Worked example: Breaking up the integral's interval

So this right over here is the graph of the function g of t. It is a function of t, and let's define a new function. Let's call it capital G of x, and it's equal to the definite integral between t is equal to negative 3 and t is equal to x. Of our lowercase g of t, g of t, dt. So, given how we have just, given how we have just defined capital G of x, let's see if we can evaluate some val, let's see if we can find some values. So, let's evaluate the function capital G. At x is equal to 4, and let's also evaluate the function capital G at x is equal to 8. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of 4 first. So this is going to be, this is going to be, well, if x is equal to 4. This, this top boundary is going to be 4, so this is going to be the definite integral of g of t between t is equal to negative 3, and t is equal to 4. g of t, g of t, d t. Now what's that going to be equal to? Well, let's see, let's look at this graph here. So, we have negative 3, this t is equal to negative 3, which is right over here. T is equal to negative 3. And we're going to go all the way to is equal to 4. Let me circle that in orange. All the way to t is equal to 4 which is right over here. So this is. One way to think about this is this is going to be the area above t-axis and below the graph of G. So, it's going to be this area, right over here, that is above the t-axis and below the graph of G of T, but, we're not going to add to it this area because this area over here. I'll shade it in yellow. This area I'm shading in yellow over here, this is going to be negative. Why's it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is, we could split it up. So, we could, actually let me just clear this out so we have more space. So, this, right over here, is going to be equal to the integral. I'll do that in this purple color. The integral from t equals negative 3 to 0, of g of dt +. I'll do this in the other color. Plus the interval from 0, t equals 0 to t is equal to 4 of g of t, dt. Now what are each of these going to be? Well this is just a triangle where the base is 3. The base has length 3. The height is length 3. So, it's going to be 3 times 3, which is 9 times one half. Because 3 times 3 would give us the area of this entire square. But this triangle is half of that. So it's, this right over here, is going to be 4.5. This is going to be 4.5. And then, what's this area in yellow? Well, let's see, we have a triangle, it's base is, or it's width here is 4, it's height is 4, 4 times 4 is 16, which would be area of this entire square. We take half of that, it's 8. Now we don't just add it to it, because, once again, this is going to be negative area. The graph over here is below, below the T axis. So this, we would say, we would, this integral is going to evaluate to negative 8. Once again, why is it negative 8? Cause the graph is below, below the T axis. And so what do we get? We get g(4) which is this area essentially minus this area. 4.5 minus 8 is going to be. Let's see 4 minus 8 is negative 4 out of .5 which is negative 3.5. Negative 3.5. Now let's try to figure out what g of 8 is. So g of 8. And if you couldn't figure it out the 1st time around, try to pause the video again, and now that we've figured out g of 4, try to figure what g of 8 is. Well g of 8 is, 1 way to think about it, it's going to be that. Minus that area and then we're going to add, then we're going to figure out the area and then we're going to have two more areas to think about. We're gonna go all the way to 8. So actually make, draw the line there. So we're gonna have to think about, we're going to think about this whole area now. This whole area now. And then we're gonna think about this one. So we could write this is going to be equal to the integral between. Actually let me do that purple color. So it's going to be this, this integral. Which is the integral between negative, t equals negative 3 and 0. G of t, dt. Plus this entire yellow region right now. The part that we looked at before plus this part over here. So, I'll just write that as, plus the definated role between t is equal to 0 and 6 of g of t dt. And then finally. Plus the definite integral between T is equal to 6 and T is equal to 8 g(t) dt. Now we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle. Whose base is length 6, height is 4, 6 times 4 is 24 times one half is going to get us 12. So this is going to be, that over there is twelve and then finally what is this area? Oh and we have to be careful. This is below the t axis and above the graph. So this is going to be a negative 12. And then finally, we have this area, which is once again going to be a positive area. It's below the graph and above the t axis. And so let's see, 2 times 4 is 8 times 0.5 is 4. So we're going to have. Plus 4. Plus 4. And so what do we have? We end up with 4.5 plus 4 is 8.5 minus 12. So this is going to be equal to. This is going to be equal to 8.8 minus 12 would be negative 4 plus 5. It is negative 3.5 again. Now why did these two things end up being the same? Well think about what happened here. When we went from capital G of 4 to capital G of 8. We subtracted, we subtracted this value right over here and we added this value right over here and you see that these two triangles have the same areas. So we subtracted and added the same amount to get the same exact value.