anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jessicaP
  • jessicaP
Ok, I'm going to try this one.
anonymous
  • anonymous
oh thanks. May be I better show you what i have done so you can find my mistakes
jessicaP
  • jessicaP
Sounds good. I'm looking at the graphs right now.

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

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