anonymous
  • anonymous
find the area between the curves f(x)=e^0.8x and g(x)=-2^0.5x +1 for the interval [-3,2]
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
f(x)-g(x) sounds familiar to me
amistre64
  • amistre64
in other words, the height of each partition is simply the distance from f(x) to g(x)
anonymous
  • anonymous
can you use a calculator or all by hand?

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amistre64
  • amistre64
|dw:1327280743692:dw|
anonymous
  • anonymous
i need to do it by hand
anonymous
  • anonymous
you will need to know which function is above and which is below.
anonymous
  • anonymous
Integrate top function minus the bottom function.
anonymous
  • anonymous
on the given interval.
amistre64
  • amistre64
\[\int f(x)-g(x)dx\] no you wont, all that changes is a sign and area is always positive for drop a sign
amistre64
  • amistre64
\[\int_{-3}^{2} e^{0.8x}-(-2^{0.5x} +1)dx\] hmm, might need to write that out better
amistre64
  • amistre64
or is that right?
amistre64
  • amistre64
you might wanna check to see that nothing cross tho
amistre64
  • amistre64
if it does then you gotta split it up at the intersection
amistre64
  • amistre64
does f(x)=g(x) at any point in the interval?
anonymous
  • anonymous
well i dont think so
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=f%28x%29%3De%5E%280.8x%29+and+g%28x%29%3D-2%5E%280.5x%29++%2B1%3B+x%3D-3+to+2
amistre64
  • amistre64
if i got your equations right, the wolf says they cross
anonymous
  • anonymous
i figured out the integral thing but I didn't get the answer 6.935
anonymous
  • anonymous
Let me say again that the graph is important. On this interval the graphs intersect and therefore the integral must be seperated.
amistre64
  • amistre64
yes, intersections are important; height? not so much :)
anonymous
  • anonymous
(-1.282, 0.359) is the intersection right?
anonymous
  • anonymous
You will integrate from -3 to the point of intersection with g(x) - f(x) and then from the point of intersection to 2 with f(x)-g(x)....
amistre64
  • amistre64
\[\int_{-3}^{i} e^{0.8x}+2^{0.5x} -1\ dx+\int_{i}^{2} e^{0.8x}+2^{0.5x} -1\ dx\] where i is the intersection of f and g
anonymous
  • anonymous
I thought you mentioned solving by hand?
amistre64
  • amistre64
and if one of those is negative, then toss out the "-" sign :)
anonymous
  • anonymous
yes I need to solve it by hand, and I did what you just said but i got the wrong answer
anonymous
  • anonymous
....can you show your work?
amistre64
  • amistre64
the wolf agrees that -1.282 is the intersect
anonymous
  • anonymous
That is correct....but that is not solving by hand as I stated earlier.
anonymous
  • anonymous
\[\int\limits_{-3}^{-1.282} -2^{0.5x}-e ^{0.8x} dx + \int\limits_{-1.282}^{2} e ^{0.8x}-(-2^{0.5x}) dx\]
amistre64
  • amistre64
its checking if her results are accurate tho.
amistre64
  • amistre64
wheres the "1"?
amistre64
  • amistre64
you might be better of with an exact intersect ....
amistre64
  • amistre64
and who is making you do this by hand?
amistre64
  • amistre64
its one thing to understand a concept; its another to torture ....
anonymous
  • anonymous
No...It is important to understand the concept and to use technology that is required in the classroom and on ap exams... I can help her learn to enter the information into the calculator and how she must set it up by hand to write the integral that represents the area. Dropping a sign because area is not negative is not good enough reasoning for a solution.
anonymous
  • anonymous
\[[-2*2^{0.5x}- 5/4e ^{0.8x}] + [5/4 e ^{0.8x}+2*2^{0.5x}]\]
anonymous
  • anonymous
Have a good night...
amistre64
  • amistre64
oh but it is; since 5-3 = 2, and 3-5 = -2 all that changes is sign, not abs value
anonymous
  • anonymous
oh
amistre64
  • amistre64
youre still missing that +1 from the top that ints up into an x
anonymous
  • anonymous
oh yess. I see may be that's why i dont get the answer
amistre64
  • amistre64
most likely :) the rest of it seems like youve got a pretty good handle on it
anonymous
  • anonymous
how can i forget that "1" OMG i'm so stupid
amistre64
  • amistre64
it happens :)
anonymous
  • anonymous
ok i'll try again. thank you so muchhhhhhhh and have a great night :)

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