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## Applying properties of definite integrals

Current time:0:00Total duration:7:25

# Worked example: Breaking up the integral's interval

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

## Video transcript

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.

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