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anonymous

  • 4 years ago

Show geometrically why ∫sqrt(2-x^2) = (pi/4)+(1/2)

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{0}^{1}\sqrt{2-x ^{2}}dx = \Pi /4+1/2\]

  2. amistre64
    • 4 years ago
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    geometrically? wouldnt that just be a graph of the sqrt part?

  3. anonymous
    • 4 years ago
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    You have to like break it into two peices

  4. anonymous
    • 4 years ago
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    a sector of a circle and a triangle

  5. amistre64
    • 4 years ago
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    oh, then the graph would be useful at anyrate

  6. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%282-x%5E2%29%2C+real%2C+x%3D-2to2%2C+y%3D0to2 the wolf hates when I try to define stuff ....

  7. JamesJ
    • 4 years ago
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    right exactly. The segment of the circle has angle pi/4, or 1/8 of a complete circle of 2pi and radius sqrt(2). Hence the segment of circle has area \[ \frac{1}{8}\pi r^2 = \frac{2}{8}\pi = \frac{\pi}{4} \] Now, what about the triangle?

  8. anonymous
    • 4 years ago
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    well the triangle wld be a special triangle

  9. JamesJ
    • 4 years ago
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    No, it would be a right angled triangle.

  10. anonymous
    • 4 years ago
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    1.1 and sqrt(2)

  11. anonymous
    • 4 years ago
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    ya a right triangle

  12. JamesJ
    • 4 years ago
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    what's the length of the base, b? And the height, h?

  13. anonymous
    • 4 years ago
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    |dw:1327361596715:dw|

  14. JamesJ
    • 4 years ago
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    yes, exactly. And its area?

  15. anonymous
    • 4 years ago
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    (b*H)/2

  16. anonymous
    • 4 years ago
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    so it is a 1/2

  17. anonymous
    • 4 years ago
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    ohhhhh I see

  18. JamesJ
    • 4 years ago
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    Nice question.

  19. anonymous
    • 4 years ago
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    Thanks :D

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