Cho \(\displaystyle\int\limits\limits_{-2}^2f(x)\mathrm{\,d}x=1\), \(\displaystyle\int\limits\limits_{-2}^4f(t)\mathrm{\,d}t=-4\). Tính \(I=\displaystyle\int\limits\limits_2^4f(y)\mathrm{\,d}y\).
 | \(I=5\) |
 | \(I=3\) |
 | \(I=-3\) |
 | \(I=-5\) |
Chọn phương án D.
Ta có
\(\displaystyle\int\limits\limits_{-2}^2{f(x)} \mathrm{\,d}x=\displaystyle\int\limits\limits_{-2}^2 f(y) \mathrm{\,d}y=1\)
và \(\displaystyle\int\limits\limits_{-2}^4 f(t) \mathrm{\,d}t=\displaystyle\int\limits\limits_{-2}^4 f(y) \mathrm{\,d}y=-4\).
Mà \(\displaystyle\int\limits\limits_{-2}^4 f(y) \mathrm{\,d}y=\displaystyle\int\limits\limits_{-2}^2 f(y) \mathrm{\,d}y+\displaystyle\int\limits\limits_2^4 f(y) \mathrm{\,d}y\).
Do đó $$\begin{align*}I=\displaystyle\int\limits\limits_2^4 f(y) \mathrm{\,d}y&=\displaystyle\int\limits\limits_{-2}^4 f(y) \mathrm{\,d}y-\displaystyle\int\limits\limits_{-2}^2 f(y)\mathrm{\,d}y\\
&=-4-1=-5.\end{align*}$$