Integration when given the exact value of an un-integratable function

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To calculate the integral ∫ e^(-((x-a)^2)/c) dx from negative infinity to infinity, a substitution can be used to relate it to the known integral ∫ e^(-x^2) dx = √π. By substituting y = (x-a)/√c, the integral can be transformed into a standard form, allowing for the exact value to be derived. The limits of integration remain unchanged due to the properties of definite integrals. This approach simplifies the problem and leverages the known result for the Gaussian integral. Understanding this substitution technique is crucial for solving similar integrals.
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given that ∫ e^(-x^2) dx=√∏, calculate the exact value of ∫ e^-((x-a)^2)/c dx

limits of integration for both are from negative infinity to infinity.

I don't even know where to start on this one. I know how to integrate regular functions. How can I find the exact value for an integral I can't even integrate?

Please Help
 
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nick.martinez said:
given that ∫ e^(-x^2) dx=√∏, calculate the exact value of ∫ e^-((x-a)^2)/c dx

limits of integration for both are from negative infinity to infinity.

I don't even know where to start on this one. I know how to integrate regular functions. How can I find the exact value for an integral I can't even integrate?

Please Help

The trick in these problems is to turn the new integral into the known one by a substitution.

Can you think of a sub to turn that integral into \displaystyle \int_{-\infty}^{\infty} e^{-y^2}dy?

I'm using y to avoid confusion here. Keep in mind that in a definite integral, the actual variable of integration doesn't matter (it's a dummy).
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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