ensure its derivative would cancel out to have no coefficient.

Also, remember the following integral rule:

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Remember, the antiderivative of a velocity function is position.

Antiderivatives include a “+C.” You are given the value of “C” for this problem in the question - read carefully.

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a) You can find the antiderivative of this velocity function by finding the total area between the function and the x-axis. Think about what the antiderivative of a velocity function represents.

b)Think about what makes a function non-differentiable. Also, remember that v’(t) is the slope of v(t).

c)To find a(t), apply the same thinking as you did in part b. (See  image below if you need help setting up a piecewise function)

d)

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Method 2

Refer to video:

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- [Instructor] What we're going to do in this video is start thinking about the position of an object traveling in one dimension. And to get our bearings there, I'm going to introduce a few ideas. So the first idea is that of displacement. So you might use that word in everyday language, and it literally means your change in position, your change in position. Now a related idea that sometimes get confused with displacement is a notion of distance traveled. You might say well doesn't that, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path, total length of path. So what are we talking about? Well let's say, and we're going to introduce a little bit of calculus now, let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget well that too can be a vector quantity. In fact this velocity is a vector quantity because you could think of it if it's positive it's moving to the right, and if it's negative it's moving to the left. So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it, or you will see it bolded, or you will see it bolded like that. I like to write an arrow in, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like, and I did that ahead of time. So you can see here, at time equals zero, let's say time is in seconds, and our velocity's in meters per second. So this is meters per second right over here, and this is seconds on this axis. At exactly time zero, this object is traveling at five meters per second. We can say to the right, it has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate, so five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things. First let's think about what is the displacement over the first five seconds, over first five seconds? Well we've seen already multiple times, if you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we can take the integral from zero to five, zero to five, of our velocity function, of our velocity function. Just like that. And we can even calculate this really fast. That would just be this area right over here, which we can just use a little bit of geometry. This is a five by five triangle, so five times five is 25, times 1/2, remember area of a triangle's 1/2 base times height. So this is going to be 12.5, and let's see this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over, over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there. So this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is gonna be negative area. And in fact this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying how can that be? After 10 seconds how do we, what why is our displacement only zero meters? This particle's been moving the entire time. Well remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, it's velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. So the particle has gone over 10 seconds 12.5 meters to the right and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed. Now you might start, you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute function, which is what the absolute value function does, in one dimension. So what would this look like if we plotted it? Well the absolute value of the velocity function would just look like that. And so if you want the distance, you would find, the distance traveled I should say, you would find the integral over the appropriate change in time of the speed function right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say I'll write it out, distance traveled over first five seconds, first five seconds, what would it be? Well it would be the integral from zero to five of the absolute value of our velocity function, which is you can just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent. Well that's because you have in this case the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it. Well that's gonna be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what? It's going to be this area plus this area right over here. So plus this area right over here. And so this is going to five times five times 1/2 plus five times five times 1/2, which is going to be 25 meters. The particle has gone 12.5 meters to the right, and then it goes back 12.5 meters to the left. Your displacement, your net change in position is a zero, but the total length of path traveled is 25 meters.

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Motion problems with integrals: displacement vs. distance (video) | Khan Academy

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Instructor] What we're going to do in this video is start thinking about the position of an object traveling in one dimension. And to get our bearings there, I'm going to introduce a few ideas. So the first idea is that of displacement.

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First try to get rid of the denominator. This will make it easier to take the antiderivative.

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Make sure to know your trig functions! Since each term is separated by an addition/subtraction sign you need to find the antiderivative of each term.

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Like #11 we must know our trig functions! In order to make this problem simpler try to distribute everything you can. What do we get once you multiply out the sinx? Once you have your two new terms you will be able to take the antiderivative of this easier.

