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Sorry this is my first time asking in forum because, please do critics how i ask question, give me some tips so I can be more clear to ask question.

If :

$\int_{-2}^2g(x)(f(x)+1)dx=8$

$\int_{-1}^2g(x)dx=5$

With : $f(x)$ is an odd function ($f(-x) = -f(x)$),

and $g(x)$ is an even function ($f(-x) = f(x) $)

So what is $\int_{-1}^0 g(x)dx=?$

Many thanks, Could you also explain ?

Chrystian N
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2 Answers2

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Here's a hint to get you started:

$$ \int_{-2}^2 g(x) (f(x) + 1) ~dx = \int_{-2}^0 g(x) (f(x) + 1) ~dx + \int_{0}^2 g(x) (f(x) + 1) ~dx = I_1 + I_2 $$ Now substitute $u = -x$ in the first integral $I_1$ to get $$ I_1 = \int_0^2 g(-u) (f(-u) + 1) (-1) ~ du $$ Can you use the facts about even-ness and odd-ness of $f$ and $g$ to simplify this expression?

John Hughes
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Since the integral of any odd function over an interval $[-a,a]$ is $0$, the first equation tells us that $$\int_{-2}^2 g(x)\,dx=8$$ Adding what we know via the second equation, it follows that $$\int_{-2}^{-1} g(x)\,dx=3$$

Now, since $g(x)$ is even we know that $$\int_{-2}^{-1} g(x)\,dx=3 = \int_{1}^{2} g(x)\,dx$$ and that $$\int_{-1}^{0} g(x)\,dx=\int_{0}^{1} g(x)\,dx$$

Can you take it from there?

lulu
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  • thanks, i know now odd function over interval -a to a will be zero, but can i asume since even function is symetric on y axis that $\int_{-2}^2g(x)dx=2\int_0^2g(x)dx$ and $\int_{-2}^0g(x)dx=\int_0^2g(x)dx$ – Chrystian N Nov 29 '17 at 13:19
  • Yes, both of those statements are correct. – lulu Nov 29 '17 at 13:23