Giả sử lim f(x) dx = 6

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Giả sử \[\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 6\] và \[\int\limits_0^5 {f\left( u \right){\rm{d}}u = 13} \]. Tổng \[\int\limits_1^3 {f\left( t \right){\rm{d}}t} + \int\limits_3^5 {f\left( z \right){\rm{dz}}} \] bằng:

\( - 6\).

\( - 12\).

\(12\).

\(7\).

Giải thích

Ta có: \[\int\limits_1^3 {f\left( t \right){\rm{d}}t + } \int\limits_3^5 {f\left( z \right){\rm{d}}z} = \int\limits_1^3 {f\left( t \right){\rm{d}}t + } \int\limits_3^5 {f\left( t \right){\rm{d}}t} = \int\limits_1^5 {f\left( t \right){\rm{d}}t} \].

Mặt khác: \[\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 6 \Rightarrow \int\limits_0^1 {f\left( t \right){\rm{dt}}} = 6\] và \[\int\limits_0^5 {f\left( u \right){\rm{d}}u} = 13 \Rightarrow \int\limits_0^5 {f\left( t \right){\rm{d}}t} = 13\].

Ta có: \[\int\limits_0^5 {f\left( t \right){\rm{d}}t = } \int\limits_0^1 {f\left( t \right){\rm{d}}t + } \int\limits_1^5 {f\left( t \right){\rm{d}}t} \] \[ \Rightarrow \int\limits_1^5 {f\left( t \right){\rm{d}}t} = \int\limits_0^5 {f\left( t \right){\rm{d}}t} - \int\limits_0^1 {f\left( t \right){\rm{d}}t} = 13 - 6 = 7\]. Chọn D