Question

integral of 3x^2-2x+3 on the bound 2,1... need to know how

Answer 1

\[ \int\limits_{1}^{2} (3-2 x+3 x^2) dx\]
Integrate the sum term by term and factor out constants:
\[3 \int\limits_{1}^{2} x^2 dx+ 3\int\limits_{1}^{2} dx-2 \int\limits_{1}^{2} x dx\]
The integral of x^2 is x^3/3:
\[x^3+ 3\int\limits_{1}^{2} dx-2 \int\limits_{1}^{2} x dx\]
The integral of x is x^2/2:
\[x^3-x^2+ 3\int\limits_{1}^{2} dx\]
The integral of dx is x:
\[[x^3-x^2+3x]\]
Now taking it from 1 to 2:
\[[(2)^3-(2)^2+3(2)]-[(1)^3-(1)^2+3(1)]=7\]
So your final answer is 7.

Answer 2

make sense?

Answer 3

yes it does...

Answer 4

but instead for going term by term couldn't you take the derivative of the function and then evaluate it for the two bound numbers

Answer 5

you can take the integral at the same time like this
integral of 3x^2-2x+3 on the bound 2,1.
=(3x^3)/3 -(2x^2)/2+3x]
=x^3 -x^2 +3x] now taking x from 2 to 1
=[2^3 -2^2 +3(2)]-[1^3 -1^2 +3(1]
=8-4+6-(3)=7 ans.......