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:5:54
CHA‑5 (EU)
CHA‑5.A (LO)
CHA‑5.A.1 (EK)

Video transcript

so we've got the function f of X is equal to x squared and what I'm concerned with is finding the area under the curve y is equal to f of X so that's my y-axis this is my my x-axis and then let me draw my function my function looks like this at least to the put in the first quadrant that's where I'll graph it for now I could also graph it obviously in the second quadrant but what I care about is the area under this curve and above the positive x-axis between between x equals 1 and x equals 4 x equals 4 and I'm tired of approximating areas I want to find the exact area under this curve above the x-axis and the way we denote the exact area under the curve this little brown shaded area is using the definite integral the definite integral from 1 to 4 of f of X DX and the way that or the way I conceptualize where this notation comes from is we imagine a bunch of infinite an infinite number of infinitely thin rectangles that we sum up to find this area and let me draw one of those infinitely thin rectangles maybe not so infinitely thin so let me draw it like this so that would be one of the rectangles that would be another rectangle this should be reminiscent of a Riemann sum in fact that's where the Riemann integral comes from taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles this is how I conceptualize it is DX and the height of this rectangle is the function evaluated at an X that's within this interval right over here and so this part right over here is the area of one of those rectangles and we are summing them all up and this is kind of an elongated s reminiscent of a sigma for summing we're summing up the infinite number of those infinitely thin erecting or the areas of those infinitely thin rectangles between 1 and 4 so that's what the notation of the definite integral comes from but we still haven't done anything we've just written some notation that says the exact area of the un-- between 1 and 4 under the curve f of and above the x-axis in order to actually do anything really productive with this we have to turn to the second fundamental theorem of calculus sometimes called part two of the fundamental theorem of calculus which tells us that if F has an antiderivative so if we have the antiderivative of F so f of X is derivative derivative of some function capital f of X or another way of saying it is F capital f of X is the antiderivative anti derivative of lowercase F of X then I can evaluate this thing and we do a whole video on conceptually understanding why this makes sense we can evaluate this by evaluating the antiderivative of F or an antiderivative of F at 4 and from that subtract the antiderivative valuated at 1 so let's do it for this particular case right over here so we are taking I'll just rewrite this statement instead of writing f of X I'll write x squared so the definite integral from 1 to 4 of x squared DX well we're just going to have to figure out what the antiderivative is so if f of X is equal to x squared what is capital f of X equal to what is the antiderivative well you might remember from your power rules that if you take the derivative with respect to X of X to the third you are going to get 3x squared which is pretty darn close to x squared except for this factor of 3 so let's divide both sides by 3 let's divide both sides by 3 and you get the derivative of X to the 3rd divided by 3 is indeed x squared or you could say that this is the same thing as the derivative with respect to X of X to the third over 3 take the derivative of this it'll be 3 times 1/3 and then you'll decrement the power will just be x squared so this right over here once again is x squared is just equal to just equal to x squared so in this case our capital f of X our antiderivative is X to the third X to the third over three and so we just have to evaluate that at four and at one and sometimes the way we would with the notation we'll use is well say that the antiderivative is X to the third over 3 and we're going to evaluate it the one I always like to just write the numbers up here at four and from that subtract it evaluated at one sometimes you'll see people write a little line here to where let's say we're going to evaluate it at 4 and then at one but I'll just do it without the line so we're going to evaluate this thing at four and from that subtracted subtract it evaluated one so this is going to be equal to 4 to the third power is 64 so it's going to be 64 over 3 let me color code it this is this right over here is this right over there and then from that we're going to subtract this business evaluated one well when you evaluate it at 1 you get 1 to the third is 1 over 3 you get 1/3 so just to be clear this is this right over there and then we are ready to just subtract these fractions 64 over 3 minus 1/3 is equal to 63 over 3 and 3 goes into 63 exactly exactly 21 times so whatever the units are the area of this brown area is 21 square units
AP® is a registered trademark of the College Board, which has not reviewed this resource.