Giá trị của biểu thức \(T = 3\int\limits_{ – 1}^1 {f\left( {2x – 1} \right)} dx + \int\limits_0^1 {f’\left( {x + 2} \right)} dx + \int\limits_1^2 {f’\left( x \right)} dx\) bằng.
A. \( – 8\)
B. \(6\).
C. \(\frac{{ – 4}}{3}\).
D. \( – 5\).
LỜI GIẢI CHI TIẾT
Ta có: \( – \int\limits_{ – 3}^{ – 1} {f\left( x \right)} dx = 6 \Rightarrow \int\limits_{ – 3}^{ – 1} {f\left( x \right)} dx = – 6\)
Xét: +) \(\int\limits_{ – 1}^1 {f\left( {2x – 1} \right)} dx = \int\limits_{ – 3}^{ – 1} {f\left( t \right)} \frac{{dt}}{2} = \frac{1}{2}\int\limits_{ – 3}^{ – 1} {f\left( x \right)} dx = \frac{1}{2}.\left( { – 6} \right) = – 3\,\,\,\,\,\,\)( đặt \(\,t = 2x – 1\)).
+) \(\int\limits_0^1 {f’\left( {x + 2} \right)} dx = \int\limits_2^3 {f’\left( u \right)} du = \left. {f\left( u \right)} \right|_2^3 = f\left( 3 \right) – f\left( 2 \right)\)( đặt \(u = x + 2\))
+) \(\int\limits_1^2 {f’\left( x \right)} dx = \left. {f\left( x \right)} \right|_1^2 = f\left( 2 \right) – f\left( 1 \right)\)
Khi đó: \(T = \int\limits_{ – 1}^1 {f\left( {2x – 1} \right)} dx + \int\limits_0^1 {f’\left( {x + 2} \right)} dx + \int\limits_1^2 {f’\left( x \right)} dx\)
\( = 3.\left( { – 3} \right) + f\left( 3 \right) – f\left( 2 \right) + f\left( 2 \right) – f\left( 1 \right) = – 9 + 3 – 2 = – 8\)
=======
Trả lời