Erweiterter Mittelwertsatz für mehrfache Integrale f und g seien R-integrierbar auf der Jordan-m

step 1 I have given this information. So let's work on the A part we have. We just work on the A part.

Erweiterter Mittelwertsatz für mehrfache Integrale f und g...

Key Concepts

Weighted Averages in Integration

This concept refers to the treatment of integrals where a function is multiplied by a nonnegative weight function. The resulting integral can be thought of as a weighted average of the function with respect to the measure determined by the weight function. When the weight function is nonnegative, it enables the formulation of an equation analog to the mean value theorem, thereby allowing one to represent the integral as the product of a scalar (an average value) and the total weight.

Continuous Functions and Compactness

When a function is continuous on a compact and connected set, several useful properties come into play. Compactness guarantees that the function attains its maximum and minimum values, while connectedness ensures that every value between these extremes is also taken by the function. This is crucial for the refinement of the extended mean value theorem, where the averaging constant can be shown to correspond directly to the function's value at some point within the domain.

Riemann Integration on Jordan Measurable Sets

This concept involves the integration of functions over sets that are Jordan measurable. A set is Jordan measurable if its boundary has measure zero, which ensures that the Riemann integral is well-defined on such domains. In the context of multiple integrals, this concept extends the idea of Riemann integration from intervals in one dimension to more complicated domains in higher dimensions.

Extended Mean Value Theorem for Multiple Integrals

This theorem generalizes the classical mean value theorem for integrals to the setting of multiple integrals. It states that for functions that are Riemann integrable on a Jordan measurable set, there exists a number (or a point, under additional conditions) such that the integral of the product of a function and a nonnegative weight function equals this number multiplied by the integral of the weight function. This concept connects the idea of an 'average value' of a function with integration.

Transcript

00:01 I have given this information.

00:02 So let's work on the a part we have.

00:08 We just work on the a part.

00:09 So a part is given as we have 0 to 3 x times 2, b2, mode of x plus mode of x negative 2, here we have dx.

00:23 So this we have got now.

00:24 So just see here, mode of x is x for x greater than equal to 0, and negative x per x less than 0.

00:33 So for 0 to 3, this will be even as positive and the mod of x negative 2 will be x negative 2 or negative x negative 2 depending on the intervals.

00:42 X2 equals then equal to, okay, we have x less than negative 2.

00:51 So just we have to 0 to 2, we just talk about here for 0 to 2, 1 will be positive.

00:59 We get here that's coming out to be x times 3 to the power x plus 2 negative x because 1 is positive.

01:10 One is negative plus 2 to 3 we have x times e to the power x plus x negative now both are both here the x this is d x you get here t squared times x2 over 2 0 to 2 so i will not do the calculations here which you can do i'll just solve the method here plus 2 to 3 that is e to the power 2x negative 2 now you can apply here ily this is first function second function here i'm sorry this will be second one there's a first function here i read algebraic function will take first make it simplify so this coming out to be e square times the two negative zero positive this will be first function 90 to the part 2x negative 2 integration over 2 from 2 to 3 and negative we have 2 to 3 and derivative of x is 1 e to the power 2x negative 2 over to 0.

02:15 So this is all this now you will get it as 5e square over 4 e square plus 1.

02:24 So 5e square plus 1 so a corresponds to s.

02:31 A corresponds to s.

02:34 Let's look at the b part now.

02:38 For the b part we can just divide in the interval we have 0 to 1.

02:41 A to the power x dx plus 1 to 2 ,000.

02:46 E to the power x and greater than d .x plus 2 to 3, e to the power x, greater than the function x, dx.

02:54 So this is going out to be greater than the function we have...