Question

How do you integrate `3x^2-5x+9` from 0 to 7?

Answer 1

You can integrate separately each function and then, once you found the anti-derivative `F(x)`, substitute `0` and `7`:
`int_0^7f(x)dx=F(7)-F(0)`
In your case:
`int_0^7(3x^2-5x+9)dx=`
`=int_0^7(3x^2)dx-int_0^7(5x)dx+int_0^7(9)dx=`

Remember that `intkx^ndx=kx^(n+1)/(n+1)+c` where `k` is a constant and `intkdx=kx+c`;

you have:

`=3x^3/3-5x^2/2+9x]_0^7=` you can now substitute the extremes of integration:

`=(7^3-5*7^2/2+9*7)-(0-0+0)=`
`=343-122.5+63=283.5`