Question

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

Answer 1

\[\int\limits_{0}^{5}f(x)dx=8\]

Answer 2

\[\int\limits_{0}^{5}g(x)=4\]

Answer 3

find\[\int\limits_{-5}^{5}[f(x)+g(x)]dx\]

Answer 4

@wio

Answer 5

\[\int\limits_{-5}^{0}[f(x)+g(x)] +\int\limits_{0}^{5}[f(x)+g(x)] dx\]

Answer 6

@abb0t

Answer 7

after that i have no idea

Answer 8

i know the 2nd part 12

Answer 9

-+12

Answer 10

well even mean f(-x)=f(x) will that mean\[\int\limits_{-5}^{0}f(x)=-8\]

Answer 11

because\[-\int\limits_{5}^{0}f(x)=-8 \]

Answer 12

even aplies if negative in f(x)? f(-x)=f(x)

Answer 13

Why should \[-\int\limits_{5}^{0}f(x)=-8\]be true? I think this should be 8 and not -8.

Answer 14

because that is the rule right to flip flop - should be their

Answer 15

\[\int_0^5 f(x)=8\]\[\int_5^0 f(x)=-8\]\[-\int_5^0 f(x)=8\]

Answer 16

\[-\int\limits_{-5}^{0}f(x)=\]

Answer 17

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.

Answer 18

I drove by to say hello to @KingGeorge!

Answer 19

Since \(f(x)=f(-x)\), we have that \[\begin{aligned}
\int\limits_{-5}^0f(x)\;dx&=\int\limits_{-5}^0f(x)\;dx \\
&=\int\limits_{-5}^0f(-x)\;dx\\
&=\int\limits_{5}^0f(-u)\;(-du)\qquad\text{this is a u-sub for }u=-x.\\
&=\int\limits_0^5f(-u)\;du\\
&=\int\limits_0^5f(u)\;du\\
&=8
\end{aligned}\]

Answer 20

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 \[-\int\limits_0^5g(u)\;du=-4\]

Answer 21

Did this all make sense?

And hello to you too @zepp!

Answer 22

i i got it i did little bit different i took u=-x and then -du=dx

Answer 23

There is a typo in my work two posts above. The second line should not be there. (The line that reads \(=\int_{-5}^0f(-x)\;dx\)).

Answer 24

That works as well.