Getting the "1/2" in int[0,t][v_0 + at] dt = v_0 + (1/2)at^2 the calculus way!

[prepforcalc]

Getting the "1/2" in int[0,t][v_0 + at] dt = v_0 + (1/2)at^2 the calculus way!

I attached a screenshot. Would someone remind me of the calculus step or rule that allows one to go from the upper line to the lower line and get 1/2? The 1/2 part is all that I'm confused about.

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. . . . .\(\displaystyle \large{ \displaystyle \int_{s_0}^s\, ds\,\, \int_0^t\, (v_0\, +\, at)\, dt }\)

. . . . .\(\displaystyle \large{ s\, -\, s_0\, =\, v_0t\, +\, \frac{1}{2}at^2 }\)

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PS - I already know how to do this the non-calculus way. Per my username, I'm sharpening my calculus skills. Thanks.

 

[Dr.Peterson]

I attached a screenshot. Would someone remind me of the calculus step or rule that allows one to go from the upper line to the lower line and get 1/2? The 1/2 part is all that I'm confused about.

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. . . . .\(\displaystyle \large{ \displaystyle \int_{s_0}^s\, ds\,\, \int_0^t\, (v_0\, +\, at)\, dt }\)

. . . . .\(\displaystyle \large{ s\, -\, s_0\, =\, v_0t\, +\, \frac{1}{2}at^2 }\)

---

PS - I already know how to do this the non-calculus way. Per my username, I'm sharpening my calculus skills. Thanks.

What is \(\displaystyle \int t \; dt\)?

Or, to check it by going backward, what is \(\displaystyle \frac{d}{dt}\left(\frac{1}{2}at^2\right)\) ?

 

[prepforcalc]

What is \(\displaystyle \int t \; dt\)?

Or, to check it by going backward, what is \(\displaystyle \frac{d}{dt}\left(\frac{1}{2}at^2\right)\) ?

Why isn't it 1/3at^3?

We start with t^2, and 2+1=3, so that makes it 1/3at^3, right? I mean, I know I'm wrong but just not sure why.

 

[Dr.Peterson]

Why isn't it 1/3at^3?

We start with t^2, and 2+1=3, so that makes it 1/3at^3, right? I mean, I know I'm wrong but just not sure why.

That's the question you should have asked in the first place: now we have an idea of what your own thinking is that needs correction.

But ... why do you say this? It looks like you're integrating t^2, not t!

To integrate t, which is your goal, use the same rule you're talking about, but apply it to t^1, not to t^2.

And if you're checking their result of 1/2 t^2, you have to differentiate that (working backward), not integrate it. What is the derivative of this? It's t, which shows that the integral of t is 1/2 t^2.

 

[prepforcalc]

That's the question you should have asked in the first place: now we have an idea of what your own thinking is that needs correction.

But ... why do you say this? It looks like you're integrating t^2, not t!

To integrate t, which is your goal, use the same rule you're talking about, but apply it to t^1, not to t^2.

And if you're checking their result of 1/2 t^2, you have to differentiate that (working backward), not integrate it. What is the derivative of this? It's t, which shows that the integral of t is 1/2 t^2.

The t from dt isn't involved in integrating (duh, sorry to overlook something so obvious). So, we have v(dt) + at(dt). Then the bold t's exponent of 1 gets another 1 added to it, and 1/2 in front of it. So then we have, v(dt) + (1/2)a(t^2)(dt). This almost works, except the bold and italic stuff seems to come out to t^3.

 

[Dr.Peterson]

The t from dt isn't involved in integrating (duh, sorry to overlook something so obvious). So, we have v(dt) + at(dt). Then the bold t's exponent of 1 gets another 1 added to it, and 1/2 in front of it. So then we have, v(dt) + (1/2)a(t^2)(dt). This almost works, except the bold and italic stuff seems to come out to t^3.

How does it come out to t^3? Can you show what you're thinking there?

Once you've changed t to 1/2 t^2, you have done the integration -- there is no dt left, and no integration to do. What you've shown, written correctly, is that [FONT=MathJax_Size1]∫ [/FONT][FONT=MathJax_Math-italic]at [/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t = 1/2 at². Period! No dt. That's exactly what you want, for that part of it. Similarly, [/FONT][FONT=MathJax_Size1]∫[/FONT][FONT=MathJax_Math-italic]v[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t = v [/FONT][FONT=MathJax_Size1]∫[/FONT][FONT=MathJax_Math-italic]1d[/FONT][FONT=MathJax_Math-italic]t = vt.[/FONT]

 

[prepforcalc]

How does it come out to t^3? Can you show what you're thinking there?

