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# 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.

1 year, 9 months 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 · 1 year, 9 months ago

Sir if the question is right can you please prove it? · 1 year, 9 months ago

@Calvin Lin @Pi Han Goh · 1 year, 9 months ago

How did you get the right hand side of the equation? · 1 year, 9 months ago

It was a proof question · 1 year, 9 months 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 $$a$$. · 1 year, 9 months ago

Can you please explain why to split the integral? · 1 year, 9 months ago

It's easier. · 1 year, 9 months ago

If u don't mind can u post the solution. I'm getting a integral of $$e^{x^2}$$. Please · 1 year, 9 months ago

Yes that's the point, you want to isolate $$e^{x^2}$$. Have you got the equation: $$ax-x^2 = -\left(x-\frac a2\right)^2- \frac{a^2}4$$? · 1 year, 9 months ago

Yes. Can you solve the integral please · 1 year, 9 months ago

Let $$y = x - \frac a 2$$. Change the upper and lower limits. Now, what's the last step? · 1 year, 9 months ago

But that does not work out. I guess the question is wrong · 1 year, 9 months ago