7b. Difference Between Differentiation and Integration
Integration Mini Video Lecture
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Let's now look at the difference between differentiation and integration.
Let's think of differentiation as going in the forward direction and integrate as going in the backwards direction.
Let's see how this works by differentiating 4 x to the power of 7 and then integrating 4 x to the power of 7 and seeing how it is different.
So, in the first one, the `d/dx` of 4x to the 7th, just remembering my rules, I do 7 down to the front.
So it is going to be 7 times 4 is 28x to the power 3 and I have a 7 here.
I have to subtract one for a differentiation so I am going to get 6.
So the derivative of 4x to the power 7 is 28x to the power 6. `(d/dx 4x^7 = 28x^6)`
Now, let's integrate 4 x to the power 7 and see the difference.
So I use the integral sign, which is this extended "S", and I do 4 x to the power 7 and I do dx. `(int 4x^7 dx)` This dx is important. Please don't forget it.
Now, I realize that this is the same as 4 integral x power 7 dx. `(4 int x^7 dx)`
All I'm doing is taking this 4 out the front so that you can see that I am just integrating the `x^7` by itself. Later on you will do this in your head.
Okay, so this is 4 and then times, now to integrate x to the 7th, what I have to do is this. I have to add one to the index, so it's going to be x to the power of 8. And then I have to divide by this new number x to the 8th over 8. And then I have to do plus K. `(4 int x^7 dx = 4(x^8)/8 + K)`
This K is a constant because if I differentiate a constant I'll get zero. So I must always include this "plus K" when I'm doing an indefinite integral.
Now, all I need to do is tidy it up. Four and the 8 cancel, so I'll get a 2 in this position so I can do x to the power 8 over 2 and then plus K. `(4 int x^7 dx = (x^8)/2 + K)`
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