Question

Let's try to correct this question.

Answer 1

Looks good to me! :-)

Answer 2

Here is the original question:

Compute the sum.\[\int_{-4}^5 e^{(x+5)^2 }dx + 3\int_{1/3}^{2/3}e^{9(x-2/3)^2}dx \]

Answer 3

Pretty sure that the limits are the only thing wrong with the question (the result is beyond the scope of what is taught to us in high school). I'm trying to find out what the correct limits are. :)

Answer 4

Now I see what is happening in the question. The exponent in the second integral is \((3x-2)^2\) and there is a 3 at the bottom. So we can substitute something like this:\[u +5= 3x-2\]\[u+5 = 2 - 3x\]

Answer 5

I'm going to post what I ended up with: had the limits on the second integral been 1/3 and -8/3, it would have been simpler.

Answer 6

Now I imagine that the guy who made the question tried to work backwards and didn't realise that the whole expression has to be positive. He first made the substitution\[u+5= 2-3x\]Which would land us with this expression:\[\int_{-4}^5 e^{(x+2)^2}dx - \int_{a}^{b}e^{(u+2)^2}du\]Now he wanted \(a\) to be \(-4\) and \(b\) to be \(5\).

Answer 7

How is that going to simplify anything ?

Answer 8

\[u = -3 - 3x\]For \(a\):\[a=-3-3(1/3)=-4\]Indeed, we get -4!

For \(b\):

\(5 = - 3 - 3x\)
Basically, he's trying to find the original upper limit in the question (given as 2/3). Now he may have made a mistake like this: forgetting the whole negative sign.\[5 = 3 + 3x\]

Answer 9

you still need to use error function to evaluate the integral right ?

Answer 10

If you look carefully, \(a = -4\) and \(b = 5\) will make the whole expression zero.

Answer 11

Oh I see now...

Answer 12

Which is why 2/3 was mistakenly placed as the upper limit.

Answer 13

Nice!
http://www.wolframalpha.com/input/?i=%5Cint_%7B-4%7D%5E5+e%5E%7B%28x%2B5%29%5E2+%7Ddx+%2B+3%5Cint_%7B1%2F3%7D%5E%7B-8%2F3%7De%5E%7B9%28x-2%2F3%29%5E2%7Ddx

Answer 14

Exactly.

So the guy who framed the question made lots of mistakes.

Answer 15

Is there a way to get W|A to solve this equation?\[\int_{-4}^5 e^{(x+5)^2 }dx + 3\int_{1/3}^{t}e^{9(x-2/3)^2}dx=10\]I just arbitrarily chose 10 there. I want to solve for \(t\).

Answer 16

Hey I'm off to watching my favourite show :)

Answer 17

http://www.wolframalpha.com/input/?i=solve+1%2F2+sqrt%28pi%29+%28erfi%281%29-erfi%282-3+t%29%29%2B1%2F2+sqrt%28pi%29+%28erfi%2810%29-erfi%281%29%29+%3D+10

Answer 18

http://www.wolframalpha.com/input/?i=solve+%5Cint_%7B-4%7D%5E5+e%5E%7B%28x%2B5%29%5E2+%7Ddx+%2B+3%5Cint_%7B1%2F3%7D%5E%7Bt%7De%5E%7B9%28x-2%2F3%29%5E2%7Ddx%3D10

Answer 19

that looks really nasty...

Answer 20

what show ?