Once you've changed t to 1/2 t^2, you have done the integration -- there is no dt left, and no integration to do. What you've shown, written correctly, is that [FONT=MathJax_Size1]∫ [/FONT][FONT=MathJax_Math-italic]at [/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t = 1/2 at². Period! No dt. That's exactly what you want, for that part of it. Similarly, [/FONT][FONT=MathJax_Size1]∫[/FONT][FONT=MathJax_Math-italic]v[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t = v [/FONT][FONT=MathJax_Size1]∫[/FONT][FONT=MathJax_Math-italic]1d[/FONT][FONT=MathJax_Math-italic]t = vt.[/FONT]

Right, the dt doesn't result in a t! Please excuse me, I'm very new to calculus. In fact the t in vt, comes from integration, as do all of the other t's in this equation! Please let me know if I'm wrong, but I think I've actually got it.

 

[Dr.Peterson]

Right, the dt doesn't result in a t! Please excuse me, I'm very new to calculus. In fact the t in vt, comes from integration, as do all of the other t's in this equation! Please let me know if I'm wrong, but I think I've actually got it.

You may. How about writing up an answer to your own original question, so we can see how well you're doing? There are a couple more details you could still have questions about, and there are probably some points of notation I could help you improve in whatever you write. That would tie up some loose ends.

 

[prepforcalc]

You may. How about writing up an answer to your own original question, so we can see how well you're doing? There are a couple more details you could still have questions about, and there are probably some points of notation I could help you improve in whatever you write. That would tie up some loose ends.

The integral of 1 is done by adding 1 to the power of zero which results in a power of 1, leaving 1 as itself. Also, in this process 1 is divided by 1, again resulting in 1. That is the procedure for the invisible 1 in front of Vo and in front of distance (on the other side of the equal sign). As part of taking the integral, the dt is transformed into a t, resulting in Vo*t and a t^2 next to a. Since the t^2 happens as part of taking the integral, we also need to divide by 2, which gives the 1/2 in front of 1/2at^2.

I think there might be a step about transforming final velocity into acceleration by the way of acceleration times time or something. This may have all come from distance = velocity times time = acceleration times time times time.

 

[Dr.Peterson]

So, we start by integrating Vo + at. We get, Vot + 1/2at^2. I'm not sure where the dt comes in. That might be my next question. Then on the other side of the equal sign we take the integral of ds, which results in s. That is the extent of my magnificence. I would love to understand this better, if anyone's willing to add more detail.

It's sufficient for your purposes to think of "dx" as simply meaning "with respect to x". So you integrated v_o + at with respect to t, and 1 with respect to s.

But this is just the indefinite integral (antiderivative), not the definite integrals in the problem. I've been waiting for some questions about that step; are you okay with it?

 

[prepforcalc]

It's sufficient for your purposes to think of "dx" as simply meaning "with respect to x". So you integrated v_o + at with respect to t, and 1 with respect to s.

But this is just the indefinite integral (antiderivative), not the definite integrals in the problem. I've been waiting for some questions about that step; are you okay with it?

Oh thanks. I actually changed my initial answer above (one post above your last post). I am still confused but feel that a lot of progress has been made.

 

[Dr.Peterson]

I'm not happy with people going back in time and totally changing what they said before someone responded to it! That makes it hard to converse sensible. But I understand it happened because we were writing at the same time.

The integral of 1 is done by adding 1 to the power of zero which results in a power of 1, leaving 1 as itself. Also, in this process 1 is divided by 1, again resulting in 1. That is the procedure for the invisible 1 in front of Vo and in front of distance (on the other side of the equal sign). As part of taking the integral, the dt is transformed into a t, resulting in Vo*t and a t^2 next to a. Since the t^2 happens as part of taking the integral, we also need to divide by 2, which gives the 1/2 in front of 1/2at^2.

I think there might be a step about transforming final velocity into acceleration by the way of acceleration times time or something. This may have all come from distance = velocity times time = acceleration times time times time.

This style of explaining what you are doing is rather confusing. It would be better to show formulas rather than using words; and talk like "the invisible 1" and "dt is transformed into a t" represents a bad kind of thinking.

What you're saying is that you applied the formula INT x^n dx = x^(n+1)/(n+1) with n=0 to find that INT dt = INT t^0 dt = t^1/1 = t, and that INT ds = s^1/1 = s in the same way. (Once you know integration well, you will just write it down immediately.) By the same rule, INT t dt = INT t^1 dt = t^2/2.

What you say about velocity and acceleration is really about the definition: acceleration is the derivative of velocity, so velocity is the integral of acceleration. When acceleration is constant, this corresponds to the fact that [change in] velocity is acceleration times time, so what you say is more or less valid as background.

Now, how are you with the last steps of the work you asked about?