## anonymous 3 years ago How do I do this?

1. anonymous

$\int\limits_{0}^{9}f(x)dx=37; \int\limits_{0}^{9}g(x)dx=16$

2. anonymous

Find: $\int\limits_{0}^{9}[2f(x)+3g(x)]dx$

3. phi

the integral is "linear" in other words, for a constant "a" $\int\limits_{0}^{9} a f(x)dx = a\int\limits_{0}^{9} f(x)dx$

4. anonymous

so in this case $2\int\limits_{0}^{9} x.dx$??

5. phi

and you can separate integrals (assuming they have the same limits $\int\limits_{0}^{9} f(x) + g(x) dx= \int\limits_{0}^{9} f(x)dx + \int\limits_{0}^{9} g(x)dx$

6. phi

not x , f(x) so separate the problem into $2\int\limits_{0}^{9} f(x)dx + 3\int\limits_{0}^{9} g(x)dx$

7. anonymous

yea right! so $37x(9,0)+16x(9,0)$

8. phi

I don't know about the x(9,0) part just replace the integral with the number they say it is equal to $2\cancel{(\int\limits_{0}^{9} f(x)dx)}37 + 3\int\limits_{0}^{9} g(x)dx$ and do the same for the other, it is 16

9. phi

I am trying to show you sub in 37 for the result of the first integral

10. anonymous

ohh alright, i got it!

11. anonymous

so its 2(37) + 3(16) = 122

12. anonymous

thanks @phi :)