Question

Integration help me please!

Answer 1

\[\int\limits_{1}^{2}(1-2x ^{2}+x ^{5})/x ^{5} dx=\]

Answer 2

ahh rearrange this first as \[\int_1^2 \frac{x^5 - 2x^ 2 + 1}{x^5}dx\]

is that okay with you?

Answer 3

thats what it is i just dont know how to write that on here

Answer 4

no i put x^5 lol..i rearranged it

Answer 5

oh ok i got it but yeah go ahead

Answer 6

okay...now are you familiar that \[\frac{a+b+c}{d} = \frac{a}{d} + \frac bd + \frac cd\]

Answer 7

yupp

Answer 8

so i'll apply that here
\[\large \int_1^2 \frac{x^5 - 2x^2 + 1}{x^5} dx \implies \int_1^2 \frac {x^5}{x^5}dx - \int_1^2 \frac{2x^2}{x^5}dx + \int_1^2 \frac{1}{x^5}dx\]

Answer 9

do you get what i did?

Answer 10

yupp

Answer 11

so what's x^5/x^5?

Answer 12

1

Answer 13

so \[\int_1^2 \frac{x^5}{x^5} dx\implies \int_1^2 dx\]
right?

Answer 14

yupp

Answer 15

now what is 2x^2/x^5?

Answer 16

im not sure how to do that one

Answer 17

hint:
\[\LARGE \frac{a^m}{a^n} \implies a^{m-n}\]

Answer 18

2x^-3

Answer 19

right so \[\large \int_1^2 \frac{2x^2}{x^5}dx \implies \int_1^2 2x^{-3}dx\]

Answer 20

thts what i got

Answer 21

now what is \[\frac{1}{x^5}\]

Answer 22

just that right?

Answer 23

well yeah but put x into the numerator..to make it look prettier

Answer 24

how do you suppose you can do that?

Answer 25

fancy 1

Answer 26

hint \[\large \frac{1}{a^m} \implies a^{-m}\]

Answer 27

multiply by x^-5

Answer 28

not multiply...

Answer 29

it really is just x^-5

Answer 30

so now let us rewrite \[\large \int_1^2 dx - \int_1^2 2x^{-3}dx + \int_1^2 x^{-5}dx\]
do you agree?

Answer 31

yes

Answer 32

so what's the integral of dx?

Answer 33

im not sure i dont know what to do from here

Answer 34

because you cant just add becuase of all the exponents right?

Answer 35

this is one of the rules in integration:
\[\large \int du = u\]

Answer 36

ok

Answer 37

so what's \(\int dx\)

Answer 38

x

Answer 39

right

Answer 40

and the next one is x^-3

Answer 41

and then x^-5

Answer 42

one by one

Answer 43

ok

Answer 44

so now what is \[\int 2x^{-3}dx\]

Answer 45

x^-3 dx

Answer 46

what did you do?

Answer 47

took away the2

Answer 48

do you know the power rule?

Answer 49

yeah but that isnt the same as the thingyou just showede me

Answer 50

\[\large \int ax^n dx \implies \frac{ax^{n+1}}{n+1}\]

Answer 51

-6x^-4

Answer 52

what you did is the power rule for derivative

Answer 53

oh

Answer 54

look at what i just posted

Answer 55

so 2x^-2/-2

Answer 56

that's right but you an cancel something out \[\frac{2x^{-2}}{-2}\]

Answer 57

-2

Answer 58

what's your final answer for that one?

Answer 59

2x

Answer 60

did you do power rule of derivatives again?

Answer 61

no

Answer 62

i canceled out the -2 in the denominator and the exponent

Answer 63

\[\large \frac{2x^{-2}}{-2} \implies \frac{2}{-2} \times x^{-2}\]
you cant cancel denominator and exponent...you can only cancel numerator and denominator

Answer 64

oh ok so

Answer 65

that makes teh 3rd on x^-4/-4

Answer 66

yes but what is the second one? what's the final answer for that one?

Answer 67

x^-2

Answer 68

right...

Answer 69

so now we're on to plugging in the limits

Answer 70

thats good i need to move on i have a few multiple choice questions that i have to get done by nine

Answer 71

\[\LARGE [x|_1^2 - [x^{-2}|_1^2 + [\; \frac{x^{-4}}{-4}|_1^2\]

Answer 72

lol dont get intimidated by how that looks =)))

Answer 73

do you know what \[\large [x|_1^2\]
means?

Answer 74

no

Answer 75

here's a hint"
\[\large [f(x)|_a^b \implies f(b) -f( a)\]

Answer 76

does that help?

Answer 77

so 2-1 for the first one?

Answer 78

correct

Answer 79

for the second one?

Answer 80

i dont know because you cant go to the ^-2

Answer 81

?

Answer 82

\[\large [x^{-2}|_1^2 \implies 2^{-2} - 1^{-2}\]
got that?

Answer 83

thats what i got but when i put it in the calculator it said error

Answer 84

maybe you typed it wrong... "syntax error" means you typed it wrong

Answer 85

TOP response to this question @lgbasallote

Answer 86

ohhhhh

Answer 87

-.75

Answer 88

correct

Answer 89

so what's the third one?

Answer 90

.234375

Answer 91

final is 1.515625

Answer 92

how did you get this?

Answer 93

doing the 3rd one and then subtracting

Answer 94

\[\large [x|_1^2 = 1\]
\[\large [x^{-2}|_1^2 = -0.75\]
\[\large [\frac{x^{-4}}{-4}|_1^2 = 0.234375\]
you agree with this right?

Answer 95

yes

Answer 96

therefore what is \[\Large [x|_1^2 - [x^{-2}|_1^2 + [\frac{x^{-4}}{-4}|_1^2\]

Answer 97

1.984375

Answer 98

yup better