How can I integrate e^{ax^{2}+bx+c} without getting tangled?

  • Thread starter Thread starter noowutah
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
To integrate the expression e^{ax^{2}+bx+c}, completing the square for the quadratic ax^{2}+bx+c is recommended. This involves rewriting it in the form a(x-h)^{2} + k, where h = -b/2a and k is the function evaluated at h. The constant term can then be factored out, simplifying the integration process to e^{a(x-h)^{2}}e^{k}. This method has proven effective for users seeking to avoid complications in their calculations. Overall, completing the square is a key technique for successfully integrating this type of expression.
noowutah
Messages
56
Reaction score
3

Homework Statement



I need to integrate an expression of the form

e^{ax^{2}+bx+c}

Homework Equations



I know that

\int_{a}^{b}e^{-y^{2}}dy=\frac{\sqrt{\pi}}{2}(\mbox{erf}(b)-\mbox{erf}(a))

The Attempt at a Solution



I tried to substitute ax^{2}+bx+c by -y^{2} but I get hopelessly tangled. (PS.: how do get the tex tags to not create an equation environment but stay inline?)
 
Physics news on Phys.org
stlukits said:

Homework Statement



I need to integrate an expression of the form e^{ax^{2}+bx+c} ← [ itex]e^{ax^{2}+bx+c}[ /itex]

Homework Equations



I know that \displaystyle \int_{a}^{b}e^{-y^{2}}dy=\frac{\sqrt{\pi}}{2}(\mbox{erf}(b)-\mbox{erf}(a)) ← [ itex]\displaystyle \int_{a}^{b}e^{-y^{2}}dy=

\frac{\sqrt{\pi}}{2}(\mbox{erf}(b)-\mbox{erf}(a))[ /itex]


The Attempt at a Solution



I tried to substitute ax^{2}+bx+c by -y^{2} but I get hopelessly tangled. (PS.: how do get the tex tags to not create an equation environment but stay inline?)
Use itex & /itex for inline LATEX. Use \displaystyle with that to keep ∫ , fractions, etc. full size. See above.

Try completing the square for ax^{2}+bx+c .
 
Last edited:
  • Like
Likes M4RC
First you need to complete the square on the quadratic. Note that the quadratic formula is derived this way so basically encodes it:
f(x) = ax^2 + b x + c = a(x-h)^2 + k
where h = -b/2a and k = f(h)=ah^2 + bh + c.

Second, note that the constant term in the exponent can be factored out:
e^{a(x-h)^2 + k} = e^{a(x-h)^2}e^k

See where that gets you.
 
  • Like
Likes M4RC
Great help. Let me try it and see where it goes.
 
Thank you, jambaugh. It worked beautifully.
 
stlukits said:
Thank you, jambaugh. It worked beautifully.
You're welcome, glad it worked out well.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
7
Views
2K