Integration Formula:
[tex]\int\limits {x}^{n} \, dx =\frac{x^{n+1}}{n+1} + C[/tex]
Integrate each term:
[tex]\int\ {x^{2}} \, dx = \frac{x^{3}}{3}[/tex]
[tex]\int\ {x} \, dx =\frac{x^2}{2}[/tex]
[tex]\int\ {5} \, dx = 5x[/tex]
[tex]\int\ {x} \, dx =\frac{x^2}{2}[/tex]
[tex]\int\ {2} \, dx = 2x[/tex]
Altogether this becomes:
[tex]\frac{\frac{x^3}{3} - \frac{x^2}{2}-5x }{ \frac{x^2}{2}+2x}[/tex]
Your limits are 1 and 2. Substitute both numbers into the equation and minus them from each other.
x = 2:
[tex]\frac{\frac{2^3}{3} - \frac{2^2}{2}-5(2) }{ \frac{2^2}{2}+2(2)}[/tex]
-
x = 1:
[tex][tex]\frac{\frac{1}{3} - \frac{1}{2}-5 }{ \frac{1}{2}+2}[/tex][/tex]
x = 2:
[tex]\frac{\frac{8}{3} - \frac{4}{2}-10 }{ \frac{4}{2}+4}[/tex]
-
x = 1: [tex]\frac{\frac{8}{3} - 2-10 }{ 2+4}[/tex]
=
-1.212317928
Now convert all the answers to integers and see which one matches -1.212317928:
Because it's a negative number, we know it can only be A or B.
(A) = -1.212317928
(B) = -1.19047619
The answer is A: [tex]-\frac{3}{2} +In\frac{4}{3}[/tex]
(A) - 3/2 = ln 4/3
