Help integration simple question (1 Viewer)

[Georgiiaa]

I got stuck at finding the point of intersection but i need help for this question



Find the area enclosed between the curve y = x3, the x-axis and the line y = -3x + 4.

THX

 

[Drongoski]

I got stuck at finding the point of intersection but i need help for this question



Find the area enclosed between the curve y = x3, the x-axis and the line y = -3x + 4.

THX

This means You want to find the area between the 2 curves:

i) y = -3x + 4

ii) y = x³

between x = 0 and x = 4/3

The most straightforward way is to find:




The 2 curves intersect when: x³ = -3x + 4
This occurs at x = 1.

 

[fluffchuck]

I'm unsure if this method is correct:

Point of intersection:
y = -3x + 4
y = x^3

If we equate these two equations, we get:
x^3 = -3x + 4
x^3 + 3x - 4 = 0

And now, we will find the integers that divide -4 and substitute it into the equation and see if we get LHS = RHS = 0.

So:
test x = -1
test x = 1
...
And we see that x = 1 gives LHS = RHS = 0.
Therefore, there is a point of intersection at x=1.

So now, you can draw out your graphs and label your point of intersection.

And now we see that the area is = int(0 to 1)(x^3)dx + int(1 to 4/3)(-3x+4)dx

 

[pikachu975]

You have 2 areas:

Area 1:
Point of intersection, equate the two equations
x^3 = -3x+4
x^3 + 3x - 4 = 0
Test x = 1 as a solution (factor theorem)
and x = 1 ends up as a correct factor

Second Area:
A = 1/2 * b * h
Sub x = 1 into either equation to get y = 1, i.e. h = 1
Make y = 0 for the line equation and get x intercept of x = 4/3, so b = 4/3 - 1 = 1/3
A = 1/2 * 1/3 * 1
A = 1/6

Therefore total Area = 1/4 + 1/6
5/12 units squared

Tell me if I made any mistakes in the working.

Edit: fixed a line

 

[fluffchuck]

You have 2 areas:

Area 1:
Point of intersection, equate the two equations
x^3 = -3x+4
x^3 + 3x - 4 = 0
(x-1)(x+3) = 0
x = -3, 1
Testing both points into the line equation we see that x = 1 is the correct one
Therefore Area = integral from 0 to 1 of (x^3) dx
= [x^4 / 4](0 to 1)
= 1/4

Second Area:
A = 1/2 * b * h
Sub x = 1 into either equation to get y = 1, i.e. h = 1
Make y = 0 for the line equation and get x intercept of x = 4/3, so b = 4/3 - 1 = 1/3
A = 1/2 * 1/3 * 1
A = 1/6

Therefore total Area = 1/4 + 1/6
5/12 units squared

Tell me if I made any mistakes in the working.

How did you factorise this?

 

[pikachu975]

How did you factorise this?

lol soz meant to be (x-1)(x+4) but I don't think it changes the answer.

 

[fluffchuck]

lol soz meant to be (x-1)(x+4) but I don't think it changes the answer.

Its x^3, not x^2

And yeah ahaha x=1 is still the answer

 

[pikachu975]

Its x^3, not x^2

And yeah ahaha x=1 is still the answer

Oh yeah oops I'm high