Find the area under the curve, of the function f ( x ) = x 2 f\left(x\right)=x^2 from x = 0 x=0 to x = 8 x=8 using limits.

Alright. So let's give this problem a try. Here, we're asked to find the area under the curve for this function, f of x equals x squared from x equals 0 to x equals 8 using limits. Now we're no longer being asked to approximate. We're asked to find an exact area. So to do this, we need to use this equation. It's going to be that our area is equal to the limit as n approaches infinity, and that's going to be the summation of our function from k equals 1 to n, and then we're going to have delta x times f of x k. So this is the equation that you need whenever you're solving these types of problems. Now I think a good place to start is to start by calculating the base, and we know that the base is going to be delta x is equal to b minus a over m. Now using this equation, b is the end of the interval, which is 8. A is the start of the interval, which is 0, and that's going to be divided by n. Now in this case, n is the number of rectangles, but since we have an infinite number of rectangles, we need to keep n in the equation. Now 8 minuteus 0 over n will just give you 8 over n, and that right there is the base. Now from here, what I'm going to do is use this base to find x of k. And x of k is going to be equal to a plus k times delta x. Now a is the start of our limit here or the excuse me, the start of the interval here, which is 0. So we're going to have 0 plus k times delta x, and we just calculated delta x to be 8 over n. So we're going to have 8 over n, meaning that our x of k is going to be equal to 8k over n. So we now have x of k and delta x, meaning we can plug things into this area equation. So we're going to keep this limit. And, by the way, this whole argument right here, I'm just going to keep put this in quotes because you typically do need to write out the limit and the summation every single time. But just to avoid just to save some space here, we're just going to say keep this argument in here, and and I'm going to focus on just the delta x times f of x k. So doing this, we're going to have delta x, which right here is 8 over n, and then we're going to have, x f of x k, which is 8 k over n, but I need to plug it into f of x. So it's going to be f of 8 k over n. Now from here, what I need to do is take this and plug it into our function. Well, I can see our function is x squared, so that means we're going to have 8 n. And then since this is the inside of our function, it's going to be 8 k over n squared. Now at this point, I can simplify everything here and square these numbers. So we're going to have 8, and that's going over n, and then it's going to be multiplied by 8 squared, which is 64, and it's gonna be times k squared all divided by n squared. Now from here, we can multiply this 8 into 64, which will give us 512. Then we're going to have k squared on top, and then that's divided by n times n squared, which is n cubed. So this right here is what happens when you simplify delta x times f of x k. So what I'm going to do is write rewrite the area right here, and I'll bring back this argument that we had in front here. So we're going to have the limit as n approaches infinity. And then keep in mind, you need to typically rewrite this argument every single time, and it's going to be the summation from k equals 1 to n. And then we're going to have what we just calculated right here, which is 512 k squared over n cubed. Right? Now at this point, what I wanna do is I wanna take anything that is not a k variable and pull it out of this summation. Because what's gonna happen is if we get a summation with just the k variables, that's going to give us a more a a form that we can actually simplify. So what I'll do is I'll take this 512 over n cubed, and I'll pull outside of the summation. And you still need to keep the limit attached, because we haven't dealt with the limit yet. But we can take this 512 over n cubed, pull it on the outside, and then we're gonna have the summation that we'll keep in here. And all we're gonna have left over inside is k squared. Now it turns out this right here is actually a form that we can use. Because whenever you see this form of our summation, this whole thing is going to become n times n plus 1 times 2 n plus 1 all over 6. So what that means is I can rewrite our entire area. We're gonna have the limit as n approaches infinity. I can rewrite this equation where we'll have 512 over n cubed. And then this whole thing is going to be multiplied by this expression right here. It's gonna be n times n plus 1 times 2 n plus 1 over 6. So this is what we get. Now at this point, what I'm trying to do is apply this limit, and it's kind of hard to see how we can do that just by looking at what we have. Have. But I can simplify a couple things because I can take this n cubed and cancel it with that n there to reduce this to n squared. Now it turns out 512 over 6 is actually going to reduce to 256 over 3, because if you just divide that as a fraction, that's what reduces. So what this whole thing is going to become if we reduce everything that we see here is we're going to have 256 over 3 n squared all multiplied by this entire thing up here. Now rather than just plugging this back in, I'm actually going to use the foil method. So n times 2 n squared or 2 n is going to give us 2 n squared, and we would have n times 1, which is n. There would have 1 times 2 n, which is 2 n, plus 1 times 1, which is 1. So this whole thing is going to become 2 n squared plus 3 n, combining these 2, plus 1. So that's what we end up getting. So plugging that into here, we'll have 2 n squared plus 3 n plus 1. Now at this point, what I can actually do is take this 256 here and distribute it into these parentheses. So going back to this area here, we're going to have the limit as n approaches infinity, and then we'll have 256 times 2. And 256 times 2 is 512. So we'll have 512 n squared, and that's going to be plus 3 times 256, which is 768. And then we're gonna add 256 times 1, which is 256. And all of that is going to be divided by this 3 n squared on bottom. Now at this point, we can now deal with this limit, because recall that there are rules for limits as n approaches infinity. Now what I'm looking for is the highest degree n on top and bottom. The highest degree n on top is this n squared. The highest degree n on bottom is also n squared. So since we have the same exponents here, what I need to do is find a ratio of the coefficients out in the front, which will be 512 over 3. So our whole area is going to become 512 over 3, and that right there is the solution to this problem. So this is how you can find the exact area underneath a a curve using limits. So hope you found this video helpful. Let's move on.

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23. Intro to Derivatives & Area Under the Curve2h 9m

23. Intro to Derivatives & Area Under the Curve

Area Under a Curve

Precalculus

23. Intro to Derivatives & Area Under the Curve

Area Under a Curve

Previous problem

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Multiple Choice

Find the area under the curve, of the function $f (x) = x^{2}$ from $x equals 0$ to $x equals 8$ using limits.

256

$512$

$\frac{1}{3}$

$512 over 3$

Previous problem

Verified step by step guidance

To find the area under the curve of the function $ f(x) = x^2 $ from $ x = 0 $ to $ x = 8 $, we need to calculate the definite integral of $ f(x) $ over the interval $[0, 8]$.

The definite integral of a function $ f(x) $ from $ a $ to $ b $ is given by $ \int_{a}^{b} f(x) \, dx $. In this case, it is $ \int_{0}^{8} x^2 \, dx $.

To evaluate this integral, we first find the antiderivative of $ x^2 $. The antiderivative of $ x^2 $ is $ \frac{x^3}{3} $.

Next, we apply the Fundamental Theorem of Calculus, which states that $ \int_{a}^{b} f(x) \, dx = F(b) - F(a) $, where $ F(x) $ is the antiderivative of $ f(x) $.

Substitute $ x = 8 $ and $ x = 0 $ into the antiderivative $ \frac{x^3}{3} $ and compute $ \frac{8^3}{3} - \frac{0^3}{3} $ to find the area under the curve.