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a) For g(3) remember that g is equal to the integral of f.  You will then be able to figure this out much easier. For g’(3) this is easier than you think! What does g’(x) equal in terms of f (x)? Once you understand what g’(x) equals in terms of f(x) you will be able to take this one step further! What do we do to get from g’(x) to g’’(x)?

b) Remember that we are trying to find the rate of change of g and not f! Use part a and the graph to help you complete this problem. The average rate of change is just looking at the difference between two points.

c) Use the value we found in part b to help you with this question. Remember we want to use that value to see where g’(c)=f(c). Look for points of intersection and make an equation out of your value in part b!

d) Remember that we are trying to find the points of inflection of g and not f. This means that we are trying to find when g”(x) does what? Use part a to guide you in going from the graph of f to g’’(x).

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Break down the fractions as much as possible using your knowledge of trig and trig reciprocals before finding the respective integrals.

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Use the law of integration (see attached photo or refer to lesson 4.6 slide 1) to separate the definite integral into two separate definite integrals .*Note the subtraction sign*

Note that the graph to the right is f(x). Think about the connection between the integral of f(x) and the graph of f(x)

utilize the antiderivative method from 4.5, OR draw a graph and find the area to find definite integrals other than f(x)

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Use rules from law of integration (as seen below) to separate the definite integral into definite integrals of single functions and remember their limits of integration are important.

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1. Start by splitting up the integral into 2 pieces that are connected by subtraction
2. For the first integral, find the antiderivative.  For the second, graph the function and use the net area method.  (Refer to question 15 on 4.5 for reference).
3. Solve.

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1. Foil completely
2. Use the power rule for antiderivatives

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1. Think about how you could get rid of the absolute value sign.
2. If you are unable to do that then watch this video to try and get you going in the right direction.
3. Be careful with your signs.

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Remember that the acceleration of the particle is the slope of the velocity. How could e^1-t be rewritten to make taking the derivative easier?

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Remember that speed and velocity are not the same. Refer to 4.1, slide 7 to see how the values of the velocity and acceleration affect the speed of an object.

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What property would change sign when the displacement changes direction? What would the value of this property be when the property changes sign? Set this equation equal to that value to find the time at which the particle changes direction. Make a number line and plug in values to make sure the critical point(s) results in a sign change at that point.

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How is distance different from displacement? How should the definite integrals be separated given what we learned in question 48c? Since the derivative of the displacement is the velocity, the antiderivative of the velocity is the displacement. To find the distance, the integrals would have to be separated into separate integrals. Refer to 4.6, slide 6, question 10 to see how to find the distance from a velocity function.

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a)

Note that the graph given is f'

Look at lesson 2.1 for a refresher on critical values of a function

Draw a number line to find intervals where f' is increasing/decreasing

b)

Look at lessons 2.2/2.3 for a refresher on inflection points

Utilize the intervals on your number line from part a) and the graph of f'. Where on the graph of f' is f"<0 and f'<0?

c)

Look at lessons 2.2/2.3 for a refresher on inflection points. Think about how concavity of f is shown on a graph of f' *remember at an inflection point there is a change in something*

d)

How can you find f(x) from a graph of f' using integrals when given a point of f(x). Refer to lesson 4.8 slides 1&2

Once you have an integral of a function, remember that a definite integral of a function is the area under the graph between the interval

to find f(4) and f(-2)

plug in the x=4 and x=2 into the expression of f(x) that you found. Think about the meaning of the integral in our expression. How can we use the given areas of the regions under the graph of the intervals [-2,1] and [1,4]?

*Look carefully at what numbers are in the top and bottom of the integral sign*

**refer to the law of integration below (also found in lesson 4.6)**

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a. Think about where relative extremum occur (critical points).  One of the functions must change signs at the critical points for a min and max to exist.  (refer to 4.7 questions 31-33 for reference).

b. Think about what f' and f'' determine in the graph. (refer to 2.1-2.5 Review 1, question 19 for a similar question).

c.  Determine what g'(x) is equal to in terms of f(x).  Then, find critical points for the function that equals g'(x).

d. Watch video linked below for a refresher on finding points of inflection.  Think about what g''(x) is equal to in terms of f(x).

