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?

C: $\displaystyle -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx = 3A$(

Is true:

$\displaystyle

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

$

by part B.

$\displaystyle

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

$

Middle integral is equal to $\displaystyle A$ is given in the question

Finally:

$\displaystyle

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

$.

So:

$\displaystyle

-\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx=3A

$.

RonL