Question

let ( f(x) ) and ( g(x) ) be functions such that ( g(x) leq f(x) ) for all ( x ).
if the area between the curves ( f(x) ) and ( g(x) ) on ( -4,1 ) is 5, and
int_{-4}^{1} f(x) d x=3,
then
int_{-4}^{1} g(x) d x
.equals:
2
0
-2
5
-5

Explanation:

Step1: Relate area to integral difference

Since \(g(x) \leq f(x)\), area = \(\int_{-4}^{1}[f(x)-g(x)]dx = 5\)

Step2: Split the integral

\(\int_{-4}^{1}f(x)dx - \int_{-4}^{1}g(x)dx = 5\)

Step3: Substitute known integral value

\(3 - \int_{-4}^{1}g(x)dx = 5\)

Step4: Solve for the unknown integral

\(\int_{-4}^{1}g(x)dx = 3 - 5 = -2\)

Answer:

-2