Question

suppose that $f$ is an even function, $g$ is odd, both are integrable on $-1,1$, and we know that $int_{0}^{1}f(x)dx = 6$, while $int_{0}^{1}g(x)dx = 1.5$. if possible, find $int_{-1}^{1}f(x)+g(x)dx$. if it is not possible, write $np$ for your answer.

Explanation:

Step1: Use integral properties

$\int_{-1}^{1}[f(x)+g(x)]dx=\int_{-1}^{1}f(x)dx+\int_{-1}^{1}g(x)dx$

Step2: Apply even - odd function integral rules

For even $f$, $\int_{-1}^{1}f(x)dx = 2\int_{0}^{1}f(x)dx=2\times6 = 12$. For odd $g$, $\int_{-1}^{1}g(x)dx = 0$.

Step3: Calculate the result

$\int_{-1}^{1}[f(x)+g(x)]dx=12 + 0=12$

Answer:

12