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Integration help needed C2

Question:

Find the areas enclosed by the curve and the straight lines
a) Y=1x21 Y=\frac{1}{x^{2}}-1
y=1 y=-1 x=12 x=\frac{1}{2} x=2 x=2



My answer
Are required = 121(x21)dx+ \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx + + Area of rectangle
First find 121(x21)dx \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx

=121(x21)dx \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx
= [x1x]121 [-x^{-1}-x]_{\frac{1}{2}}^{1}
= (11)(1120.5)=0.5(-1-1)-(-\frac{1}{\frac{1}{2}}-0.5) = 0.5

Area of rectangle

= width=20.5=32width= 2-0.5 = \frac{3}{2}
Area = 32×1=32 \frac{3}{2} \times 1 = \frac{3}{2}

therefore area required = 32+0.5=2 \frac{3}{2} +0.5 =2
However I was wrong the book gave an answer of 32 \frac{3}{2}

Please help me understand where I went wrong

Thank you very much
(edited 8 years ago)
Original post by bigmansouf
x


Think carefully about the area of the rectangle and what the question requires - you're almost there though :smile:
Reply 2
Original post by SeanFM
Think carefully about the area of the rectangle and what the question requires - you're almost there though :smile:

pleasecan i ask for a hiny
Original post by bigmansouf
pleasecan i ask for a hiny


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.
as SEAN pointed out you have included a non-enclosed bit of area...
Reply 5
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 121(x21)dx \int_{\frac{1}{2}}^{1} (x^{-2}-1)dx and add 12(x21)dx \int_{1}^{2} (x^{-2}-1)dx
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
(edited 8 years ago)
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
Reply 7
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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