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The foe’s Entreeng’s defense has risen! The foe used an INFORMATION! People are entering and leaving the gym at the same time 24 hours a day. The rate at which people enter is given by the function: On the other hand, the rate at which people exit is given by:
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[object Object],[object Object],First we must take the integral of R(t) from 0 to 18 since we are only looking for how many people have entered at 6pm. Since people start entering at 0 hours, we can assume that 6pm will be equivalent to 18 (Simply think of it as following the 24 hour clock.
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[object Object],[object Object],Assuming the fact that we’re not allowed to use a calculator, we will need to integrate by hand. Remember the  First Fundamental Theorem of Calculus , which states that the integral of a derivative = total change in the parent function. So anti differentiating R(t) should give us: Which simplifies to:
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[object Object],[object Object],Derivee used SUBSTITUTION! Its super effective! Entreeng takes 40 damage!
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[object Object],[object Object],In order to find how many people are in the gym at 4am, we need to integrate the difference between the rate at which people are entering and the rate at which people are leaving from 0 to 4. Normally, you would be given a calculator to solve these types of questions, but we’ll go through the steps in anti differentiating the fraction only since the steps have already been shown in part a.
[object Object],[object Object]
[object Object],[object Object]
Derivee used ANTI DIFFERENTIATE! REMINDER: We’re only showing the anti differentiation of the fraction in this part. Start by using the “u substitution” technique by taking the most complicated part of the fraction and letting it equal “u”. Then take the derivative so that we can find a way to substitute out the dt, in order to make everything in terms of “u”.
We can’t continue with this alone since not everything is in terms of “u”. As you can see, we still have a “t” left in the numerator. First let’s bring out the 15 so that we just have “t” at the top. To get rid of the “t”, first remember in our first step that we made “u = 1 + 3t”. From here, we can solve for “t”, leaving us with an integral in terms of “u”.
Now substitute in the newly found value of “t”. Simplify a bit.. We can now start anti differentiating the function in order to get our parent function to work with.
Let’s just take a look at the integral by itself and worry about the outside fraction later. This is actually pretty easy to anti diffirentiate. The hardest part being the fraction since some confuse its derivative/anti derivative pairing. Now we can bring our focus back to the whole thing.
Let’s just take a look at the integral by itself and worry about the outside fraction later. This is actually pretty easy to anti diffirentiate. The hardest part being the fraction since some confuse its derivative/anti derivative pairing. Let’s not forget the fraction we had on the outside of the integral.
And there’s your anti derivative! Now we can get back on track to the original question, which was to find how many people are in the park at 4 am. Remember what we found out originally: REMINDER: We’re subtracting the exiting function from the entering one and integrating since that represents how many people are currently in the gym. Now let’s apply the fundamental theorem of calculus since we know the anti derivatives, or parent functions, of each.
Solving for that using a calculator, or if you have the time to do it by hand, we should come up with an answer of 48.6082. Since we are counting in terms of people, we will round down to 48 since you can’t have 48 and about a half people. ∴   There are 48 people in the gym at 4 am.
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[object Object],[object Object]
[object Object],This question is very similar to the last one. The only difference is that you may have to think of this question in terms of parts and then putting everything together.
Let’s take a look at the first part of the integration first and then try to make some sense of what is going on in that specific part. Reading the question again, we are told that people are continuously entering from 12 am to 3pm and then from 7 pm to 8 pm. We are also told that people are continuously leaving along the whole time interval of 12 am to 8 pm. Converting the times in terms of our interval, we’ll see that from 12 am to 3 pm is 0 to 15. Since there are still people entering, we end up with the integral above.
Moving along we have: This part represents the period of time in which people are no longer entering the gym. Knowing this, we only need to integrate along the specific time interval and only the exiting function. We end up subtracting this integral from the previous one since this represents the people who have left the gym.
Lastly we have: This integral represents the part at which trainers have resumed entering the gym, which is basically the same procedure as the first part but a different interval since this is the second period of time trainers are entering. Also keep in mind that you are adding this to the previous two since this represents the total people in the specific time interval.
