Calculus Problem by LOGA 4

Lemma : Let $f$ be a positive and continuous function on the interval $[a,~b]$ . Then $\lim_{n\to\infty}\left(\int_{a}^{b}\left[f(x)\right]^{n}dx\right)^{\frac{1}{n}} =M$ where $M$ is its maximum value on the interval $[a,~b]$ .

Proof : Let $c$ be any number between $0$ and $M$ . Since $f$ is continuous, there exist $\alpha_c$ and $\beta_c$ such that $\alpha_c\leq x\leq\beta_c~\Rightarrow~f(x)\geq c$ Therefore, $c^{n}(\beta_c-\alpha_c)\leq\int_{\alpha_c}^{\beta_c} [f(x)]^{n}dx\leq\int_{a}^{b} [f(x)]^{n}dx\leq M^{n}(b-a)$ $c(\beta_c-\alpha_c)^{\frac{1}{n}}\leq\left(\int_{a}^{b} [f(x)]^{n}dx\right)^{\frac{1}{n}}\leq M(b-a)^{\frac{1}{n}}$ $c\leq\lim_{n \rightarrow \infty}\left(\int_{a}^{b} [f(x)]^{n}dx\right)^{\frac{1}{n}}\leq M$ As $c$ is any number between $0$ and $M$ , we can conclude that $\lim_{n\to\infty}\left(\int_{a}^{b}\left[f(x)\right]^{n}dx\right)^{\frac{1}{n}} =M$

By the lemma, we can know that $g(t)$ is the maximum value of $f(x)$ on the interval $[t-1,~t+1]$ . Since $f'(x)=-x^{2019}e^{-x}(x-2020)$ , $f$ attains its maximum value at $2020$ . Therefore $a_0=2019,~b_0=2021,$ and $a_0+2b_0=6061$ .