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Calc HW

Anonymous Tue 09 Jul 2024 20:07:04 No.1474605 View View Reply Original Report

Quoted By: >>1474608 >>1474674 >>1475046

Can someone visualize this somehow on a graph, Im having trouble understand the explanation chatgpt gives mem I was asking it why setting y to zero gives me the limits to the integral
>Setting y=0 helps us find the points where the curve intersects the x-axis. These points are the boundaries of the area we are trying to find.

Also the steps are, find the limits, integrate the polynomial, solve the antiderivative at each limit and add the two for the answer?

Anonymous

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Anonymous Tue 09 Jul 2024 21:14:53 No.1474608 Report

Quoted By: >>1474615

>>1474605
For the area to be enclosed between the x-axis and y(x), y(x) must intersect the x-axis at the boundaries of the enclosed region. The x-axis is the line where y = 0, so the limits of the enclosed areas are x where y(x) = 0. These roots, {-2, 0, 1}, are trivially found from the first form of y(x).
Because the area should be positive, it is given by the integral of the absolute difference of the enclosing functions. In this case, it is given by the integral of the first region (between -2 and 0), minus the integral of the second region (between 0 and 1), because the second integral is negative. By the fundamental theorem of calculus, the integral between two points is the difference (NB: not the sum) of the antiderivatives at these points.
The steps therefore are:
- find the limits, i.e., the points y(x) = 0, which is easy for a polynomial given in factorised form;
- integrate the regions between the intersections by taking the difference in antiderivative;
- add the absolute value of the integrals for the regions.

Anonymous

Anonymous Tue 09 Jul 2024 22:34:16 No.1474615 Report

Quoted By: >>1474617

>>1474608
thank you
in the event the second integral is positive is it still subtracted? Im confused about the end

The antiderivative is x^4/4 + x^3/3 -x^2/ So I plug in 1, get an answer and -2 and get an answer and I add the absolute values? Or I subtract the negative region which end ups being the same but that could be just this problem, does that happen every time?

Anonymous

Anonymous Tue 09 Jul 2024 23:26:16 No.1474617 Report

Quoted By: >>1474622

>>1474615
In general, you integrate the absolute difference |f(x) - g(x)| between your limits. The absolute value is |a| = a if a is positive and |a| = -a if a is negative, so you split your integral into parts where f(x) - g(x) switches sign. After computing these integrals, you add the results back together.
In this problem, f(x) = y(x) and g(x) = 0, so you get int y(x) dx where y is positive and int -y(x) dx = - int y(x) dx where y is negative.
So:
- obtain the antiderivative;
- integrate the first region by plugging in the upper bound (0) and the lower bound (-2) and subtracting the latter answer from the former;
- integrate the second region by plugging in 1 and 0, and subtracting the latter answer from the former;
- add the absolute values of these integrals. The first integral is already positive, but the second is negative, so its sign needs to be flipped. If the second integral were positive, it would be added as is.

In essence: integrate each region between sign changes as usual, by subtracting the antiderivative at the lower bound from the antiderivate at the upper bound of the region, then make all the obtained integrals positive, and add them.

Anonymous

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Anonymous Tue 09 Jul 2024 23:55:00 No.1474622 Report

Quoted By: >>1474674 >>1474694

>>1474617
I think im more cpnfused now so is pic rel wrong?

Anonymous

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Anonymous Wed 10 Jul 2024 05:11:15 No.1474674 Report

Quoted By:

>>1474605
Use Desmos, dumbass.
>why setting y to zero gives me the limits to the integral
The question says "x-axis". The x axis is y=0. The function is a 3rd order polynomial. You're already given its easier to digest form of x(x-1)(x+2), telling you that it intersects with the x axis (the values of x which make the function equal to 0) at x=0, x=1, x=-2.
>>1474622
Yes. You have 2 regions, one above the x axis and one below. If you integrate the one above, you'll get a positive result. If you integrate the one below, you'll get a negative result. If you integrate the whole region, you'll get the one subtracted by the other. You're asked to find the area enclosed. Area is a positive measure. You need to integrate between -2 and 0, and between 0 and 1, separately, and take their absolute value. That makes the area above the axis and the area below the axis. Sum them up.

Anonymous

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Anonymous Wed 10 Jul 2024 09:53:39 No.1474694 Report

Quoted By:

>>1474622
You calculated the absolute value of the integral, instead of the integral of the absolute value. They only commute when the sign does not change, because the absolute value reduces to constant multiplication by 1 or -1.
See image.

Anonymous

Anonymous Fri 12 Jul 2024 13:32:42 No.1475046 Report

Quoted By: >>1475400

>>1474605
Anon, if you're in calculus you should be able to graph this with no issue.

Anonymous

Anonymous Sun 14 Jul 2024 01:38:40 No.1475400 Report

Quoted By:

>>1475046
very helpful thank you

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