Evaluate the integral by computing the limit of Riemann sums. ∫1^3(x^2-3) d x

Name: Problem 9, Chapter 4: Calculus: Early Transcendental Functions
Uploaded: 2020-06-10T18:44:00-08:00
Duration: 9 min 56 s
Channel: Ishita J.
Description: Evaluate the integral by computing the limit of Riemann sums. ∫1^3(x^2-3) d x

Evaluate the integral by computing the limit of Riemann...

Key Concepts

Definite Integral

A definite integral represents the accumulation of a quantity, typically interpreted as the net area under a curve between two bounds. It is defined as the limit of the sum of small contributions (areas) as the number of subintervals increases infinitely, providing an exact measure of the total accumulation.

Riemann Sum

A Riemann sum is a method for approximating the definite integral of a function. It involves dividing the integration interval into subintervals, selecting sample points in each subinterval, and then summing the product of the function value at these points and the width of the subintervals. This method lays the groundwork for the formal definition of the integral.

Partition of an Interval

Partitioning an interval involves dividing it into a finite number of subintervals. This process is essential for computing Riemann sums because it determines how the function is sampled over the interval, directly affecting the accuracy of the approximation.

Limit Process

The limit process is the mathematical procedure of taking the number of subintervals to infinity while shrinking their width to zero. This approach ensures that the Riemann sum converges to the exact value of the definite integral, thereby providing a rigorous definition of integration.

Transcript

00:01 We're given the integral from 1 to 3 for the function x squared minus 3.

00:05 And we know that to solve this, our formula is going to be right over here.

00:12 The limit as an approach is infinity, since we're having an infinite amount of rectangles of sigma with i equals 1 on the bottom for a right -handed room and sum, which is what we're using in this case.

00:25 And then inside the sigma, we have delta x times f of a plus delta.

00:30 Xr.

00:32 So as you can see, to actually plug numbers into this formula, we're going to have to find what delta x is.

00:39 So delta x is always b minus a over n, where b is the upper limit of the integral, and a is lower limit of the integral.

00:49 So in this case, delta x would be 3 minus 1 over n, and we're just going to leave the n as it is.

00:56 So we end up getting 2 over n for delta x.

00:59 So now we can plug this into our limit over here.

01:05 So we still have to rewrite the sigma.

01:09 And inside instead of delta x, we can write 2 over n.

01:14 And then we know that f of x, our original function, is x squared minus 3.

01:19 So we're going to have a plus delta x i squared and then all of that minus 3.

01:27 So a plus delta xi is our lower limit of the integral, which is 1 plus delta xi is 2 over n times, or 2 over n, which is 2i over n.

01:39 So all of that's squared, and i'll just add another parentheses here, minus 3.

01:48 Or, yeah, just like that.

01:50 So now that we have this, we can simplify this down and make it.

02:03 We're still going to have 2 over here.

02:06 And then 1 plus 2 i over n squared equals 1 plus 4 i squared over n squared, and then we still have our minus 3 over here.

02:30 So 1 minus 3 is equal to 2 or negative 2.

02:34 So if we rewrite this, we can multiply 2 over n times all of these different things over here.

02:50 So 2 over n times negative 2 is equal to negative 4 over n.

02:58 And that gets rid of our 1 and our negative 3.

03:03 Then 2 over n times 4 i squared over n squared is equal to 8 i squared over nq.

03:12 Or sorry, this should be written as plus that, not times that, plus 8 i squared over n cubed, plus 2 over n times 4i over n is equal to 8 i over n squared.

03:38 So now that we have all three of these things being added to each other, we can separate this into three different sigmas to make it easier for us to get rid of the sigmas and solve the limit.

03:51 So we still have the limit of all of this as an approaches infinity, but now we can split this into three different sigmas.

04:00 So our first sigma will just have negative 4 over n.

04:06 Our second sigma will have 8 i squared over n cubed, and our third sigma will have 8 i over n squared.

04:32 So now that we have this, we need to isolate all of our eyes or make it and leave it so that it is sigma of i squared, sigma of i, or sigma of one.

04:43 So to do this for our first one, we can take out the negative 4 over n outside the sigma.

04:50 So once we do that, we're left with just sigma of 1.

04:59 And then for our second one, we can take away 8 over n cubed.

05:09 And then we're just left with sigma of i squared.

05:15 And then for our last one, we can take away 8 over n squared.

05:20 And we are left with sigma of i...

Please give Ace some feedback