Câu hỏi:
Nếu \(\int\limits_1^3 {f\left( x \right){\rm{d}}x = – 2} \) thì \(\int\limits_1^3 {\left[ {2f\left( x \right) – 3{x^2} + 1} \right]\,} {\rm{d}}x\) bằng
A. \( – 30\).
B. \( – 28\).
C. \( – 26\).
D. \( – 27\).
GY:
Ta có \(\int\limits_1^3 {\left[ {2f\left( x \right) – 3{x^2} + 1} \right]\,} {\rm{d}}x = 2\int\limits_1^3 {f\left( x \right){\rm{d}}x + \int\limits_1^3 {\left( { – 3{x^2} + 1} \right){\rm{d}}x} } \).
+ Ta có \(2\int\limits_1^3 {f\left( x \right){\rm{d}}x} = – 4\).
+ Mặt khác \(\int\limits_1^3 {\left( { – 3{x^2} + 1} \right){\rm{d}}x} = \left. {\left( { – {x^3} + x} \right)} \right|_1^3 = – 24\).
Vậy \(\int\limits_1^3 {\left[ {2f\left( x \right) – 3{x^2} + 1} \right]\,} {\rm{d}}x = – 4 – 24 = – 28\).
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