Question

help again? ;D just setting it up..

Answer 1

\[\int\limits_{1}^{2} (\frac{ x }{ 2 }-\frac{ 2 }{ x })dx\]

Answer 2

change it x^-2?

Answer 3

2^(-x)

Answer 4

no you cannot do that

Answer 5

\[\int\limits_{1}^{2} \left(\frac{ x }{ 2 }-\frac{ 2 }{ x }\right)\text dx\]\[=\int\limits_{1}^{2} \frac{ x }{ 2 }\text dx-\int\limits_{1}^{2}\frac{ 2 }{ x }\text dx\]\[=\frac12\int\limits_{1}^{2} x\text dx-2\int\limits_{1}^{2}\frac{ 1 }{ x }\text dx\]

Answer 6

oh okay ;/

Answer 7

do you know the integral of 1/x ?

Answer 8

I do not...

Answer 9

well, its a special one,

the integral of 1/x

is the natural logarithm of x

\[\int \frac1x\text dx=\ln x+c\]

Answer 10

If it was x^(-n) and n was any number apart from 1, then you could apply the standard polynomial antiderivative. But 1/x is kind of a special case because... there is no polynomial whose derivative is 1/x = x^(-1).

Answer 11

\[\frac{ 1 }{ 2 }\int\limits_{1}^{2} (\frac{ x^2 }{ 2 })dx-2\int\limits_{1}^{2}\frac{ lnx^2+C }{ 2 }) ?\]

Answer 12

You don't want those anti-derivatives inside the integral, because, this processes of finding these new functions from the ones inside the integral is a way of getting away from the integral. Instead of those integral signs, you wanted to wrote those same functions you just wrote, but the symbol "evaluated at 2 and -1".

Answer 13

huh?

Answer 14

\[[\frac{1}{2}\frac{x^2}{2} - 2 \ln x] |_{1}^2\]

Answer 15

note that d/dx (ln x + c) = 1/x. So the anti derivative of 1/x is ln x + c. But when you have these integrals over a specified interval ( [1, 2] ) like here, then the constant c of the anti-derivative is unnecessary.