Definite Integrals Question 20

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 $\int\limits_0^3 (f(x)-g(x)) dx = 18$ ?

4

8

12

16

Approach

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}