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f(x) is even and g(x) is odd

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Other answers:

\[\int\limits_{-5}^{0}[f(x)+g(x)] +\int\limits_{0}^{5}[f(x)+g(x)] dx\]
after that i have no idea
i know the 2nd part 12
well even mean f(-x)=f(x) will that mean\[\int\limits_{-5}^{0}f(x)=-8\]
because\[-\int\limits_{5}^{0}f(x)=-8 \]
even aplies if negative in f(x)? f(-x)=f(x)
Why should \[-\int\limits_{5}^{0}f(x)=-8\]be true? I think this should be 8 and not -8.
because that is the rule right to flip flop - should be their
\[\int_0^5 f(x)=8\]\[\int_5^0 f(x)=-8\]\[-\int_5^0 f(x)=8\]
But you're certainly on the right path to solving this. You know\[\int\limits_{0}^{5}[f(x)+g(x)] dx=12\]So you just need to find \[\int\limits_{-5}^{0}[f(x)+g(x)]dx =\int\limits_{-5}^0f(x)\;dx+\int\limits_{-5}^0g(x)\;dx\]First, lets start with the f(x) part.
I drove by to say hello to @KingGeorge!
Since \(f(x)=f(-x)\), we have that \[\begin{aligned} \int\limits_{-5}^0f(x)\;dx&=\int\limits_{-5}^0f(x)\;dx \\ &=\int\limits_{-5}^0f(-x)\;dx\\ &=\int\limits_{5}^0f(-u)\;(-du)\qquad\text{this is a u-sub for }u=-x.\\ &=\int\limits_0^5f(-u)\;du\\ &=\int\limits_0^5f(u)\;du\\ &=8 \end{aligned}\]
This is basically the same thing you do for \(g(x)\). Instead we have \(-g(x)=g(-x)\).If we repeat the above argument, we get to the point \[\int\limits_0^5g(-u)\;du\]Substitute for \(-g(u)\), and we get \[-\int\limits_0^5g(u)\;du=-4\]
Did this all make sense? And hello to you too @zepp!
i i got it i did little bit different i took u=-x and then -du=dx
There is a typo in my work two posts above. The second line should not be there. (The line that reads \(=\int_{-5}^0f(-x)\;dx\)).
That works as well.
ohk thx a lot
You're welcome.
i will work on this if i have nay questions can i post on this wall

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