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If you were paying close attention in the last video, an interesting question might have popped up in your brain. We have talked about the intervals over which the function is concave downwards. And then we talked about the interval over which the function is concave upwards. But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing and then the slope started increasing. So that's one way to look at it. Right here in our function we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name you're probably thinking. And you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has a extrema point, or the point at which our second derivative switches signs like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u. Or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive. But we could have also switched from being positive to negative. So inflection point, your second derivative f prime prime of x switches signs. Goes from being positive to negative or negative to positive. Switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards, like that, then this inflection point up until that point, the slope was increasing. So the second derivative would to be positive. And then the slope is decreasing, so your second derivative would be negative. So here your second derivative is going from positive to negative. Here your second derivative is going from negative to positive. In either case, you are talking about an inflection point.

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Inflection points introduction (video) | Khan Academy

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Inflection points are points where the function changes concavity, i.e. from being "concave up" to being "concave down" or vice versa. They can be found by considering where the second derivative changes signs. In similar to critical points in the first derivative, inflection points will occur when the second derivative is either zero or undefined.

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1. Use the law of integration(posted below) to split the integration into two separate integrals; connect them by subtraction

2. Draw the a graph of y = sqrt(25-x^2) to help find the total area from the integral 0 to 5 of sqrt(25-x^2)dx

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1. refer to 4.6 Properties of Integrals where the Law of Integrations are listed

2. find the area of f(x) from (-1,2)

3. refer to 4.4 Antiderivatives if you need help finding the antiderivative of 3x^2

4. plug in 2 and -1, then subtract

5. Solve

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A) Make a velocity number line
B) Use your velocity number line that you drew in part A and test all critical points (including endpoints because it is on a closed interval).
C) Remember that you are taking into account TOTAL distance so you're going to have to use absolute value. Also, use your values in part B to help answer part C so you do not have to redo any of your previous work.
D) To find a(t), use the formula for rate of change of v(t) from the interval (7,10). If still stuck, refer to 4.3, slide number 3, question 5a to help find the acceleration equation. To find v(t), remember that velocity is the antiderivative of acceleration. Use the power rule for antiderivatives to get v(t) and solve for C with the points given by the velocity graph.
To find x(t), consider that you only have to find the squirrel's distance starting from the t=7 because the interval is from 7 < t < 10. Then, position is the antiderivative of velocity, so use the v(t) equation you had just found earlier to form an integral and add that to the squirrel's position when t=7.

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Split up the integral into two parts (see the image below)

Find the critical points of the inside of absolute value, negate the negative parts of the equation to always have positive value

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8c/v/definite-integral-of-absolute-value

Draw a possible graph that would resemble f(-x) = -f(x). How will this impact the area under the curve?

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A) Remember that in this scenario g’(x) = f(x)
B) Draw a g’ number line to find when g is positive/negative
C) Use the 3 C’s checklist on 4.7 Slide 4 to accurately justify your answer
D) Recall that the point of inflection in a formula is when its second derivative changes signs. f’(x) = g”(x) as well

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- Remember a(t) = v’(t)
- Use the equation v(b)-v(a)/ b-a to get the acceleration of the bike

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V(t)= derivative of the position function

Use the Khan Academy video for a refresher on finding definite integrals using area formulas (start at 1:06 end at 3:30)

Remember that the absolute value of something is always positive

See the video lesson in the following post for a refresher on distance vs. displacement

