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Studying for a test? Prepare with these 4 lessons on Area & arc length using calculus.
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What I want to do in this video is find the area of this region that I'm shading in yellow. And what might seem challenging is that 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, 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 0 and 1. y equals-- when x is equal to 1, this function is equal to 1. 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 subregion 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 from that, we want to subtract our lower function-- square root of x 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 part right over here, the difference between these two functions is essentially this height. Let me do it in a different color. 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 approaches 0. So as you get ultra, ultra thin rectangles, and you have an infinite number of them. And 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 could figure out the area of the right region. The right region-- and 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 to x is equal to 2, 1 to 2. The upper function is 2 minus x. And from that, we're going to subtract the lower function, which is 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 the definite integral from 1 to 2 of 2 minus x, minus x squared over 4. Then subtracting a negative is a positive 3-- or a positive 1. We could just add it to this 2. And so this 2 just becomes a 3. I said 2 minus negative 1 is 3, dx. And now we just have to take 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. Increment the power by 1, you get x to the 3/2, and then multiply by the reciprocal of the new exponent-- so it's 2/3 x to the 3/2. Minus-- the antiderivative of x squared over 4 is x to the third, divided by 3, divided by 4, so divided by 12, plus x. That's the antiderivative of 1. We're going to evaluate it at 1 and 0. And then here the antiderivative is going to be 3x minus x squared over 2 minus x to the third over 12. Once again, evaluate it at-- or not once again. Now we're going to evaluate at 2 and 1. So over here, you evaluate all of this stuff at 1. You get 2/3 minus 1/12 plus 1. And then from that, you subtract this evaluated at 0. 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 mauve, or 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. And then from that, you're going to subtract this evaluated at 1. So it's going to be 3 times 1-- that's 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 8/12 minus 1/12 plus 12/12. So this simplifies to-- what's this? This is 19/12, the part that we have in yellow. And 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 48/12-- that's 4-- minus 8/12. And then we're going to have to subtract a 3, which is 36/12. Then we're going to add to 1/2, which is just plus 6/12, and then we're 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 11/12. 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 add these two. 19 plus 11 is equal to 30/12. Or if we want to simplify this a little bit, we can divide the numerator and the denominator by 6. This is equal to 5/2, or 2 and 1/2. And we're done. We figured out the area of this entire region. It is 2 and 1/2.