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

- anonymous

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- katieb

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- jessicaP

Ok, I'm going to try this one.

- anonymous

oh thanks. May be I better show you what i have done so you can find my mistakes

- jessicaP

Sounds good. I'm looking at the graphs right now.

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## More answers

- anonymous

|dw:1327278455526:dw|

- anonymous

is that what you get?

- anonymous

\[\int\limits_{-1.259}^{0.62} 2^{0.4x}-[(3x+1)^{2}-7]dx \]

- 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

\[\left[ 2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right]\]

- jessicaP

Sorry, it's taking me some time to check the algebra/arithmetic. :) But so far your approach looks good.

- anonymous

it's okay :) I can't find what's wrong here :(

- anonymous

oh btw did I get the interval right? [-1.259, 0.62]

- jessicaP

Did you get those values by setting the equations equal to themselves and solving for x?

- anonymous

No. I got it from calculator. :)

- 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

well i just look at the intersections of f(x) and g(x) on calculator. I'm confused now.

- jessicaP

Ok, never mind what I said if that confuses you. When I look at my calculator I get [-0.64, 0].

- anonymous

oh my bad. I typed a wrong equation f(x)=(3x+1)^2 -7 sorry sorry sorry

- jessicaP

No problem! Glad you found your mistake. Let's try it with these limits.

- anonymous

:) I'm so sorry.

- 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

hehe this problem takes me like more than one hour to find the answer that my teacher stated.

- anonymous

my answer is 9.913 do you get the same one?

- jessicaP

Almost have it . . .

- 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

you didn't change the equation, did you? cuz it's supposed to be 9x+6x+1-7

- 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

I mean question.

- anonymous

i just said that I typed a wrong equation f(x)=(3x+1)^2-7 :)

- 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

Is this right?

- anonymous

yup but the interval is different cuz you changed the equation. You should check the intersections with your calculator too.

- jessicaP

Oh yes

- anonymous

hey wait it is \[2^{0.4x}-9x ^{2}-6x+6\]

- jessicaP

You are right, I meant 6x. So are you saying that the minus sign is already distributed?

- anonymous

Yes

- jessicaP

So I should take it out when I plug it in to my calculator.

- anonymous

Yup. Did you find the interval?

- jessicaP

Here we go: [-1.277,0.63] . However, the first number is not exact.

- anonymous

I got [-1.259, 0.62]

- jessicaP

hmmm this is why I think an equation is better if only I could solve it correctly.

- jessicaP

But I think we are in the ballpark. This was tricky with the x in the exponent.

- anonymous

you're trying to find the interval by solving the equations? I can't find it by hand, you know.

- jessicaP

Ok, well I was estimating at first; I get the same answer that you did using the calc function.

- anonymous

yes then we just need to integrate that equation. I started over and get 9.913

- jessicaP

The integration is fine. I used all of the digits for the intersection and got 6.702.

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