Gaussian integral 2016, Part II
The Gaussian integral states that ∫ − ∞ + ∞ e − x 2 d x = π .
Hence or otherwise, evaluate the integral
∫ − ∞ + ∞ e − x 2 + 2 x + 2 0 1 6 d x
If the above integral can be expressed in the form a e b π , where a , b are positive integers, find a + b .
The answer is 2018.
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2 solutions
Similar answer with @Vignesh S 's:
I = ∫ − ∞ + ∞ e − x 2 + 2 x + 2 0 1 6 d x = ∫ − ∞ + ∞ e − ( x 2 − 2 x + 1 ) + 2 0 1 7 d x = e 2 0 1 7 ∫ − ∞ + ∞ e − ( x − 1 ) 2 d x = e 2 0 1 7 ∫ − ∞ + ∞ e − u 2 d u = e 2 0 1 7 π Let u = x − 1 ⟹ d u = d x
⟹ a + b = 1 + 2 0 1 7 = 2 0 1 8
Write − x 2 + 2 x + 2 0 1 6 as − ( x − 1 ) 2 + 2 0 1 7 and substitute x − 1 = t , the limits of the integral will not be affected. Therefore take e 2 0 1 7 outside, since its a constant , then the integral is nothing but the Gaussian integral