Find the area of the region bounded by the curves f(x)=(3x+1)^2 g(x)=2^0.4x i checked my work a lot of times but I couldn't get 8.3522

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Find the area of the region bounded by the curves f(x)=(3x+1)^2 g(x)=2^0.4x i checked my work a lot of times but I couldn't get 8.3522

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Ok, I'm going to try this one.
oh thanks. May be I better show you what i have done so you can find my mistakes
Sounds good. I'm looking at the graphs right now.

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|dw:1327278455526:dw|
is that what you get?
\[\int\limits_{-1.259}^{0.62} 2^{0.4x}-[(3x+1)^{2}-7]dx \]
\[\left[ (2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right] \]\[\int\limits_{-1.259}^{0.62} 2^{0.4x}-9x ^{2}-6x+6 dx\]\[\int\limits_{-1.259}^{0.62}2^{0.4x}-[9x ^{2}+6x+1-7]dx\]
\[\left[ 2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right]\]
Sorry, it's taking me some time to check the algebra/arithmetic. :) But so far your approach looks good.
it's okay :) I can't find what's wrong here :(
oh btw did I get the interval right? [-1.259, 0.62]
Did you get those values by setting the equations equal to themselves and solving for x?
No. I got it from calculator. :)
Well, I have to admit that I was having difficulty solving the equation, but what I did was put it into wolframalpha.com\[9x ^{2}+6x+1=2^{0.4x}\] It gave me x=0 and 1.246 E -8. This is the link: http://www.wolframalpha.com/input/?i=9x%5E2%2B6x%2B1%3D2%5E%28.4x%29
well i just look at the intersections of f(x) and g(x) on calculator. I'm confused now.
Ok, never mind what I said if that confuses you. When I look at my calculator I get [-0.64, 0].
oh my bad. I typed a wrong equation f(x)=(3x+1)^2 -7 sorry sorry sorry
No problem! Glad you found your mistake. Let's try it with these limits.
:) I'm so sorry.
Don't be! I just started this too and have made errors that are much worse :) It has helped me to work through this with you.
hehe this problem takes me like more than one hour to find the answer that my teacher stated.
my answer is 9.913 do you get the same one?
Almost have it . . .
This is my equation:\[\int\limits_{-0.64bj }^{0}2^{0.4x}-(9x+6x+1) dx\] My answer is way off, I got 3.018
you didn't change the equation, did you? cuz it's supposed to be 9x+6x+1-7
My bad it is supposed to be\[9x ^{2}\], but where did the -7 come from? I didn't see it in the original equation.
I mean question.
i just said that I typed a wrong equation f(x)=(3x+1)^2-7 :)
Ok now I get it so it is\[\int\limits\limits_{-0.64}^{0}2^{0.4x}-( 9x-3x+1-7) dx \]which equals\[\int\limits_{-0.64}^{0}2^{0.4x}-9x+3x+6dx\]
Is this right?
yup but the interval is different cuz you changed the equation. You should check the intersections with your calculator too.
Oh yes
hey wait it is \[2^{0.4x}-9x ^{2}-6x+6\]
You are right, I meant 6x. So are you saying that the minus sign is already distributed?
Yes
So I should take it out when I plug it in to my calculator.
Yup. Did you find the interval?
Here we go: [-1.277,0.63] . However, the first number is not exact.
I got [-1.259, 0.62]
hmmm this is why I think an equation is better if only I could solve it correctly.
But I think we are in the ballpark. This was tricky with the x in the exponent.
you're trying to find the interval by solving the equations? I can't find it by hand, you know.
Ok, well I was estimating at first; I get the same answer that you did using the calc function.
yes then we just need to integrate that equation. I started over and get 9.913
The integration is fine. I used all of the digits for the intersection and got 6.702.

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