Câu hỏi:
Cho \(\int\limits_{ – 1}^2 {f\left( x \right){\rm{d}}x} = 2\) và \(\int\limits_{ – 1}^2 {g\left( x \right){\rm{d}}x} = – 1\). Khi đó \(I = \int\limits_{ – 1}^2 {\left[ {x + 2f\left( x \right) – 3g\left( x \right)} \right]{\rm{d}}x} \) bằng
A. \(I = \frac{{17}}{2}\).
B. \(I = \frac{7}{2}\).
C. \(I = \frac{5}{2}\).
D. \(I = \frac{{11}}{2}\).
GY:
Ta có: \(I = \int\limits_{ – 1}^2 {\left[ {x + 2f\left( x \right) – 3g\left( x \right)} \right]{\rm{d}}x} \)
\( = \left. {\frac{{{x^2}}}{2}} \right|_{ – 1}^2 + 2\int\limits_{ – 1}^2 {f\left( x \right){\rm{d}}x} – 3\int\limits_{ – 1}^2 {g\left( x \right){\rm{d}}x} \)
\( = \)\(\frac{3}{2} + 2.2 – 3\left( { – 1} \right)\)
\( = \)\(\frac{{17}}{2}\)
Vậy \(I = \frac{{17}}{2}\).
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