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Main content
Current time:0:00Total duration:6:54
AP.CALC:
CHA‑5 (EU)
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CHA‑5.A (LO)
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CHA‑5.A.1 (EK)

Video transcript

what I want to do with this video is find the area of this region that I'm shading in yellow and what might seem challenging is is throughout this region I have the same lower function or I guess the lower boundary is y is equal to x squared over 4 minus 1 but I have a different upper boundary and the way that we can tackle this is by dividing this area into two sections and or dividing this region into two regions the region on the left and the region on the right where for this first region which I'll do I'll color even more in yellow for this first region over that entire interval in X and it looks like X is going between zero and at one y equals when X is equal to one this function is equal to 1 when y when X is equal to 1 this function is also equal to 1 so this is the point 1 comma 1 that's where they intersect so for this section this sub region right over here y equals square root of x is the upper function the entire time and then we can have a we can set up a different we can separately tackle figuring out the area of this region from X is equal to 1 to X is equal to 2 where y equals 2 minus X is the upper function so let's do it so let's first think about this first region well that's going to be the definite integral from X is equal to 0 to X is equal to 1 and our upper function is square root of x so square root of x and then we're going to from that we want to subtract our lower function square root of x minus x squared over 4 minus 1 so minus x squared over 4 minus 1 and then of course we have our DX so this right over here this is describing the area in yellow and you could imagine it that this this part right over here the difference between these two functions is essentially this height we do in a different color is essentially this height this height and then you multiply it times DX you get a little rectangle with width DX and then you do that for each X each X you get a different rectangle and then you sum them all up and you take the limit as your change in X or which is zero so as you get ultra ultra thin rectangles and you have an infinite number of them and that's our that's our definition or the riemann definition of what a definite integral is and so this is the area of the left region and by the exact same logic we can figure out the area of the right region the right region and with then we could just sum the two things together the right region we're going from X is equal to 0 to X sorry X is equal to 1 X is equal to 2 1 to 2 the upper function is to minus X 2 minus X and from that we're going to subtract from that we're going to subtract the lower function which is x squared over 4 minus 1 x squared over 4 minus 1 and now we just have to evaluate so let's first simplify this right over here this is equal to the definite integral from 0 to 1 of square root of x minus x squared over 4 plus 1 DX I'm going to write it all in one color now plus plus the definite integral from 1 to 2 of 2 minus X minus x squared over 4 then subtracting a negative 1 is a positive 3 we could a positive 1 we could just add it to this 2 and so this 2 is just becomes a 3 I said 2 minus negative 1 is 3 D X and now we just have to take n the antiderivative and evaluate it at 1 and 0 so the antiderivative of this is well this is X to the 1/2 increment it by 1 and comment' the power by 1 you get X to the three halves and then multiply by the reciprocal of the new exponent so it's 2/3 X to the three halves minus the antiderivative of x squared over four is X to the third divided by 3/4 so divided by 12 plus X that's the antiderivative of one we're going to evaluate it at 1 and zero and then here the antiderivative is going to be 3x minus x squared over two minus X to the third over 12 once again evaluated at or not once again now we're going to evaluate at two and one so over here you evaluate all this stuff at one you get two thirds minus 1/12 plus one and then from that you subtract this evaluated at zero but this is just all 0 so you get nothing so this is what the yellow stuff simplified to and then this purple stuff or this magenta stuff or mo for whatever color this is first you evaluate it at 2 you get 6 minus let's see 2 squared over 2 is 2 minus 8 over 12 minus 8 over 12 and then from that you're going to subtract this evaluated at 1 so it's going to be it's going to be 3 times 1 is 3 minus 1/2 minus 1 over 12 and now what we're essentially left with is adding a bunch of fractions so let's see if we can do that it looks like 12 would be the most obvious common denominator so here you have 8 over 12 minus 1 over 12 plus 12 over 12 so this simplifies to assistance is 19 over 12 the part that we have in yellow then this business let me do it in this color so 6 minus 2 this is just going to be 4 so we can write this as 48 over 12 that's 4 minus 8 over 12 and then we're going to have to subtract a 3 which is 36 over 12 36 over 12 then we're going to add 1/2 which is just plus 6 over 12 and then we're going to add a 1 12 I'm going to add a 1 12 so this is all going to simplify to let's see 48 minus 8 is 40 minus 36 is 4 plus 6 is 10 plus 1 is 11 so this becomes plus 11 over 12 let me make sure I did that right 48 minus 8 is 40 minus 36 is 4 10 11 so that looks right and then we're ready to take to add these to 19 plus 11 is equal to 30 over 12 or if we want to simplify this a little bit we can divide the numerator and the denominator by 6:00 this is equal to five halves or two and a half and we're done we figured out we figure it out the area of this entire region it is two and a half
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