Originally Posted by

**scorpion007** Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

$\displaystyle f$ is a continuous, differentiable function with $\displaystyle f(x) = f(-x)$ such that $\displaystyle f(x) \ge 0$ for $\displaystyle x \in [0, a]$ and $\displaystyle f(x) < 0$ for $\displaystyle x \in (a, b]$. It is also known that $\displaystyle \int^{b}_{0} f(x) dx = 0$ and $\displaystyle \int^{a}_{0} f(x) dx = A$. Which of the following is false?

A: $\displaystyle \int^{b}_{-b} f(x) dx = 0$(

Is true, as:

$\displaystyle

\int^{b}_{-b} f(x) dx=\int^{b}_{0} f(x) dx+\int^{0}_{-b} f(x) dx

$

$\displaystyle

=\int^{b}_{0} f(x) dx-\int^{0}_{b} f(-u) du

$

by change of variable $\displaystyle u=-x$ in the second integral

$\displaystyle

=\int^{b}_{0} f(x) dx-\int^{0}_{b} f(u) du

$

by symetry of $\displaystyle f(u)$ about $\displaystyle u=0$, ie $\displaystyle f(u)=f(-u)$

$\displaystyle

=\int^{b}_{0} f(x) dx+\int^{b}_{a} f(u) du=2\int^{b}_{0} f(x) dx=0

$

by interchanging the upper and lower limits of integration.

RonL