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d92292

  • 2 years ago

find the area enclosed between the curve y=x^3 and the line y=x

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  1. Algebraic!
    • 2 years ago
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    did you sketch the region?

  2. d92292
    • 2 years ago
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    im not sure on how to

  3. Algebraic!
    • 2 years ago
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    you're not sure how to sketch x^3 ?

  4. d92292
    • 2 years ago
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    no i mean i sketched it, i dont know what to do after

  5. Algebraic!
    • 2 years ago
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    say, for example, you're asking about the area enclosed for positive values of x..: |dw:1355190249043:dw|

  6. Algebraic!
    • 2 years ago
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    you can see that it's the area under the line x (from 0 to 1) minus the area under x^3 (from 0 to 1)

  7. Algebraic!
    • 2 years ago
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    aka:\[\int\limits_{0}^{1} x-x ^{3} dx\]

  8. Algebraic!
    • 2 years ago
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    can you evaluate that integral?

  9. d92292
    • 2 years ago
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    and then how do we find the anti derivitve of that hat

  10. Algebraic!
    • 2 years ago
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    the usual way:\[\int\limits_{ }^{ } x^n dx = \frac{ x^{n+1} }{ n+1 }\]

  11. Algebraic!
    • 2 years ago
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    so for \[\int\limits_{ }^{} x^1 dx\] \[\int\limits_{ }^{} x^1 dx = \frac{ x^{1+1} }{ 1+1}\]

  12. Algebraic!
    • 2 years ago
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    try it for x^3...

  13. d92292
    • 2 years ago
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    x^4/4

  14. Algebraic!
    • 2 years ago
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    yes

  15. Algebraic!
    • 2 years ago
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    so \[\frac{ x^2 }{ 2 } -\frac{ x^4 }{4 } \]

  16. Algebraic!
    • 2 years ago
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    ok?

  17. d92292
    • 2 years ago
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    and whats next?

  18. Algebraic!
    • 2 years ago
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    plug in the limts

  19. Algebraic!
    • 2 years ago
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    for the region on the interval x =0...+inf ( 1/2 -1/4 ) - (0/2 -0/4)

  20. d92292
    • 2 years ago
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    kk

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