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Integration Help, Please (1 Viewer)

xMaNx

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Apr 7, 2009
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1. Find the area bounded by the curve y = 9 - x^2 and the x axis between x = 0 and x = 6

2. Find the area of the region bounded by the curve y = 2x - x^2, the line y = 4-2x and the y axis.

3. Find the area enclosed by the curve x = -y^2 + 7y - 10 and the y axis.

4. Find the area bounded by y = sqrt (x - 4), the y axis and the lines y = 1 and y = 2.


I've been trying for the last hour but am stuck on these questions, any assistance such as a method to tackle them or worked solutions is greatly appreciated.

Thanks
 

obliscape

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May 5, 2010
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Well no point giving you answers since you won't be learning.
Finding the area is the practically the same as definite integrals.
So you take your equation, add 1 to the power, divide by the new power, put it in square brackets, write your limits, then you sub your limits in. Upper limit - lower limit.
Hence
y = 9-x^2 => In b/w 0 and 6 of 9-x^2 dx
= [ 9x - x^3/3]0...6
= [9(6) - (6)^3/3] - [9(0) - 0^3/3]
=...

For question 2 you have 2 equations. So you need to equate them to find their points of intersection to find your limits - but you get x limits so sub into equations to find y-values. Then you sketch both graphs and you take the area below the top curve MINUS the area below the bottom curve. But again since your question asks for y-axis, you need to make x the subject (rearrange both equations) and then integrate as usual, using y limits.

Question 3 it says the curve and the y-axis, questions like that require you to factorise the equation so that you can find your limits.
Factorise x = -y^2 + 7x - 10 => (y-2) (5-y)
So you do:
In b/w -5 and 2 of -y^2 + 7x - 10 dx (x is already the subject, finding about y-axis)
Integrate as normal

Question 4 basically the same, draw your curve to see. Make x the subject and integrate.
With some questions you have to be careful that the curve goes above and below the x axis or whatever, when they do you need to do 2 separate areas with 2 different sets of limits.

Hope this somewhat helped...
 

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