2007
Problem - 4757
Let function $f(x)$ satisfy: $$\int^1_0 3f (x) dx +\int^2_1 2f (x) dx = 7$$
and $$\int^2_0 f (x) dx + \int^2_1 f (x) dx = 1$$
Find the value of $$\int^2_0 f (x) dx$$
Let $u = \int_0^1f(x)dx$ and $v=\int_1^2f(x)dx$. Then the given relations are equivalent to $3u + 2v = 7$ and $(u+v) +v =1$. Solving this system leads to $u=3$ and $v=-1$. Hence, $$\int^2_0 f (x) dx= u+v = \boxed{2}$$