If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:4:48

the continuous function f is defined on the interval negative 4 is less than or equal to X is less than or equal to 3 the graph of f consists of two quarter circles and one line segment so this right here is one quarter circle then we have another quarter circle and then it has this line segment over here as shown in the figure above I put the figure on the side so that we have some screen real estate down here let G of X be equal to 2x plus the definite integral from 0 to X of F of T DT all right that's a little bit interesting now let's do Part A Part A find G of negative 3 so we want to find G of negative 3 so first of all they tell us what G of X is so G of negative 3 every time we see an X we just put a negative 3 there so is going to be 2 times negative 3 2 times negative 3 plus the definite integral from 0 to negative 3 F of T F of T DT so this first part is pretty straightforward 2 times negative 3 is negative 6 and then this part right over here is the definite integral from 0 to negative 3 of F of T DT so from 0 to negative 3 f of T DT so this is essentially this is essentially the area under under F of T or under f of X if you want to view it that way from 0 to 3 and this is pretty straight forward to figure out but we have to be careful here because this area this area right over here this would be the integral let me do this in a color that you can see that area right over there would be the integral from negative 3 to 0 F of f of T or we can use a F of T DT or f of X DX either way DT that would be this area right over here they've swapped it they have the larger number as the lower bound they have 0 as the lower bound so what we could do this we could rewrite this this is the same thing this is equal to negative 6 minus if you swap the bounds of integration you swap the sign on the integral so we could say minus from negative 3/2 is zero of F of T DT and now this thing right over here this expression right over here is the area it is the area under this quarter circle and the area under this quarter circle we can just use a little bit of geometry to figure that out we know it's radius the radius here is three radius is equal to three the entire circle is PI R squared so the entire circle the area if this was an entire circle right over here if this was an entire circle that area would be PI R squared so pi times three squared so nine pi so it would be nine PI but this is only one-fourth of the entire circle so we want to divide it by four so this right over here will evaluate to nine PI over four so Part A we get it's equal to negative six that's that right over there minus nine PI minus nine PI over four and that's at least the first part of Part A then they say find G prime of X so let's find G prime of X so G prime of X we did the first part of Part A right over there then we need to find G prime of X G prime of X well it's just going to be the derivative of G of X the derivative of 2x is just going to be two and then the derivative and this is the fundamental theorem of calculus right over here the derivative of the definite integral from 0 to X of F of T DT that's just going to be f of X that is just going to be f of X so that is G prime of X we did this second part right over there that is G prime of X and then we need to evaluate it at negative 3 so G prime of negative 3 G prime of negative 3 is going to be equal to 2 plus F of negative 3 which is equal to 2 plus and let's look at our let's see what F of negative 3 is this is our function definition when X is equal to negative 3 our function is at 0 so f of negative 3 is just 0 so it's just 2 plus 0 which is equal to 2 so G of negative 3 is equal to 2 and we're done with Part A and the trickiest part of this is just real the bounds of integration that you might be tempted to say oh the integral between 0 and negative 3 that's just this area over here because we're in between 0 and negative 3 but you got to realize that we actually have to swap them and make them negative if you really want to consider this area right over here that this integral is going to be the negative of this area and that's what we did when we went to this step right over there

AP® is a registered trademark of the College Board, which has not reviewed this resource.