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

Explanation:

Step1: Use integral property for sum

For \(\int_{0}^{2}(f(x)+g(x))dx\), by \(\int_{a}^{b}(u(x)+v(x))dx=\int_{a}^{b}u(x)dx+\int_{a}^{b}v(x)dx\), we have \(\int_{0}^{2}f(x)dx+\int_{0}^{2}g(x)dx\). Substitute \(\int_{0}^{2}f(x)dx = - 4\) and \(\int_{0}^{2}g(x)dx=-6\), so \(-4+( - 6)=-10\).

Step2: Use integral property for difference and sum

For \(\int_{0}^{3}(f(x)-g(x))dx=\int_{0}^{3}f(x)dx-\int_{0}^{3}g(x)dx\). First, find \(\int_{0}^{3}g(x)dx=\int_{0}^{2}g(x)dx+\int_{2}^{3}g(x)dx=-6+( - 2)=-8\). Then \(\int_{0}^{3}f(x)dx=-11\), so \(-11-( - 8)=-3\).

Step3: Use integral linearity

For \(\int_{2}^{3}(3f(x)+2g(x))dx = 3\int_{2}^{3}f(x)dx+2\int_{2}^{3}g(x)dx\). Since \(\int_{0}^{3}f(x)dx=\int_{0}^{2}f(x)dx+\int_{2}^{3}f(x)dx\), then \(\int_{2}^{3}f(x)dx=\int_{0}^{3}f(x)dx-\int_{0}^{2}f(x)dx=-11-( - 4)=-7\). And \(\int_{2}^{3}g(x)dx=-2\). So \(3\times(-7)+2\times(-2)=-21 - 4=-25\).

Answer:

a. \(-10\)
b. \(-3\)
c. \(-25\)