# Inverse operations

The exams had ended and there were no classes. Sara and Johnny had to wait for a week before their summer jobs would start. Sara asked Johnny if they could go for a stroll. Johnny agreed readily and they went to a nearby park. It was relatively quiet and they talked and talked about everything. When it was time to leave, Sara asked him to come to her home. At Sara’s home, they nibbled away at some snacks Nana had prepared for them.

Sara then suddenly asked Johnny if he knew that in math most operations have opposites. Johnny gave several examples like subtraction vs. addition, division vs, multiplication, sine of an angle vs. angle for sine inverse (arcsine) and so on. Then he asked her why she had brought this up.

Sara: We are going to take Calculus 2, and yesterday’s conversation with your mom reminded me that we should start working on it.

Johnny: I remember what you did to cheer her up. Mom was looking at a decrease in the number of hours she and dad together per year and you summed it over the years and showed how their total time spent with each other was increasing. That was clever. It was like saying how much total distance I rode my bike in a race rather than how fast.

# Calculus 1 – derivatives, Calculus 2 – anti-derivatives

Sara: Remember in Calculus 1 we learned about derivatives which gave you the rates. In Calculus 2, we are going learn about anti-derivatives. In Calculus, differentiation is about getting rates but integration is about getting anti-derivatives which is like summing up or getting areas under the curves.

Johnny: That’s it. You make this sound very simple. All you did yesterday was added the number of hours per year over each year to get the cumulative sums. You just wrote this series for different number of years and then just summed up. For example:

Cumulative hours in the year they meet and the next 4 years = Hours in year 0 + hours in year 1 + hours in year 2 + hours in year 3 + hours in year 4.

I saw this somewhere in algebra where they summed up a series and wrote it like:

Sara: You can do this for your bike race too, and also for just about anything. These are examples of summing up discrete values. Remember we talked about functions and continuity. Integration is more about summing up values of continuous functions to get areas under the curves. Do you remember, McLinton road?

Johnny: Yes, it was curved. The curvature of this road was y = x^{2}/10 and we said that meant the slope of the curvature was x/5 (see the story Dinner Date).

Sara: If we only knew the slope of the curvature, we could write the curvature equation to be

# y = ∫ slope dx = ∫ (x/5) dx = x^{2}/10 + constant

here ∫ is the symbol of integral or antiderivative. So in differentiation we went from the curve to the slope and in integration we go backwards. The only thing is that we don’t know if there was some slope even at x=0 and this rate is just adding to that slope. So we added a constant to the integral.

Johnny: I also remember that v = ds/dt where v is the velocity and s is the distance.

Sara; That means you could also write s = ∫vdt. Here the symbol ∫ is for an integral which is the same as an anti-derivative. However, there is one problem. This type of an integrals are indefinite integrals because we didn’t fix any time period. This integral includes all the distance cycled unless we mention otherwise. Remember the derivative of a constant is zero.

Therefore, for an indefinite integral we write

s = ∫vdt + c where is a constant.

Johnny: Okay, I get it. From what we did in Calculus 1, we can also say what to do in Calculus 2.

dx^{3}/dx = 3x^{2}. Therefore ∫3x^{2 }= x^{3 }+ c,

d(ax^{2}+bx+c)/dx = 2ax + b. Therefore ∫(2ax + b)dx = ax^{2}+bx+c,

d (sin x)/dx = cos x. Therefore ∫(cos x)dx = sin x + c.

I could whiz by all of this. Is that all?

Sara: Lover boy, you are sharp. Give me a kiss.

Johnny gave her a kiss and then asked Sara: You said that these were indefinite integrals, are there any other kind?

Sara: I see you were listening. Yes, there also the definite integrals.

After that he had to leave because his dad wanted him home. Sara also had a gabbing session with Nana. Later she told her mom about her conversation the previous day with Johnny’s mom.

Johnny called Sara again the next day. Hey, there was no school! Johnny went over to Sara’s home.

Sara said: Let us say you were bike riding on McLinton road and you wanted to know how much North you went when you travelled West from 5 km to 10 km. Then you will use the definite integrals. Remember, for indefinite integral we wrote

y = ∫ slope dx = ∫ (x/5) dx = x^{2}/10 + c

# For definite integral you will write

In this the term c will cancel out so that it will become (x^{2}/10 at x=10) – (x^{2}/10 at x=5)

which is 100/10 – 25/10 = 7.5 kilometers (see picture).

Johnny: Wow, I got to check it out with my bike. Remember, with the bike I can measure Westward and Northward distances. I bet you the value of this definite integral is much higher for the next 5 kilometer.

This is neat. I will check the integrals for a few more functions.

Sara: We thought of the integral only to determine distance covered over time if we knew the speed. Integration can also be used for determining area under a curve, volume of a cube, sphere, cone, cylinder etc. I guess we will see many more examples of the applications of integrals in our course.

That ended the day and the two went home.

*Calculus Joke*

John and Jim were in a restaurant when they started to argue. John said, “Most people know enough Math to get by” but Jim disagreed.

After a short time Jim went to the washroom. John called the waitress, gave her a ten dollar tip and told her, “When my friend comes, I will ask you a question and just say x cube.”

Jim came back. John called the waitress and asked, “What is the integral of three x square?”

The waitress said, “x cube” and started walking away.

After a few steps she looked back, smiled and said, “Plus a constant!”

*Challenge*

Johnny can ride his bike with the velocity V(t) = 1 + 4t + 10t^{2} meters/minute where t is the time from his start. How much distance can he cover in first 10 minutes?

*Solution*

V(t) = ds/dt = 1 + 4t +10t^{2} meters/minute.

The distance covered in first 10 minutes will be

=(10+200+10000) – 0 =10210 meters or 10.21 kilometers.