[[Christmas]] New Year Streak 85/88: Calculus Rush #3

Calculus Level 5

A function defined over all reals, f ( x ) , f(x), satisfies the below conditions:

1. For − 1 ≤ x < 4 -1\le x < 4 we have f ( x ) = { 2 x e x 2 − 1 + 2 ( − 1 ≤ x < 1 ) 2 x + 2 ( 1 ≤ x < 2 ) 6 ∣ x − 3 ∣ ( 2 ≤ x < 4 ) . f(x)=\begin{cases}2x e^{x^2-1}+2& (-1\le x < 1) \\ 2x+2& (1\le x <2) \\ 6|x-3| & (2\le x<4) \end{cases}.

2. For all integer k , k, we have f ( x + 5 k ) − f ( x ) = 6 k ( − 1 ≤ x < 4 ) . f(x+5k)-f(x)=6k~(-1\le x<4).

For a real t , t, let g ( t ) g(t) be the sum of all the real roots for x x for the equation f ( x ) = t . f(x)=t.

Then, ∑ n = 1 5 ∫ 6 n − 4 6 n − 2 g ( t ) d t = p + q e , \displaystyle \sum_{n=1}^{5} \int_{6n-4}^{6n-2}g(t)dt = p+\frac{q}{e}, for two rational numbers p , q . p,~q.

Find the value of p + q . p+q.

This problem is a part of <Christmas Streak 2017> series .


The answer is 370.0.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Dec 31, 2017

Let h ( x ) = 2 x e x 2 − 1 + 2 ( 0 ≤ x < 1 ) , h(x)=2xe^{x^2-1}+2~(0\le x <1), then g ( t ) = h − 1 ( t ) + 6 ( 2 ≤ t < 4 ) . g(t)=h^{-1}(t)+6~(2\le t<4). Therefore,

∑ k = 1 5 ∫ 6 n − 4 6 n − 2 g ( t ) d t = ∫ 2 4 { 5 g ( t ) + 150 } d t = 5 ∫ 2 4 h − 1 ( t ) d t + 360 = − 5 [ e x 2 − 1 + 2 x ] 0 1 + 380 = 365 + 5 e . \sum_{k=1}^{5} \int_{6n-4}^{6n-2}g(t)dt \\ = \int_{2}^{4} \{5g(t)+150\}dt \\ = 5 \int_{2}^{4} h^{-1}(t)dt + 360 \\ =-5\left[e^{x^2-1}+2x\right]_0^1 + 380 \\ = 365+\frac{5}{e}.

∴ p + q = 365 + 5 = 370 . \therefore~p+q=365+5=\boxed{370}.

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