# Isn't the question wrong?

We have to prove that

$$\displaystyle \int _{ 0 }^{ a }{ { e }^{ ax-{ x }^{ 2 } } } dx\quad ={ e }^{ \frac { { a }^{ 2 } }{ 4 } }\int _{ 0 }^{ a }{ { e }^{ \frac { -{ x }^{ 2 } }{ 4 } } } dx$$

I used the property:

$\displaystyle \int _{ 0 }^{ a }{ f(x) } dx\quad =\quad \int _{ 0 }^{ a }{ f(a-x) } dx$

for RHS.

Therefore I got

$\displaystyle { e }^{ \frac { { a }^{ 2 } }{ 4 } }\int _{ 0 }^{ a }{ { e }^{ \frac { -{ x }^{ 2 } }{ 4 } } } dx\quad =\quad \int _{ 0 }^{ a }{ { e }^{ \frac { 2ax-{ x }^{ 2 } }{ 4 } } } dx$ Please correct me if I'm wrong. If I'm right please do mention it.

6 years, 1 month ago

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You haven't proven or disproven the statement.

E.g. If you are asked to show that $2 + 2 = 2 \times 2$, then saying that $2 + 2 = 4$ doesn't mean that the initial statement must be wrong.

Staff - 6 years, 1 month ago

Sir if the question is right can you please prove it?

- 6 years, 1 month ago

- 6 years, 1 month ago

How did you get the right hand side of the equation?

- 6 years, 1 month ago

It was a proof question

- 6 years, 1 month ago

No not right. Consider completing the square for $ax- x^2$. Then split the integral into two: one from $0$ to $\frac a2$ and the other from $\frac a2$ to $a$.

- 6 years, 1 month ago

Can you please explain why to split the integral?

- 6 years, 1 month ago

It's easier.

- 6 years, 1 month ago

If u don't mind can u post the solution. I'm getting a integral of $e^{x^2}$. Please

- 6 years, 1 month ago

Yes that's the point, you want to isolate $e^{x^2}$. Have you got the equation: $ax-x^2 = -\left(x-\frac a2\right)^2- \frac{a^2}4$?

- 6 years, 1 month ago

Yes. Can you solve the integral please

- 6 years, 1 month ago

Let $y = x - \frac a 2$. Change the upper and lower limits. Now, what's the last step?

- 6 years, 1 month ago

But that does not work out. I guess the question is wrong

- 6 years, 1 month ago

What have you tried? Tell me where you got stuck.

- 6 years, 1 month ago