Cho hai tích phân - 2^5 f( x )dx = 8 và tícch phân 5^ - 2 g( x )dx= 3. Tính I = tích phân - 2^5 f(x)

2/30

Cho hai tích phân \(\int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}} x = 8\) và \(\int\limits_5^{ - 2} {g\left( x \right){\rm{d}}} x = 3\). Tính \(I = \int\limits_{ - 2}^5 {\left[ {f\left( x \right) - 4g\left( x \right) - 1} \right]{\rm{d}}} x\)

\(13\).

\(27\).

\( - 11\).

\(3\).

Giải thích

Chọn A

\(I = \int\limits_{ - 2}^5 {\left[ {f\left( x \right) - 4g\left( x \right) - 1} \right]{\rm{d}}} x\)\( = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} - \int\limits_{ - 2}^5 {4g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \)\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} - 4\int\limits_{ - 2}^5 {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]

\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} + 4\int\limits_5^{ - 2} {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]\[ = 8 + 4.3 - x\left| \begin{array}{l}5\\ - 2\end{array} \right.\]\[ = 8 + 4.3 - 7\]\[ = 13\].