Question

I know I just asked a similar question, but this time I'm not sure how to handle the 2 in the integration?
\[\int\limits_{-1}^{0}(2+ {1 \over x-1})dx\]

Answer 1

Common denominaotor, but here you dont need it

Answer 2

So, how do you handle it then?

Answer 3

2x+ln(x-1) answer

Answer 4

you can integrate them seperately, so 2 becomes 2x, and 1/x-1 becomes ln(x-1)

\[\int\limits\limits_{-1}^{0} 2 + (1/x-1) dx\] = 2x + ln(x-1) from -1 to 0

Answer 5

Why is that?

Answer 6

Remember the defintion of the integral?

its just the Riemann's Sum, and so addition properties apply

Answer 7

(2(-1)+ln(-1-1))-(ln(0-1))
(-2+ln(2))-(ln(1))

Answer 8

Yes, Sam that's the answer... but how did you integrate it in the first place? I don't know Riemann's Sum..

Answer 9

break the problem down

\[\int\limits_{-1}^{0} 2 dx + \int\limits_{-1}^{0} 1/(x +1) dx\]

1st integral gives F(x) = 2x 2nd Integral is G(x) ln(x+1)

the its F(0) - F(-1) + G(0) - G(-1)

Answer 10

Oh, That breaks it down nicely :)

Answer 11

The Definition of the Integral is just a summation of all the area under a curve.

Which means the Integral of a Sum is the same as the Sum of two integrals:

∫2+(1/x−1)dx=∫2dx+∫(1/x-1)dx

Answer 12

Example
\[\huge \int\limits_{}^{}(2A)dx=2Ax+c\]
c= constant

Answer 13

Yup. Thank you all! :) I understand this topic now :)

Answer 14

Lol math is all about definitions, if you remember your definitions, it makes math so much easier.