Cho \(\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x=-2\) và \(\displaystyle\int\limits_1^5 2f(x)\mathrm{\,d}x=6\), khi đó \(\displaystyle\int\limits_0^5 f(x)\mathrm{\,d}x\) bằng
Ta có \(\displaystyle\int\limits_1^5 2f(x)\mathrm{\,d}x=2\displaystyle\int\limits_1^5 f(x)\mathrm{\,d}x=6\) \(\Leftrightarrow\displaystyle\int\limits_1^5 f(x)\mathrm{\,d}x=3\).
Do đó $$\begin{align*}\displaystyle\int\limits_0^5 f(x)\mathrm{\,d}x&=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x+\displaystyle\int\limits_1^5 f(x)\mathrm{\,d}x\\ &=-2+3=1.\end{align*}$$