Answer:
Given that \( f(x) \) is an odd function, we know that \( f(-x) = -f(x) \).
We are given:
\[ \int_a^b f(x) \, dx = -3 \]
To find \( \int_{-a}^{-b/2} f(x) \, dx \), we can use substitution.
Let \( u = -x \), then \( du = -dx \) and the limits of integration change as follows:
\[ u(-a) = -(-a) = a \]
\[ u(-b/2) = -(-b/2) = b/2 \]
Thus, the integral becomes:
\[ \int_{-a}^{-b/2} f(x) \, dx = -\int_a^{b/2} f(-u) \, du \]
Because \( f(-x) = -f(x) \), we have \( f(-u) = -f(u) \).
So, the integral becomes:
\[ -\int_a^{b/2} (-f(u)) \, du = \int_a^{b/2} f(u) \, du \]
Now, using the given integral:
\[ \int_a^{b/2} f(u) \, du = -3 \]
Thus, \( \int_{-a}^{-b/2} f(x) \, dx \) equals \( -3 \).
