Area Integration Question (1 Viewer)

[Dragie]

Hey - I just can't figure out how to do these - it's taken me ages and it is quite urgent so if you can, please help me with them:

1. Find the area bounded by y=x(x^2-1); x=-1 to x=2
(Answer to this question is 2 and 3/4.)

2. Show that the x-axis and the curve y=4x(x-1)(x-2) encloses two regions. Find their areas.
(Answer to this is 1 square unit and 1 square unit)

 

[e7aine]

1.

2
∫ x^3 - x dx (i expanded the orginal equation)
-1

= [x^4/4 - x^2/2]

= [4 - 2] - [1/4 - 1/2]
= 4 1/4 sq. units

hmm...my answer is diff from urs...maybe i got it wrong...

2. for this question...it wud rili help if u draw the graph first

for the equation... when y= 0, x = 0, 1, 2,....u will end up with 2 regions, one region over the x axis and the other under the x-axis..

so first i wud find the area from x= 0, x= 1

1
∫ 4x^3 - 12x^2 + 8x dx (i expanded the orignal equation)
0

= [x^4 - 4x^3 + 4x^2]
= [1] - [0]
= 1 sq unit

1
∫ 4x^3 - 12x^2 +8x dx
2

= [x^4 - 4x^3 + 4x^2]
= [1] - [0]
= 1 sq. unit

 

[PLooB]

Hey - I just can't figure out how to do these - it's taken me ages and it is quite urgent so if you can, please help me with them:

1. Find the area bounded by y=x(x^2-1); x=-1 to x=2
(Answer to this question is 2 and 3/4.)

2
∫ x³-x =
-1
2
[x⁴/4 - x²/2]
-1

= 4 - 2 - (1/4 - 1/2) = 4.25

 

[SoulSearcher]

1. Find the area bounded by y=x(x^2-1); x=-1 to x=2
(Answer to this question is 2 and 3/4.)

what you have to be careful of here is where the curve cuts the x-axis, in this case, x = -1, 0 and 1. since the curve is under the x-axis between x = 0 and x = 1, then you have to take that into consideration.

thus answer is
0 1 2
∫ x^3 - x dx +| ∫ x^3 - x dx |+ ∫ x^3 - x dx
-1 0 1

which simplifies to 1/4 + 1/2 + 2 = 2 + 3/4, which is your answer

 

[Dragie]

Hey thanks alot - now I can kill myself with more integration