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# Integration help needed C2 watch

1. Question:

Find the areas enclosed by the curve and the straight lines
a)

Are required = + Area of rectangle
First find

=
=
=

Area of rectangle

=
Area =

therefore area required =
However I was wrong the book gave an answer of

Thank you very much
Attached Images

2. (Original post by bigmansouf)
x
Think carefully about the area of the rectangle and what the question requires - you're almost there though
3. (Original post by SeanFM)
Think carefully about the area of the rectangle and what the question requires - you're almost there though
4. (Original post by bigmansouf)
I'll say the same as my last hint but a bit more. Look at the green shaded area in one of your attachments - you have more of the rectangle than you need.
5. as SEAN pointed out you have included a non-enclosed bit of area...
6. (Original post by the bear)
as SEAN pointed out you have included a non-enclosed bit of area...
I have tried again to find it but this time I calculated the area from coordinates x=0.5 to x=1 and then add the intergal vaule for and add
I got a result of 7/4
I have attached the area I calculated as a picture
Please point in the right of the direction this question is quite confusing
Attached Images

7. to find the area between two curves ( one of which may be a straight line such as y = -1 ) you just find ∫ ( top curve - bottom curve ) dx

here ∫ ( 1/x2 - 1 - ( -1 )) dx

with limits 1/2 and 2

ANOTHER WAY

if you imagine the curves translated upwards by 1 you would find the area by integrating 1/x2 - 1 + 1 or just 1/x2
8. (Original post by the bear)
to find the area between two curves ( one of which may be a straight line such as y = -1 ) you just find ∫ ( top curve - bottom curve ) dx

here ∫ ( 1/x2 - 1 - ( -1 )) dx

with limits 1/2 and 2

ANOTHER WAY

if you imagine the curves translated upwards by 1 you would find the area by integrating 1/x2 - 1 + 1 or just 1/x2
thank yuo very much i mistakenly got x=1/2 and x=-2 to be the straight lines

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