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- [Instructor] We're told to find the following integrals, and we're given the graph of f right over here. So this first one is the definite integral from negative six to negative two of f of x dx. Pause this video and see if you can figure this one out from this graph. All right we're going from x equals negative six to x equals negative two, and the definite integral is going to be the area below our graph and above the x-axis. So it's going to be this area right over here. And how do we figure that out? Well this is a semicircle, and we know how to find the area of a circle if we know its radius. And this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared. So it would be pi times our radius which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only 1/2 the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out. All right let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the, below the function and above the x-axis. But here the function is below the x-axis. And so what we can do is, we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here? Well there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid or you can just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1/2. So this has an area of one. This rectangle right over here has an area of two times one, so it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two, so it's one times two times 1/2. Remember the area of a triangle is 1/2 base times height. So it's one. So if you add up those areas, one plus two plus one is four, and so you might be tempted to say oh is this going to be equal to four? But remember our function is below the x-axis here, and so this is going to be a negative four. All right let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out. So we're gonna go from here to here, and so it's gonna be this area right over there. So how do we figure that out? Well it's just the formula for the area of a triangle, base times height times 1/2. So or you could say 1/2 times our base, which is a length of, see we have a base of three right over here, go from one to four, so 1/2 times three times our height, which is one, two, three, four, times four. Well this is just going to get us six. All right last but not least, if we are going from four to six of f of x dx. So that's going to be this area right over here, but we have to be careful. Our function is below the x-axis, so we'll figure out this area and then it's going to be negative. So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared. That would be the area if we went all the way around like that, but this is only half of the circle, so divided by two. And since this area is above the function and below the x-axis, it's going to be negative. So this is going to be equal to negative pi over two. And we are done.

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Finding definite integrals using area formulas (video) | Khan Academy

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Instructor] We're told to find the following integrals, and we're given the graph of f right over here. So this first one is the definite integral from negative six to negative two of f of x dx. Pause this video and see if you can figure this one out from this graph.

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Here is the video lesson on distance vs. displacement:

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Get access to files anywhere through secure cloud storage and file backup for your photos, videos, files and more with Google Drive.

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- When does the velocity change direction? **Think negative or positive**
- Remember when the velocity is 0, that means Caren is not moving
--> This question is asking when Caren turns around not when she stops moving

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You need to find both Caren and Larry’s displacement in miles from the school

For Larry:

When finding the integral, don’t forget to multiply by the reciprocal of the “inside part” of the trig functions in order to make up for the chain rule

For Caren:

This is similar to b) but remember that you don’t use absolute values when finding displacement

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Get access to files anywhere through secure cloud storage and file backup for your photos, videos, files and more with Google Drive.

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Remember that the x^3 is in the denominator. Keep in mind how this will effect the antiderivative and your answer. Pay attention to signs!

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a) Keep in mind the area is below the x-axis. Remember that the graph of f is equal to the graph of g'. Take into account differentiability.
b) Remember a point of inflection is where the slope of g' changes signs. Keep in mind g'=f.
c)  Pay attention to the placement of x in the integral. Look for points where h(x)=0. Make a table with points on the graph and find the integrals where the areas will cancel out.

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Remember the relationship that exists between velocity and acceleration. When taking the derivative of an expression with "e" in it do not forget to use the chain rule.

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Find the velocity at t=3 and compare it to acceleration at t=3. Remember what it means when something is speeding up, are velocity and acceleration in the same direction or are they in opposite directions?

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A critical point of the function v(t) does not always signify a change in direction, use a number line to see when the particle changes direction. Remember to explain your answer using words!

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Method 1:
To refresh your memory about working from velocity to position, refer to 4.3 Question 6. Draw a wiggle diagram to help find the total distance traveled by the particle over the time interval [0,3].

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d)

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Keep in mind the inverse trig functions (ArcSin, ArcCos, ArcTan)

Break down, simplify, and separate any fractions that you can

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multiply all aspects and simplify it prior to taking the integral

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a) after finding your velocity equation for particle P, make a number line
b) place the P and Q velocity timelines on top of each other and use your knowledge of the relationship between velocity signs and direction.
c) think about how the relationship between velocity and acceleration affects a particles change in speed
d) There are 2 ways to go about this problem

1. think about ways to create a position equation and find the constant if one exists

2. set up your integral with the given constant and plug in what we know

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Simplify the quadratic equation using division

1. (Use long division or synthetic division)

1. If you need help on how to long divide: https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/v/polynomial-division

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1. With subtraction, we can separate the two components of the integral and find the antiderivative of them separately

2. If your stuck on how to find the antiderivative of the second component, think about the anti-derivatives of trig functions

