Question

let
int_{0}^{2} f(x) d x=-4, quad int_{0}^{3} f(x) d x=-11,
int_{0}^{2} g(x) d x=-6, quad int_{2}^{3} g(x) d x=-2.
use these values to evaluate the given definite integrals.
a. ( int_{0}^{2}(f(x)+g(x)) d x= )
b. ( int_{0}^{3}(f(x)-g(x)) d x= )
c. ( int_{2}^{3}(3 f(x)+2 g(x)) d x= )
d. find the value ( a ) such that
int_{0}^{3}(a f(x)+g(x)) d x=0.
( a= )

Explanation:

Step1: Apply integral properties

Use the properties \(\int_{0}^{3}(af(x)+g(x))dx=a\int_{0}^{3}f(x)dx+\int_{0}^{3}g(x)dx\).

Step2: Substitute known values

We know \(\int_{0}^{3}f(x)dx = - 11\) and \(\int_{0}^{3}g(x)dx=\int_{0}^{2}g(x)dx+\int_{2}^{3}g(x)dx=-6+( - 2)=-8\).
So the equation becomes \(a\times(-11)+(-8)=0\).

Step3: Solve for \(a\)

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Answer:

\(a =-\frac{8}{11}\)