Let \(f\) and \(g\) be continous functions such that \(\int\limits_0^5 f(x)   dx = 18 \), \(\int\limits_0^5 \dfrac{1}{3}g(x)   dx = 4 \), and \(\int\limits_3^5 (f(x)-g(x))   dx = 2 \). What is the value of \(\int\limits_0^3 (f(x)-g(x))   dx = 18 \) ?

Summary Submit

Use properties of integrals:

$$ \int\limits_0^5 \frac{1}{3}g(x)   dx = 4 $$ $$ \frac{1}{3}\int\limits_0^5 g(x)   dx = 4 $$ $$ \int\limits_0^5 g(x)   dx = 12 $$ $$ \int\limits_0^5 f(x)   dx - \int\limits_0^5 g(x)   dx = \int\limits_0^5 (f(x)-g(x))   dx = 18-12=6 $$ $$ \int\limits_0^3 (f(x)-g(x))   dx = \int\limits_0^5 (f(x)-g(x))   dx - \int\limits_3^5 (f(x)-g(x))   dx $$ $$ \int\limits_0^3 (f(x)-g(x))   dx = 6-2 $$ $$ =\boxed{4} $$