Help on intergration (1 Viewer)

[Bored Of Fail 3]

rearrange and complete the square

-x = y^2+5y+6
y^2 +5y= -x -6
y^2 +5y + (5/2)^2 = -x -6 + (5/2)^2
(y+(5/2) )^2 = -x + (1/4)

(y+(5/2) )^2 = - (x -(1/4) )

so it is sideways ( from the y^2) , opening to the left ( because we have negative out the front of the bracket on the RHS ) .

 

[kittyful]

oh right thanks

 

[kittyful]

How abt this question: Find the area enclosed between the line y=2 and the curve y = x^2 +1

For working i got intergral of 2x- x^3/3 +x from 1 to -1 ... is this correct ?

 

[deterministic]

How abt this question: Find the area enclosed between the line y=2 and the curve y = x^2 +1

For working i got intergral of 2x- x^3/3 +x from 1 to -1 ... is this correct ?

you mean then you are correct.

 

[kittyful]

you mean then you are correct.

yes that was what i meant, but who come when i but the values into it, the answer doesn't match the book :S but when i arranged the equation the other way, i get the correct answer but its negative :S

 

[deterministic]

thats what i got??

 

[kittyful]

thats what i got??

oh i get

 

[TheTutorBot]

yes that was what i meant, but who come when i but the values into it, the answer doesn't match the book :S but when i arranged the equation the other way, i get the correct answer but its negative :S

If yuo get a negative answer, it means your area is under the curve in most circumstances..thats why you should draw a graph first and break the question down into area below the curve and area above the curve.

I think you might have done the subtraction the wrong way around

 

[kittyful]

How about this question: Find the area enclosed between the curve y=x^2 and the line y=-6x+16

Is the intergral between -8 and 2 ?
And the equation after integration is (-3x^2 +16x+x^3/3)

 

[bored of fail 1]

fairly sure the x^3 / 3 at the end should be -x^3/ 3

 

[kittyful]

is possible to factorise x^3 +3x+4=0

 

[bored of fail 1]

is possible to factorise x^3 +3x+4=0

yes, using 3unit methods

P(-1) =0

so (x+1) is a factor , and you can polynomial division, but u dont do 3unit, do you?

 

[ohexploitable]

is possible to factorise x^3 +3x+4=0

Yes but only in 3/4U.

 

[kittyful]

yes, using 3unit methods

P(-1) =0

so (x+1) is a factor , and you can polynomial division, but u dont do 3unit, do you?

no i do 2unit

 

[kittyful]

I had this question: Find the area enclosed between the curve y=x^3 the x axis and the line y=-3x+4

I simulataneous the two equ
and got x^3 +3x+4 = 0
so i factorised it to this (x+1) (x^2 - x +4)

is it possible to find the end points with it ? to find the integral ?

 

[bored of fail 1]

the discriminant of the second bracket is (-1)^2 -4(1)(4) < 0

the second bracket has no real solutions

are u sure u read the question correctly?

 

[bored of fail 1]

u made a mistake, should be x^3 +3x -4 =0

then you have a different factorisation

gives (x-1)(x^2 +x + 4) =0

but again the second bracket has a negative discriminant, no real solution, so question is not correct

 

[kittyful]

oh right thanks for picking up the errors

 

[kittyful]

How do i find the area enclosed by the curvees y= x^2 and x=y^2

 

[Drongoski]

Why don't you get bored of fail 1 as your tutor. He is good.