Area Question: Solving for the Area Between a Parabola and Axes

[transgalactic]

i added 2 files with the question and the way i tried to solve it
it messes up and nothing come out

if my handwriting is problematic to you
the question is:

parabula y=x^2 +b*x+c cuts the X axes in two points
one of them is (1,0)
the area between the parabula the X and Y axes equals to the area between the parabula and theX axes

i showed it all in the file it includes a graph

please help

 

[HallsofIvy]

What, exactly, are you asking?

 

[transgalactic]

I am asking how can isolve this question
and find the b, c parameters?

 

[HallsofIvy]

Okay, I didn't see your last question on "page 2".

I notice that you are integrating from 0 to 1. Why? Clearly the "other" point at which the parabola crosses the x-axis is NOT "0". You are told that one x-intercept is x= 1 so you know that the function factors as (x-1)(x- a) for some number a. a is the other x-intercept and the area is the integral from a to 1 or 1 to a, depending upon which is positive.

(x-1)(x- a)= x²-(1+a)x+ a= x²+ bx+ c so b= -1-a and c= a.

The "area between the parabola and the x and y axes" is either
[tex]/int_0^1 x^2- (1+a)x+ a dx[/tex]
if 1< a or
[tex]/int_0^a x^2- (1+a)x+ a dx[/tex]
if a< 1.

Similarly, the area between the parabola is either
[tex]\int_1^a x²-(1+a)x+ a dx[/itex]
if 1< a or
[tex]\int_a^1 x²-(1+a)x+ a dx[/itex]
if a< 1.

Try both possiblilites and see if you can solve for a. Then of course use b=-1-a and c= a.

 

[transgalactic]

you devided the answers into 2 possiblities
depending if the parbula is positive or negative
if a>1
or a<1
however the direction of the parabule is always towards the positive
part of the Y axes

because the coefficient of X^2 is 1
it cannot flip to the opposite side

?

 

[transgalactic]

what is that sigh
<sup>or int??

regarding the question why i have an integral from 0 to 1
it is one of the areas

the second area is between 1 and the second point

what is that sigh
or int??</sup>

 

[happyg1]

(x-1)(x- a)= x^2-(1+a)x+ a= x^2+ bx+ c so b= -1-a and c= a.

The "area between the parabola and the x and y axes" is either
[tex]\int_0^1 x^2- (1+a)x+ a dx[/tex]
if 1< a or
[tex]\int_0^a x^2- (1+a)x+ a dx[/tex]
if a< 1.

Similarly, the area between the parabola is either
[tex]\int_1^a x^2-(1+a)x+ a dx[/tex]
if 1< a or
[tex]\int_a^1 x^2-(1+a)x+ a dx[/tex]
if a< 1.

Try both possiblilites and see if you can solve for a. Then of course use b=-1-a and c= a.

The int and sup are tex code to make it easy to read. maybe this will be clearer now

 

[transgalactic]

on the second possibility when i check the area between 1 and a
this area is under the X axes line
aren"t we suppose to do [ 0-f(x) ] dx

if i do mesure the are by this [ 0-f(x) ] dx method
i get that the area equals to zero
??

why it is wronge?