Cho hàm số $y=f(x) = x^{2025} + x$ và $y= g(x) = x^{2025 }+ x^{2024}+x + 2025$. Khi đó
a) $\displaystyle\int\limits_{-1000}^{1000} f(x) \mathrm{d} x =0$.
b) $\displaystyle \int\limits_0^1 g'(x) \mathrm{d} x =4$.
c) $\displaystyle \int\limits_{0}^{1000} f^{\prime}(x) \mathrm{d} x= 1000^{2025}+1000$.
d) $\displaystyle \int\limits_{-2024}^{2024} \left[g(x) – f(x) \right]\mathrm{d} x= 2 \cdot \left(\dfrac{2024^{2025}}{2025} + 2025 \cdot 2024\right).$.
Lời giải:
(Đúng) $\displaystyle\int\limits_{-1000}^{1000} f(x) \mathrm{d} x =0$ (Vì): Ta có hàm số $y= f(x)$ là hàm số lẻ trên $\mathbb{R}$ nên $\displaystyle\int\limits_{-1000}^{1000} f(x) \mathrm{d} x =0$. (Sai) $\displaystyle \int\limits_0^1 g'(x) \mathrm{d} x =4$ (Vì): Ta có $\displaystyle \int\limits_0^1 g'(x) \mathrm{d} x = g(x) \biggl|_0^1 = g(1)- g(0)= 2028 – 2025 =3.$ (Đúng) $\displaystyle \int\limits_{0}^{1000} f^{\prime}(x) \mathrm{d} x= 1000^{2025}+1000$ (Vì): Ta có $\displaystyle \int\limits_{0}^{1000} f^{\prime}(x) \mathrm{d} x= f(1000) – f(0) = 1000^{2025}+1000$. (Đúng) $\displaystyle \int\limits_{-2024}^{2024} \left[g(x) – f(x) \right]\mathrm{d} x= 2 \cdot \left(\dfrac{2024^{2025}}{2025} + 2025 \cdot 2024\right).$ (Vì): Ta có $\begin{array}{l} \displaystyle \int\limits_{-2024}^{2024} \left[g(x) – f(x) \right]\mathrm{d} x = \displaystyle \int\limits_{-2024}^{2024} (x^{2024}+ 2025) \mathrm{d} x\\ = 2 \displaystyle \int\limits_{0}^{2024} (x^{2024}+ 2025) \mathrm{d} x\\ = 2 \cdot \left(\dfrac{2024^{2025}}{2025} + 2025 \cdot 2024\right). \end{array}$ PHẦN III.