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allsmiles

  • 3 years ago

Evaluate the following integrals by first reversing the order of integration. ∫(0,8)∫(y^(1/3),2) 8e^x^2 dxdy. Why is the limit when you reverse 0<=x<=2, 0<=y<=x^3 and not x^3<=y<=8?

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  1. allsmiles
    • 3 years ago
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    I'm confused with finding the limits for each when you reverse the order anyone got any tips?

  2. TuringTest
    • 3 years ago
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    drawing out the region of integration is almost always a good idea

  3. TuringTest
    • 3 years ago
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    |dw:1352232985691:dw|

  4. allsmiles
    • 3 years ago
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    Yeah I got that too, but can't you say y^(1/3) to 8 is the limit? Go along the curve y=x^3 and stop at 8?

  5. TuringTest
    • 3 years ago
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    well what would the area between y^(1/3) and y=8 look like?

  6. TuringTest
    • 3 years ago
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    or rather x^3<=y<=8

  7. allsmiles
    • 3 years ago
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    ohh

  8. allsmiles
    • 3 years ago
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    it'd be the shaded area under y=8 but above y=x^3?

  9. allsmiles
    • 3 years ago
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    which we're not looking for right

  10. TuringTest
    • 3 years ago
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    |dw:1352233420315:dw|you got it :)

  11. TuringTest
    • 3 years ago
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    so in the area we want we want y bounded above by the function, not the line

  12. allsmiles
    • 3 years ago
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    it's so hard to see sometimes but got it thanks so much for your help!!

  13. allsmiles
    • 3 years ago
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    so whenever you want it bounded by the curve you'd take it from 0 to the curve?

  14. TuringTest
    • 3 years ago
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    welcome, and never underestimate the power of making sure your drawings are correct

  15. allsmiles
    • 3 years ago
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    or where ever it starts at

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