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Main content
Current time:0:00Total duration:5:25
CHA‑5 (EU)
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Video transcript

what we're going to do using our powers of calculus is find the area of this yellow region and if at any point you get inspired I always encourage you to pause the video and try to work through it on your own so the key here is you might recognize hey this is an area between curves a definite integral might be useful so I'll just set up the definite integral sign and so first we need to think about what are our left and right boundaries of our region well it looks like the left boundary is where the two graphs intersect right over here and the right boundaries were they intersect right over there well what is this point of intersection it looks like it is negative 1 comma negative 2 we can verify that in this red curve if X is negative 1 let's see you square that she'll get 1 minus 3 you do indeed get Y is equal to negative 2 and in this blue function is X is equal to negative 1 you get 1 minus 4 plus 1 once again you do indeed get Y is equal to negative 2 and the same thing is true when X is equal to 1 1 minus 3 negative 2 1 minus 4 plus 1 negative 2 so our bounds are indeed we're going from x equals negative 1 to x equals positive 1 and now let's think about our upper and lower bounds over that interval this blue graph is our upper bound and so we would subtract the lower bound from the upper bound so we would have X to the 4th minus 4 x squared plus 1 and from that we will subtract x squared minus 3 DX and in many other videos we have talked about why you do this why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them but now we just have to evaluate this definite integral so let's just get down to business all right so we have the integral from negative 1 to 1 and so we have X to the fourth X to the fourth and now we have minus 4 x squared and then when you distribute this negative sign you're gonna have you're gonna subtract another x squared you're gonna have minus 5 x squared and then you have plus 1 and then you're gonna subtract a negative 3 so it's going to be one plus three so it's going to be plus four DX DX and just to be clear I should put parentheses right over there because it's really the DX is being multiplied by this entire expression and so let's see let's find the antiderivative of this this should be pretty straightforward we're just going to use the reverse power rule multiple times so this is going to be the antiderivative of X to the fourth is X to the fifth over five we just incremented the exponent and divided by that incremented exponent - same idea here five X to the third over three plus four X and then we are going to evaluate it at 1 and then subtract from that it evaluated at negative one so let's first evaluate it at one we're gonna get 1/5 minus 5/3 plus four and now let us evaluate it at negative one so - let's see if this is negative one we're gonna negative one-fifth negative one-fifth and this is going to be plus 5/3 plus 5/3 and then this is going to be minus four minus four but then when you distribute the negative sign this so we're going to distribute this over all of these terms and so this is going to be four make this positive this will be positive this one will be negative and then this one will be positive so you have 1/5 plus 1/5 which is going to be 2/5 that and that and then minus 5/3 minus 5/3 so minus 10 over 3 and then 4 plus 4 so plus 8 and though so we just need to simplify this this is going to be let's see it's going to be 8 and then if I write so plus I'm gonna write these two with a denominator of 15 because that's the common denominator of 3 and 5 I'll see 2/5 is 6 15 less yeah that's right 5 times 3 is 15 2 times 3 is 6 and then 10 thirds let's see if we multiply the denominator times 5 after multiply numerator times 5 so there's gonna be 50 15 and so what's 6 15 minus 50 15 so this is going to be equal to 8 minus 6 minus 50 is minus 44 minus 44 over 15 and so what is 44 over 15 44 over 15 is equal to 2 and 14 15 so that's really what we're subtracting we're gonna subtract 2 and 14 15 so if you subtract 2 from this you would get 6 minus 14 over 15 because we still have to subtract the 14 15 and then 6 minus 14 15 is going to be equal to 5 and 1 15th so just like that we were indeed able to figure out this area
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