Question

the values of several definite integrals are given as follows:
$\int_{0}^{2} f(x) dx = 5$, $\int_{0}^{3} f(x) dx = 7$, $\int_{0}^{2} g(x) dx = -3$, $\int_{2}^{3} g(x) dx = 5$
use these values and properties of definite integrals to evaluate the indicated definite integral.
find a formula for $a$ in terms of $b$ such that $\int_{0}^{3} (af(x) + bg(x)) dx = 0$.

Explanation:

Step1: Split the integral using linearity

$\int_{0}^{3}[af(x)+bg(x)]dx = a\int_{0}^{3}f(x)dx + b\int_{0}^{3}g(x)dx$

Step2: Substitute given values

$\int_{0}^{3}f(x)=7$, $\int_{0}^{3}g(x)=\int_{0}^{2}g(x)+\int_{2}^{3}g(x)=-3+5=2$
So: $a(7) + b(2) = 0$

Step3: Solve for $a$ in terms of $b$

$7a = -2b \implies a = -\frac{2b}{7}$

Answer:

$a = -\frac{2b}{7}$