Question

solve (x-x^2)^2 plssss

Answer 1

There's nothing there to solve. Do you mean expand it?

Answer 2

yes

Answer 3

mmm loooks hard

Answer 4

lke multiply its square

Answer 5

dont know if i can do it

Answer 6

helpppppppp

Answer 7

multiply its square

Answer 8

is it (x-x^2)(X-X^2)

Answer 9

is it x^4-2x^3+x^2

Answer 10

\[(x-x^2)(x-x^2) = x(x-x^2) - x^2(x-x^2)\]by the distributive property. Now do that again and collect like terms.

Answer 11

\[\left( a-b \right)^2=a^2+b^2-2ab\]
\[\left( x^2 \right)^2=\left( x \right)^{2*2}=x ^{4}\]

Answer 12

Are you familiar with the acronym FOIL? First, Outside, Inside, Last?

Answer 13

FOIL is a bad crutch — useless as soon as you start multiplying anything other than binomials. Better to think of it as distributive property, repeatedly applied.

Answer 14

\[ x^4-2x^3+x^2\] is correct, good job!

Answer 15

i reather distribute

Answer 16

@whpalmer4 Good point.

Answer 17

wait now

Answer 18

antiderive that

Answer 19

pls

Answer 20

Now you want to integrate it?

Answer 21

anti

Answer 22

the antiderivative is the integral. do you want the integral, or the derivative?

Answer 23

1/5x^5-2x^4+1/3x^3 what i got

Answer 24

n yes integral

Answer 25

Well, this is an indefinite integral, so you forgot the arbitrary constant of integration. Not optional!

Answer 26

pi interal from 0 to 1 the function

Answer 27

antiderrive the function

Answer 28

But we'll pretend you had it there, and check answer by taking derivative:

\[\frac{d}{dx}[\frac{1}5 x^5-2x^4+\frac{1}{3}x^3+C] = 5*\frac{1}{5}x^{5-1}-4*2x^{4-1}+3*\frac{1}{3}x^{3-1}+0 \]\[= x^4-8x^3+x^2\]That's not correct...

Answer 29

you know to much swag. just checking though

Answer 30

umm the function is x^4-2x^3+x^2. sorry if i sound rude

Answer 31

i want to antiderrive

Answer 32

\[\int x^4-2x^3+x^2 \,dx = \frac{1}{4+1}x^{4+1}-\frac{2}{3+1}x^{3+1} + \frac{1}{2+1}x^{2+1} + C \]\[=\frac{1}{5}x^5-\frac{1}{2}x^4+\frac{1}{3}x^3+C\]

Yes, and I was pointing out that you made a mistake, because when we take the derivative of what we get when we take the antiderivative, we should end up with the same thing!

Answer 33

so (1/5)x^5-2x^4+(1/3)x^3 should

Answer 34

Break it up into separate integrals and take the constants outside the integrands.

Answer 35

ohhh ok

Answer 36

ok1/2

Answer 37

well then that was rathe silly of me. all this time lol

Answer 38

When in doubt, check your work. When not in doubt, check it anyhow :-)

Answer 39

ok now multiply all of that by pi and plug 1 into x

Answer 40

go i race u

Answer 41

now hold on, you're evaluating a definite integral now? what are the limits?

Answer 42

yup

Answer 43

0 to 1

Answer 44

lol

Answer 45

\[\frac{\pi}{30}\]

Answer 46

rady

Answer 47

what

Answer 48

how what when where

Answer 49

You said evaluate it from x = 0 to x = 1 and multiply by 30. I did. That's my answer.

Answer 50

u go the right answer

Answer 51

of course I did! :-)

Answer 52

ok dont u need a comon denominator for the number in ()

Answer 53

30

Answer 54

lol just did the math. damm ur fast

Answer 55

i got it

Answer 56

im also stuck on another problem which i have no idea how to do start or end

Answer 57

\[\pi \int_0^1 (x^4-2x^3+x^2)\,dx = \pi((\frac{1}{5}(1)^5 - \frac{1}{2}(1)^4+\frac{1}{3}(1)^3)- 0)= \pi(\frac{1}{5}-\frac{1}{2}+\frac{1}{3 } )\]\[= \pi\frac{6+10-15}{30} = \frac{\pi}{30}\]

Answer 58

yea i got it but thanks

Answer 59

can i just give you the question???

Answer 60

its similar but with trig

Answer 61

\[\pi \int\limits_{0}^{1}(\tan(\pi/4)x)^2dx\]

Answer 62

so how do we square the tan????????

Answer 63

Isn't that supposed to be \[\pi \int_0^1 (\tan(\frac{\pi}{4}x))^2\,dx\]

Answer 64

yea

Answer 65

oh, probably take \[u = \frac{\pi}{4}x\] as the substitution, and rewrite \(\tan^2 u\) as \(\sec^2u -1\)

Answer 66

ok

Answer 67

give it a try, I'll stick around and help if needed

Answer 68

ok

Answer 69

is sec^2u-1 the antiderrivative or just rewritten?

Answer 70

???

Answer 71

Yes?

Answer 72

rewritten?

Answer 73

I'm a little unclear on the question you are asking me here :-)

Answer 74

is sec^2u-1 the antiderrivative or just rewritten?

Answer 75

could you do it im kinda tired.not to be mean or anything :)

Answer 76

it is the identity you use to find the anti derivative since the anti derivative of \(\sec^2(x)\) is known to be \(\tan(x)\) and the anti derivative of \(-1\) is \(-x\)

Answer 77

satellite meet whpalmer, whpalmer meet satellite . you guys are smart lol

Answer 78

ok

Answer 79

if it is \(\tan^2(\frac{\pi}{4}x)\) make the substitution
\[u=\frac{\pi}{4}x\] and change the limits of integration while you are at it

Answer 80

\[\frac{4}{\pi}\times \pi\int_{0}^{\frac{\pi}{4}}(\sec^2(u)-1)du\]

Answer 81

why did you do that???

Answer 82

change limits

Answer 83

why? so you don't have to go back and replace \(u\) with \(\frac{\pi}{4}x\)
just stick with the \(u\)'s

Answer 84

ok i see

Answer 85

you get
\[4\tan(u)-u\] which you then evaluate at \(\frac{\pi}{4}\) and at \(0\) except you can ignore the zero part

Answer 86

4-pi

Answer 87

correctamundo...

Answer 88

you found the sound

Answer 89

yes indeed i did

Answer 90

ok guys thanks for all your help ill be back b4 i take my test next week

Answer 91

thank you <3

Answer 92

If I don't see you before then, best of luck on the test!