Question

integrate x-4/x^2 dx from 1 to 2

Answer 1

i have u=x^2
du=2x
u(1)=1
u(2)=4

Answer 2

\[\frac{x - 4}{x^2}??\]

Answer 3

Split them into 2 integrals

Answer 4

if so.. break it down \[\large \int_1^2 \frac{x}{x^2}dx - \int_1^2 \frac{4}{x^2}dx\]

Answer 5

alright, and use u substitution?

Answer 6

Simply, there's no substitution!

Answer 7

just simplify them algebraically

Answer 8

then integrate

Answer 9

ok

Answer 10

how do i integrate it? massive migraine, can't remember

Answer 11

what did you get from simplifying?

Answer 12

1/x and 4/x^2

Answer 13

so where did you get stuck?

Answer 14

integrating them. is 1/x =logx after integrated

Answer 15

right

Answer 16

for the second one pull out the constant \[4\int_1^2 \frac{1}{x^2} dx\]

Answer 17

or is it ln

Answer 18

whichever of log or ln can be used

Answer 19

ok so the second would be 4lnx?

Answer 20

no...ln x is only used when it's 1/x

Answer 21

oh yeah.. idk then

Answer 22

change \(\large \frac{1}{x^2}\) into \(\large x^{-2}\)

Answer 23

and then what i dont get it, i just keep thinking derivatives

Answer 24

@jackiefml should you memorize the formula or have it handy?

Answer 25

power rule \[\Large x^n = \frac{x^{n+1}}{n +1}\] you MUST memorize these

Answer 26

i have short term memory loss. i can't memorize things very well. I do have cards with the formulas in front of me.

Answer 27

well anyway try it now

Answer 28

its x-1/-1?

Answer 29

x^-1 not x - 1

Answer 30

so -1/x?

Answer 31

yup! now multiply to 4

Answer 32

-4/x

Answer 33

so if i do lnx from 1 to 2 its ln2-ln1?

Answer 34

good..now we evaluate the limits

\[ [\ln x |_1^2 + [\frac{4}{x^2}|_1^2\] notice how it became plus because originally it was minus (-4/x^2)

Answer 35

yes..but ln 1 = 0 remember that

Answer 36

oh yeah

Answer 37

so it's \[\large \ln 2 + [\frac{4}{x^2}|_1^2\]

Answer 38

now evaluate the second term

Answer 39

6?

Answer 40

let's do this slowly... \[[\frac{4}{x^2}|_1^2 = \frac{4}{(2)^2} - \frac{4}{(1)^2}\]

Answer 41

why is it squared? i thought i had to use the integrated part

Answer 42

oops my mistake :P

Answer 43

all good.

Answer 44

sorry bout that lemme rewrite

Answer 45

do i use the -4/x equation?

Answer 46

\[\large [\frac{4}{x}|_1^2 = \frac{4}{2} - \frac{4}{1}\] and no..not -4/x remember i said it became positive now because it was \[\large \ln x - \int \frac{4}{x^2} = \ln x - (-\frac{4}{x}) = \ln x+ \frac{4}{x}\]

Answer 47

@lgbasallote Now I understand why you use that face icon :P

Answer 48

hahahah exactly :P

Answer 49

wow.

Answer 50

anyway back to the \(\large [\frac{4}{x}|_1^2 = \frac{4}{2} - \frac{4}{1}\)

Answer 51

nvm. im done ill do it myself.

Answer 52

just split it up....
can you do both integrals?