This question definitely wouldn’t require you to do manually. If for some strange reason that does happen though, you already know how to apply the fundamental theorem of calculus and we’ve already shown how to anti differentiate both functions. So using a calculator we should come with an answer of approximately 963.1848. Like the previous question stated, we can’t have a decimal of a person, so let’s round down to 963. ∴  There are 963 people in the gym at 8 pm . Gym Leader Pat Wa is a busy guy o_o
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Gym Leader - Pat Wa

  • 1.  
  • 2.
  • 3.
  • 4. The foe’s Entreeng’s defense has risen! The foe used an INFORMATION! People are entering and leaving the gym at the same time 24 hours a day. The rate at which people enter is given by the function: On the other hand, the rate at which people exit is given by:
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19.
  • 20. Derivee used ANTI DIFFERENTIATE! REMINDER: We’re only showing the anti differentiation of the fraction in this part. Start by using the “u substitution” technique by taking the most complicated part of the fraction and letting it equal “u”. Then take the derivative so that we can find a way to substitute out the dt, in order to make everything in terms of “u”.
  • 21. We can’t continue with this alone since not everything is in terms of “u”. As you can see, we still have a “t” left in the numerator. First let’s bring out the 15 so that we just have “t” at the top. To get rid of the “t”, first remember in our first step that we made “u = 1 + 3t”. From here, we can solve for “t”, leaving us with an integral in terms of “u”.
  • 22. Now substitute in the newly found value of “t”. Simplify a bit.. We can now start anti differentiating the function in order to get our parent function to work with.
  • 23. Let’s just take a look at the integral by itself and worry about the outside fraction later. This is actually pretty easy to anti diffirentiate. The hardest part being the fraction since some confuse its derivative/anti derivative pairing. Now we can bring our focus back to the whole thing.
  • 24. Let’s just take a look at the integral by itself and worry about the outside fraction later. This is actually pretty easy to anti diffirentiate. The hardest part being the fraction since some confuse its derivative/anti derivative pairing. Let’s not forget the fraction we had on the outside of the integral.
  • 25. And there’s your anti derivative! Now we can get back on track to the original question, which was to find how many people are in the park at 4 am. Remember what we found out originally: REMINDER: We’re subtracting the exiting function from the entering one and integrating since that represents how many people are currently in the gym. Now let’s apply the fundamental theorem of calculus since we know the anti derivatives, or parent functions, of each.
  • 26. Solving for that using a calculator, or if you have the time to do it by hand, we should come up with an answer of 48.6082. Since we are counting in terms of people, we will round down to 48 since you can’t have 48 and about a half people. ∴ There are 48 people in the gym at 4 am.
  • 27.
  • 28.
  • 29.
  • 30.
  • 31.
  • 32. Let’s take a look at the first part of the integration first and then try to make some sense of what is going on in that specific part. Reading the question again, we are told that people are continuously entering from 12 am to 3pm and then from 7 pm to 8 pm. We are also told that people are continuously leaving along the whole time interval of 12 am to 8 pm. Converting the times in terms of our interval, we’ll see that from 12 am to 3 pm is 0 to 15. Since there are still people entering, we end up with the integral above.
  • 33. Moving along we have: This part represents the period of time in which people are no longer entering the gym. Knowing this, we only need to integrate along the specific time interval and only the exiting function. We end up subtracting this integral from the previous one since this represents the people who have left the gym.
  • 34. Lastly we have: This integral represents the part at which trainers have resumed entering the gym, which is basically the same procedure as the first part but a different interval since this is the second period of time trainers are entering. Also keep in mind that you are adding this to the previous two since this represents the total people in the specific time interval.
  • 35. This question definitely wouldn’t require you to do manually. If for some strange reason that does happen though, you already know how to apply the fundamental theorem of calculus and we’ve already shown how to anti differentiate both functions. So using a calculator we should come with an answer of approximately 963.1848. Like the previous question stated, we can’t have a decimal of a person, so let’s round down to 963. ∴ There are 963 people in the gym at 8 pm . Gym Leader Pat Wa is a busy guy o_o
  • 36.