Area of a region between curves Find the area of the region bounded by the curves y=(x^2)/(x^3-3

step 1 Section 7 .1 problem 59. So they ask us to find the area that's bounded between two curves for t

Area of a region between curves Find the area of the region...

Key Concepts

Area between Curves

This concept involves finding the region enclosed by two or more curves by integrating the difference between the functions over a given interval. It is a central application of integration in calculus, allowing one to compute the size of a region bounded by curves on a coordinate plane.

Definite Integral

A definite integral is used to compute the accumulation of a quantity, such as area, over a specified interval. Its evaluation, based on the Fundamental Theorem of Calculus, connects the process of integration with the antiderivatives of the function being integrated.

Determining Upper and Lower Functions

Before setting up the integral for the area, it is crucial to identify which function forms the upper boundary and which forms the lower boundary over the interval of interest. This ensures that the integrand correctly represents the positive difference between the functions, corresponding to the area of the region.

Algebraic Simplification

In many problems involving integrals, especially when the functions share common components, simplifying the integrand through algebraic manipulation can make it easier to evaluate the integral. This may involve combining like terms, factoring, or canceling factors to streamline the integration process.

Transcript

00:01 Section 7 .1 problem 59.

00:04 So they ask us to find the area that's bounded between two curves for the x values from zero to four.

00:12 So what i wanted to do first is just take a look at both of these curves.

00:17 So the first function we had was what you see here, x squared over x cubed minus 3x.

00:24 So that is the function you see here in red.

00:27 And then if you look at the next curve we see right here, 1 over x cubed minus 3x that is the curve that you see in blue and we're integrating from the value of 2 over to the value of 4 so over that particular x integral so what you can see is if i knew the area you know beneath the red curve and then i subtracted the area that was beneath the blue curve i would have the area between the red and the blue curve so what it tells me is the area between those two curves is going to be the integral x squared over x cubed minus 3x d x minus the integral of 1 over x cube minus 3x d x so i just need to find the integration and subtract those and then evaluate all of that from 2 to 4 to get my final answer so let's go work that out so that was our graphical exploration here and now we're just going to go back and do the algebraic and the calculus calculation.

01:31 So i've just determined here that the area between these two curves is going to be the integral from 2 to 4 of x squared over x cubed minus 3x minus 1 over x squared minus 3x dx.

01:57 So the way i'm going to do is since i've got to work on these integrals let's just work on these individually and then bring everything back together.

02:04 So i've got to come up with two anti -derivatives.

02:07 So how do i find the anti -derivative of x squared over x -cube minus 3x dx? and then how do i find the anti -derivative of 1 over x squared minus 3x d x.

02:31 So let's just take a look at these.

02:33 In the first case, i could write this as the integral, so x squared.

02:39 And here i could fact out an x and i'm left with what x squared minus 3 dx that is helpful because then this turns into the integral of x over x squared minus 3 dx and now i can make a substitution i can say let u equal x squared minus 3 then you know that du is going to be 2x d x so in order to find a 2x i would need to modify this integral in that way and now i see this is in the form of one half and what you see here is just one over u d u now that's pretty straightforward so this is going to give me one half natural log absolute value of you plus a constant of integration so that's just going to be one half natural log of u which is x squared minus three plus a consumm integration.

03:53 So that's the anti -derivative for the first part of this problem.

03:57 Now let's go look at the second part of this problem...

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