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AP.CALC:

CHA‑5 (EU)

, CHA‑5.B (LO)

, CHA‑5.B.1 (EK)

the base of a solid is the region enclosed by the graphs of y is equal to negative x squared plus 6x minus 1 and Y is equal to 4 cross sections of the solid perpendicular to the x-axis are rectangles whose height is X express the volume of the solid with a definite integral so pause this video and see if you can have a go at that all right now what's interesting about this is they just given us the equations for the graphs but we haven't visualized them yet and we need to visualize them or at least I like to visualize them so I can think about this region that they're talking about so maybe a first thing to do is think about well where do these two lines intersect so when do we have the same y value or another way to think about it is when does this thing equal 4 so if we set them equal to each other we have negative x squared plus 6x minus 1 is equal to 4 this will give us the X values where these two lines intersect and so we will get if we want to solve for X so we can subtract 4 from both sides negative x squared plus 6x minus 5 is equal to 0 we can multiply both sides by negative 1 we will get x squared minus 6x plus 5 is equal to 0 and then this is pretty straightforward to factor 1 times 5 is 5 or actually I say negative 1 times negative 5 is 5 and negative 1 plus negative 5 is negative 6 so it's going to be X minus 1 times X minus 5 is equal to 0 and so these intersect when X is equal to 1 or X is equal to 5 since we have a negative out front of the second-degree term right over here we know it's going to be a downward-opening parabola and we know that we intersect y equals 4 when X is equal to 1 and x equals 5 and so the vertex must be right in between them so the vertex is going to be at x equals 3 so let's actually visualize this a little bit so it's going to look something like this I draw it with some perspective because we ramp to think about a three-dimensional shape so that's our y-axis this is our x-axis and let me draw some y-values so one two three four five six seven eight this is probably sufficient now we have y is equal to four which is going to look something like this so that is y is equal to four and then we have y is equal to negative x squared plus 6x minus 1 which we know intersects y equals 4 at x equals 1 or x equals 5 so let's see 1 2 3 4 5 so x equals 1 so we have that point right over there 1 comma 4 and then we have 5 comma 4 and then we know the vertex is when x is equal to 3 so it might look something like this we could substitute 3 back in here so see Y will equal to negative 9 3 squared plus 18 minus 1 and so what is that going to be that's going to be Y is going to be equal to 8 so we have the point 3 comma 8 this is 5 6 7 8 yeah right about there and so we are dealing with a situation we're dealing with a situation that looks something like this this is the region in question so that's going to be the base of our solid and they say cross sections of the solid perpendicular to the x axis so let me draw one of those cross sections so this is a cross section perpendicular to the x axis our rectangles whose height is X so this is going to have height X right over here hi X now what is this the width I guess we could say of this rectangle well it's going to be the difference between these two functions it's going to be these this upper function minus this lower function so it's going to be that right over there it's going to be negative x squared plus 6x minus 1 and then minus 4 minus the lower function so that could be simplified as negative x squared plus 6x - five and so if we want to figure out the volume of this little section right over here we'd multiply x times this and then we would multiply that times an infinitesimal small infinitesimally small depth DX and then we can just integrate from x equals 1 to x equals 5 so let's do that the volume of just this little slice over here is going to be the base which is negative x squared plus 6x minus 5 times the height times X times the depth times DX and then what we want to do is we want to sum up all of these and so you could imagine right over here you would have or like right over here you would have a cross-section that looks like this X is now much larger the height is X so now it looks something like this so I'm just drawing two cross sections just so you get the idea so these are the cross this is any one cross section for given X but now we want to integrate our X is going from x equals 1 to x equals 5 x equals 1 to x equals 5 and there you have it we have expressed the volume of that solid as a definite integral and it's worth noting that this is that this definite integral if you distribute this X if you multiply it by all of these terms it's very solvable you don't need a or is very solvable without a calculator you're just going to get a polynomial over here that you have to take the antiderivative of in order to evaluate the definite integral

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