Given a function \(f(x)\), you are told that \[\int_0^1 3f(x)\,dx+\int_1^2 2f(x)\,dx=7,\] \[\int_0^2 f(x)\,dx+\int_1^2 f(x) \,dx=1.\] It follows that \(\int_0^2 f(x) \,dx\) equals

  1. \(-1\),

  2. \(0\),

  3. \(\dfrac{1}{2}\),

  4. \(2\).

Note that \(\int_0^1 af(x)\,dx=a\int_0^1 f(x)\,dx\), where \(a\) is a constant, and \(\int_0^2 f(x)\,dx=\int_0^1 f(x)\,dx +\int_1^2f(x)\,dx\).

Write \(A=\int_0^1 f(x)\,dx\) and \(B=\int_1^2f(x)\,dx\). So our two equations become the simultaneous equations \[3A+2B=7\] and \[A+2B=1.\] Subtracting the second from the first gives \(2A=6\), so \(A=3\) and \(B=-1\).

Now \(\int_0^2 f(x) \,dx=A+B=2\).

The answer is (d).