Not the same

Calculus Level 4

Let f ( x ) = ( ∫ 0 x e − t 2 d t ) 2 \displaystyle f(x) = \left( \int_0^x e^{-t^2} \, dt \right)^2 and g ( x ) = ∫ 0 1 e − x 2 ( t 2 + 1 ) t 2 + 1 d t \displaystyle g(x) = \int_0^1 \dfrac{e^{-x^2 (t^2 + 1)}}{t^2 + 1} \, dt . Find f ( x ) + g ( x ) f(x) + g(x) to 3 decimal places.


The answer is 0.7853.

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

Mark Hennings
Apr 27, 2017

Differentiating inside the integral, and then changing variables, g ′ ( x ) = − 2 x ∫ 0 1 e − x 2 ( t 2 + 1 ) d t = − 2 x e − x 2 ∫ 0 1 e − x 2 t 2 d t = − 2 e − x 2 ∫ 0 x e − t 2 d t = − f ′ ( x ) g'(x) \; = \; -2x \int_0^1 e^{-x^2(t^2+1)}\,dt \; = \; -2xe^{-x^2} \int_0^1 e^{-x^2t^2}\,dt \; = \; -2e^{-x^2}\int_0^x e^{-t^2}\,dt \; = \; -f'(x) so that f ′ ( x ) + g ′ ( x ) = 0 f'(x) + g'(x) = 0 for all x x , and hence f ( x ) + g ( x ) = f ( 0 ) + g ( 0 ) = ∫ 0 1 d t t 2 + 1 = 1 4 π f(x) + g(x) \; = \; f(0) + g(0) \; = \; \int_0^1 \frac{dt}{t^2+1} \; = \; \boxed{\tfrac14\pi}

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