Google
Câu hỏi: Giả sử $\int_0^1 f(x) \mathrm{d} x=6$ và $\int_0^5 f(u) \mathrm{d} u=13$. Tổng $\int_1^3 f(t) \mathrm{d} t+\int_3^5 f(z) \mathrm{dz}$ bằng
A. -12 .
B. 12 .
C. 7.
D. -6 .
A. -12 .
B. 12 .
C. 7.
D. -6 .
Ta có: $\int_1^3 f(t) \mathrm{d} t+\int_3^5 f(z) \mathrm{d} z=\int_1^3 f(t) \mathrm{d} t+\int_3^5 f(t) \mathrm{d} t=\int_1^5 f(t) \mathrm{d} t$.
Mặt khác: $\int_0^1 f(x) \mathrm{d} x=6 \Rightarrow \int_0^1 f(t) \mathrm{dt}=6$ và $\int_0^5 f(u) \mathrm{d} u=13 \Rightarrow \int_0^5 f(t) \mathrm{d} t=13$.
Ta có: $\int_0^5 f(t) \mathrm{d} t=\int_0^1 f(t) \mathrm{d} t+\int_1^5 f(t) \mathrm{d} t \Rightarrow \int_1^5 f(t) \mathrm{d} t=\int_0^5 f(t) \mathrm{d} t-\int_0^1 f(t) \mathrm{d} t=13-6=7$.
Mặt khác: $\int_0^1 f(x) \mathrm{d} x=6 \Rightarrow \int_0^1 f(t) \mathrm{dt}=6$ và $\int_0^5 f(u) \mathrm{d} u=13 \Rightarrow \int_0^5 f(t) \mathrm{d} t=13$.
Ta có: $\int_0^5 f(t) \mathrm{d} t=\int_0^1 f(t) \mathrm{d} t+\int_1^5 f(t) \mathrm{d} t \Rightarrow \int_1^5 f(t) \mathrm{d} t=\int_0^5 f(t) \mathrm{d} t-\int_0^1 f(t) \mathrm{d} t=13-6=7$.
Đáp án C.