# Integration Question C2Watch

#1
The curve . The point P has co-ordinates (3,27) and PQ is a tangent to the curve at P . Find area of the region enclosed between the curve, PQ and the x- axis.

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What I done is to first find the equation of PQ which would be:
? as it doesn't tell you any other thing other than it has the same gradient.

Then equate, , so x^3 - 9x .

Integrating that, using limits as 0 to 3 gives:
x^4/4 -9x^2/2 = -81/4

0
8 years ago
#2
Your tangent line is wrong. It passes through (3,27) but you'll need the gradient at P, so use dy/dx.
1
8 years ago
#3
I remember the days when i used to think this stuff was hard.
2
8 years ago
#4
y -y1 = m(x - x1)
y - 27 = 27(x - 3)

For the area required you will need to know where this tangent crosses the x-axis...sketch useful!
0
#5
Okay so I end up with the tangent equation as:
y=27x - 54 , which will cross the x axis when x=2

Will this mean the limits are from 0 to 2 ?

I subbed in 2 as x after integrating, I get 158 , the answer is 27/4
0
8 years ago
#6
diagram here:
http://www.wolframalpha.com/input/?i=plot+y%3Dx^3++and++y%3D27x-54++for+x%3D0+to+x%3D3

You need to integrate the curve from x=0 to x=3, then subtract the area of the triangle formed by the tangent and the line x=3.
0
8 years ago
#7
(Original post by WildBerry)
Okay so I end up with the tangent equation as:
y=27x - 54 , which will cross the x axis when x=2

Will this mean the limits are from 0 to 2 ?

I subbed in 2 as x after integrating, I get 158 , the answer is 27/4
The limits are 3 and 0 because (3,27) is where the point P lies. However if you integrate and find the area enclosed by the curve and the x-axis between the limits 0 and 3, then you will get an area that is too large to be the area between the curve, PQ and the x-axis. This is because PQ is not a vertical line that crosses the x-axis at 3.

Let the point (2,0), where PQ crosses the x-axis, be Q. Also let the point (3,0), the point vertically below P, be T:

The solution would be to integrate and find the area of the region enclosed by y = x^3 and the x-axis between the limits 0 and 3 (i.e. the area enclosed by the curve, the x-axis and the line PT), then subtract the area of the right-angled triangle PTQ. The resulting area will be the area between the curve, x-axis, and PQ.
0
8 years ago
#8
(Original post by vc94)
diagram here:
http://www.wolframalpha.com/input/?i=plot+y%3Dx^3++and++y%3D27x-54++for+x%3D0+to+x%3D3

You need to integrate the curve from x=0 to x=3, then subtract the area of the triangle formed by the tangent and the line x=3.
Lol, beat me to it You got right in there while I was typing that long post
1
#9
Thanks everyone , finally got it !!!
0
8 years ago
#10
(Original post by WildBerry)
Thanks everyone , finally got it !!!
You are welcome
0
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