Express the following limit as a definite integral on the interval [0,10][0,10][0,10].limn→∞∑k=1n(xk∗−3)2Δx\lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta xlimn→∞∑k=1n(xk∗−3)2Δx

∫010 ⁣(x−3)2 dx\$$int_0$$^{10}\!\left(x-3\$$right)^2$$\,dx∫010(x−3)2dx

Express the following limit as a definite integral on the interval [ 0 , 10 ] [0,10] [ 0 , 10 ] . lim n → ∞ ∑ k = 1 n ( x k ∗ − 3 ) 2 Δ x \lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\right)^2\Delta x lim n → ∞ ∑ k = 1 n ( x k ∗ − 3 ) 2 Δ x

So in this problem we're asked to express the following limit as a definite integral, and that's gonna be on the interval from zero to 10. So let's go ahead and see how we can solve this. Now what I can see is this is the function we have right here. We have x of k star minus three squared multiplied by delta x, and then we have this limit as n approaches infinity for this summation right here. Now what this in essence is is a Riemann sum that has no end to it. We have unlimited amounts of subintervals, and that's going to give us a truly accurate area under the curve of the function. Now recall that there's another way that we can represent this. What we can do is represent this as an integral of our function right here. Now we don't have to worry about this k stars thing since we're not incrementing up anymore. We're finding a true area. So this whole thing is just gonna be x minus three squared as the function all with respect to x. Now of course, what what's gonna be on the bounds of this integral? Because if we're doing a definite integral, we can't just leave it like this. We have to put some numbers on here. And notice our interval goes from zero to 10. So since that's the place we're finding the area, we're going to go at a low bound of zero to a high bound of 10. And this right here would be the solution. That is how we could represent the following limit as a definite integral given this interval right here. So that is how you can solve these problems where you need to set up a definite integral. Hope you found this video helpful, and let's move on.

Express the following limit as a definite integral on the interval [0,10][0,10][0,10].lim⁡n→∞∑k=1n(xk∗−3)2Δx\lim_{n\to\infty}\sum_{k=1}^{n}\left(x_{k}^{\ast}-3\$$right)^2$$\Delta xlimn→∞∑k=1n(xk∗−3)2Δx

∫010 ⁣(x−3)2 dx\$$int_0$$^{10}\!\left(x-3\$$right)^2$$\,dx∫010(x−3)2dx

∫100 ⁣(x−3)2 dx\int_{10}^0\!\left(x-3\$$right)^2$$\,dx∫100(x−3)2dx

∫0∞ ⁣(x−3)2 dx\$$int_0$$^{\infty}\!\left(x-3\$$right)^2$$\,dx∫0∞(x−3)2dx

∫∞0 ⁣(x−3)2 dx\int_{\infty}^0\!\left(x-3\$$right)^2$$\,dx∫∞0(x−3)2dx

8. Definite Integrals

Introduction to Definite Integrals

Business Calculus

8. Definite Integrals

Introduction to Definite Integrals

Struggling with Business Calculus?

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Multiple Choice

Express the following limit as a definite integral on the interval $[0,10]$ .
$$\lim$_{n$\to\]\infty$}$\sum$_{k=1}^{n}$\left$(x_{k}^{$\ast$}-3$\right$)^2$\Delta$ x$

$$\int$_{10}^0\!$\left$(x-3$\right$)^2\,dx$

$$\int$_0^{10}\!$\left$(x-3$\right$)^2\,dx$

$$\int$_0^{$\infty$}\!$\left$(x-3$\right$)^2\,dx$

$$\int$_{$\infty$}^0\!$\left$(x-3$\right$)^2\,dx$

Verified step by step guidance

Recognize that the given limit represents a Riemann sum, which is a method for approximating the value of a definite integral. The general form of a Riemann sum is: lim⁡n→∞∑k=1nf(xk∗)Δx, where Δx is the width of each subinterval and xk∗ is a sample point in the k-th subinterval.

Identify the components of the Riemann sum in the problem: f(xk∗) = (xk∗−3)^2 and Δx is the width of each subinterval. The interval of integration is [0, 10], so Δx = (10−0)/n = 10/n.

Understand that xk∗ represents the sample points in the interval [0, 10]. For a uniform partition, xk∗ can be expressed as xk∗ = 0 + kΔx = k(10/n).

Substitute the expressions for xk∗ and Δx into the Riemann sum: lim⁡n→∞∑k=1n((k(10/n)−3)^2)(10/n). This represents the Riemann sum for the function f(x) = (x−3)^2 over the interval [0, 10].

Conclude that the limit of the Riemann sum as n→∞ is the definite integral of f(x) = (x−3)^2 over [0, 10]. Therefore, the integral is ∫_0^10 (x−3)^2 dx.

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Struggling with Business Calculus?

Express the following limit as a definite integral on the interval [0,10][0,10][0,10].limn→∞∑k=1n(xk∗−3)2Δx\(\lim\)_{n\(\to\]\infty\)}\(\sum\)_{k=1}^{n}\(\left\)(x_{k}^{\(\ast\)}-3\(\right\))^2\(\Delta\) xlimn→∞∑k=1n(xk∗−3)2Δx

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Introduction to Definite Integrals practice set

Express the following limit as a definite integral on the interval $[0,10]$ .
$\(\lim\)_{n\(\to\]\infty\)}\(\sum\)_{k=1}^{n}\(\left\)(x_{k}^{\(\ast\)}-3\(\right\))^2\(\Delta\) x$