1. If you don’t remember, look at 3.4 ⇒  pages 1-4 (specifically question 13)

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a) Look for particular sign change in f': Note this is a local minimum and the graph shown is of f'. (Refer to lessons 2.1-2.3 on Behaviors of Functions)

b)
- Think about where critical points and endpoints are (Lesson 2.1)
- For the minimum, think of where  f' changes negative to positive,
- For the maximum, think of where f' changes from positive to negative
- Remember laws of integration, and how this may affect the equation when solving (Refer to lesson 4.6)

c) Refer to Lesson 2.2 on concavity and Lesson 4.2 on particle motion to understand rules and behaviors of functions.

d) Remember to use Chain Rule (Refer to Lesson 1.5 - Chain Rule)

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A)

1. Hint: the graph of f(x) should be a piecewise function divided by f>0 and f<0

2. Refer to 1.1 for a refresher on the derivative formula

3. Determine derivative as h approaches 0 from the left side

4. Determine the derivative as h approaches 0 from the right side

5. Determine if the derivatives from the left and right side of 0 are equal

B)

1. Hint: The average rate of change of f is equal to 0 when the change in y is equal to zero

2. Find where the change in y = 0 on the graph of f’(x)

3. Determine the amount of points where the change in y = 0 on the graph of f’(x)

C)

1.  Hint: the interval of (-4,6) can be divided into a piecewise functions

2. Refer to 2.7 to see the Mean Value Theorem

3. Input intervals where the average rate of change is 0 by using the mean value theorem

4. Find the interval where f’(c) = ⅓ (the given output of the mean value theorem)

D)

1.  Hint: if g’ = f, then g’’ = f’

2. Refer to 2.2 Concavity

3. Recall that a graph is concave up when its second derivative is increasing.

4. Determine the intervals where the function g(x) is concave up using concepts form 2.2

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1. Part A.

1. Open the link, Velocity is the speed of something in a given direction. Plug in a positive and negative velocity to see the difference.

1. https://phet.colorado.edu/sims/cheerpj/moving-man/latest/moving-man.html?simulation=moving-man

2. Recognize the relationship between velocity and direction.

1. If the particle moves right, velocity is ________. When the particle moves left velocity is ________.

3. Use a number line to find when the particle moves to the left

2. Part B

1. There is a difference between total distance and what the integral gives you. Can total distance include negatives? How can you get the integral to include total distance rather than displacement?

3. Part C

1. Think about the relationship between our velocity equation and acceleration equation. We learned these on 4.1. On the bottom of page 4 of 4.1. You should have these relationships.

2. How do we know when a particle is spending up or slowing down? If you're unsure, go back to the moving man link given in part a).

1. Acceleration: When positive we are pushing the man to the right. When negative we are pushing him to the left.

2. Speed: absolute value of velocity/ directionless velocity

3. Speeding up= signs of velocity and acceleration are the same

4. Slowing down= songs of velocity and acceleration are different

3. It is recommended to make a velocity number line and acceleration number line to compare the signs.

4. Part D

1. Recognize the relationship between velocity and position (also on bottom of 4.1)

1. Furthermore, take the antiderivative of your velocity equation. Remember that you need to multiply by the inverse of the constant attached to “t”). Remember to use proper integral notation.

2. A time and position is given to us, using this we can find a numerical value for “c” in our antiderivative equation.

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Split the question into g(x) and -2f(x)

Do not forget to add values for the split integrals at the end

**QUESTION 36 IN LESSON 4.5 WILL BE OF USE

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a)

Steps:

1. Plug 3 into g(x)

2. Find the area in between the graph and the x-axis from -3 to 3

b)

HINT: f is the derivative of g… In order to find when f is increasing what does the derivative graph have to look like? and in order to find when f is concave down what does the derivative graph have to be doing?

c)

Steps:

1. find h’(x) by finding the derivative of g(x)/5x (Quotient Rule!)

2. Plug in 3 for x

3. Solve

d)

Steps:

1. Find p’(x) (Chain Rule!)

2. Plug in -1 for x

3. Solve