Question

let
int_{0}^{8} f(x) d x=4, quad int_{0}^{3} f(x) d x=11,
int_{0}^{8} g(x) d x=6, quad int_{0}^{3} g(x) d x=4
use these values to evaluate the given definite integrals.
a. ( int_{0}^{8}(f(x)+g(x)) d x= )
b. ( int_{0}^{3}(f(x)-g(x)) d x= )
c. ( int_{0}^{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: Use integral properties

For part a:
$$\int_{0}^{3}(f(x)+g(x))dx=\int_{0}^{3}f(x)dx+\int_{0}^{3}g(x)dx$$
Given $\int_{0}^{3}f(x)dx = 11$ and $\int_{0}^{3}g(x)dx = 4$, so $\int_{0}^{3}(f(x)+g(x))dx=11 + 4=15$.

Step2: Use integral properties

For part b:
$$\int_{0}^{3}(f(x)-g(x))dx=\int_{0}^{3}f(x)dx-\int_{0}^{3}g(x)dx$$
Substitute the values: $11-4 = 7$.

Step3: Use integral properties

For part c:
$$\int_{0}^{3}(3f(x)+2g(x))dx=3\int_{0}^{3}f(x)dx+2\int_{0}^{3}g(x)dx$$
$3\times11+2\times4=33 + 8=41$.

Step4: Use integral properties

For part d:
$$\int_{0}^{3}(af(x)+g(x))dx=a\int_{0}^{3}f(x)dx+\int_{0}^{3}g(x)dx$$
Set $a\times11+4 = 0$. Then $11a=-4$, so $a=-\frac{4}{11}$.

Answer:

a. $15$
b. $7$
c. $41$
d. $-\frac{4}{11}$