HELP:Calculus (area region)

  • #1
1) Find the area of the region bounded by the graphs of
f(x) = x^2 - 5x + 3
and
H(x) = -x

Is H(x) an antiderivative? how do I solve this?

2) Find the area of the region bounded by the graphs of
x - y^2 + 2y + 4 = 2
and
2y - 2 - x = 0

so, x = y^2 - 2y - 2 and x = 2y - 2
I, then, substitute the x: y^2 - 2y -2 = 2y -2 and got y^2 - 4y=0
so, y=0,4
then I solve with antiderivatives and got -32/3 but the correct answer is
-16/3
 

Answers and Replies

  • #2
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1
buffgilville said:
1) Find the area of the region bounded by the graphs of
f(x) = x^2 - 5x + 3
and
H(x) = -x

Is H(x) an antiderivative? how do I solve this?

It looks like you're meant to assume that H(x) is just another function, like the quadratic you're given. What you need to do is find out where H(x) = f(x) and then find the areas under H(x) and f(x) between these limits and subtract one from the other. Finding the area under H(x) won't even require integration, as it's a straight line.
 
  • #3
cepheid
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1)

Did you draw a diagram so that you could see what was going on? You could simply plug the two functions into your computer or graphing calculator and see what bounded region the question is talking about. But it's not hard to graph it by hand, right?

h(x) = -x --a line passing through the origin with slope -1

The second graph is obviously parabolic, and it's easy to see what it's like by completing the square:

[tex] f(x) = x^2 - 5x + 3 = x^2 - 5x + 25/4 + 3 - 25/4 [/tex]

[tex] = (x - 5/2)^2 -13/4 [/tex]

So the graph is just that of y = x2 with a vertex shifted 5/2 to the right and 3 1/4 down. Now that you're looking at the graph, you can see that it's bound by h(x) on top and f(x) below. Can you see then that the area will be given by:

[tex] \int_a^b {(h(x) - f(x))}dx [/tex]

But what are these points bounding the region on the left and right, with x-coordinates a and b, respectively? They are the intersection points of the graphs of the two functions! They must satisfy both expressions, so that:

[tex] -x = x^2 - 5x + 3 [/tex]

[tex] 0 = x^2 - 4x + 3 = (x-3)(x-1) [/tex]

[tex] x = 1, x = 3 [/tex]

so we have intersection points (1, -1) and (3, -3)

The area becomes:

[tex] \int_1^3 {(-x - x^2 + 5x -3)}dx [/tex]

[tex] = \int_1^3 {(- x^2 + 4x -3)}dx [/tex]

I get -4/3 by hand, which agrees with the TI-83 answer -1.33333333333333....

Try #2 now
 
  • #4
cepheid, I followed what you did and still got -32/3 for #2, but the correct answer is -16/3
 
  • #5
cepheid
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Well, I gave it a shot, but really you should post YOUR work from now on. so that people can see what you did and help you along...don't just say..."Oh I did such and such, but it didn't work."

In this case, I'm at a loss, unless there's a typo in the original question. I didn't get -16/3. Here's what I did:

1) Solved for x in terms of y for both functions (it happens to be easier that way in this problem)

2) Plotted the graphs (x vs y) for both functions, and determined which curve formed the lower and upper boundary of the region

3) Equated the expressions for the functions to solve for the intersection points.

That gave me:

[tex] \int_0^4 {(-y^2 + 4y)dy} = \left[-\frac{y^3}{3} + 2y^2\right]_0^4 [/tex]

[tex] = -\frac{64}{3} + 32 = -\frac{64}{3} + \frac{96}{3} = \frac{32}{3} [/tex]

According to you, that's not the answer! I don't know what's going on with this problem.
 
  • #6
shmoe
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buffgilville said:
cepheid, I followed what you did and still got -32/3 for #2, but the correct answer is -16/3

Where did you get the "correct" answer from? Does it bother you that your question asked for an area and this solution is negative? (you might want to check cepheid's anwser for #1 as well)
 

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