Cho hàm số $y=f(x)=\left\{\begin{array}{l} 2x^2-3 \text{ khi }x\ge 1\\ 1-x \text{ khi }x{<}1.\end{array}\right.$
a) $\displaystyle\int\limits_1^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_1^{2024}(2x^2-4)\mathrm{d}x$.
b) $\displaystyle\int\limits_{-2024}^{1}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}(2-x)\mathrm{d}x$.
c) $\displaystyle\int\limits_{-2024}^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}(1-x)\mathrm{d}x+\displaystyle\int\limits_{1}^{2024}(2x^2-3)\mathrm{d}x$.
d) $\displaystyle\int\limits_{-2024}^{0}f(x)\mathrm{d}x=2050312$.
Lời giải:
(Sai) $\displaystyle\int\limits_1^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_1^{2024}(2x^2-4)\mathrm{d}x$ (Vì): Vì $f(x)=2x^2-3$ khi $1\le x\le 2024$. (Sai) $\displaystyle\int\limits_{-2024}^{1}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}(2-x)\mathrm{d}x$ (Vì): Vì $f(x)=1-x$ khi $-2024\le x{<}1$. (Đúng) $\displaystyle\int\limits_{-2024}^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}(1-x)\mathrm{d}x+\displaystyle\int\limits_{1}^{2024}(2x^2-3)\mathrm{d}x$ (Vì): Vì $\displaystyle\int\limits_{-2024}^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}f(x)\mathrm{d}x+\displaystyle\int\limits_{1}^{2024}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{1}(1-x)\mathrm{d}x+\displaystyle\int\limits_{1}^{2024}(2x^2-3)\mathrm{d}x$. (Đúng) $\displaystyle\int\limits_{-2024}^{0}f(x)\mathrm{d}x=2050312$ (Vì): Vì $\displaystyle\int\limits_{-2024}^{0}f(x)\mathrm{d}x=\displaystyle\int\limits_{-2024}^{0}(1-x)\mathrm{d}x=\left(x-\dfrac{x^2}{2}\right)\bigg|_{-2024}^0=2050312$.