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Without a graphing utility and short of graphing both functions very accurately to tell this, is there any other way to see that this is what the question was asking for?

- Thread starter daigo
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- Thread starter
- #1

Without a graphing utility and short of graphing both functions very accurately to tell this, is there any other way to see that this is what the question was asking for?

- Mar 1, 2012

- 249

What's the actual question? I'm trying to think it through but it'll be easier with a real example.I was given a problem with two functions and two x-values for boundaries, so I found the points of intersection (there were two) and attempted to find the area between those functions, but I didn't get to finish. In any case, I would have gotten it wrong, because when graph the two functions and then look at the boundaries, there are 3 separate areas that needed to be added up. I just thought it was another problem with a parabolic curve and a line going through it, and there was only one area in between them, but it was asking me to also include the areas that were not in both functions, i.e.

<snip>

Without a graphing utility and short of graphing both functions very accurately to tell this, is there any other way to see that this is what the question was asking for?

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- Feb 1, 2012

- 57

First step is find where they intersect.y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

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In my mind, I am thinking:

- Feb 1, 2012

- 57

- Mar 1, 2012

- 249

They meet at: $x = \dfrac{1\pm\sqrt{21}}{2}$.y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

Let $\alpha = \dfrac{1}{2}(1-\sqrt{21}) \text{ and } \beta = \dfrac{1}{2}(1+\sqrt{21})$

Let $f(x) = x+10$ and $g(x) = x^2+5$. Work out the points at the lower and upper bounds so we know which order to subtract the integral from (so we don't have a negative answer)

- $f(-7) = ?$
- $ f(6) = ?$
- $g(-7) = ?$
- $g(6) = ?$

You want to find the area under g(x) between the lower bound (x=-7) and the negative intersection to the x-axis. If you do the same for f(x) between the same two limits then you can subtract the area under f(x) from the area under g(x).

i.e. $\displaystyle \int^{-7}_{\alpha} (x^2+5)dx - \int^{-7}_{\alpha}(x+10)dx$

Then you do it between the points of intersection as you know how

You then do the same thing for the area between the positive intersection and the upper bound

i.e. $\displaystyle \int^{6}_{\beta} (x^2+5)dx - \int^{6}_{\beta}(x+10)dx$

Then you add all three values to get the final answer.