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ta123

need help on the attachment @Calculus1

  • 2 years ago
  • 2 years ago

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  1. ta123
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    • 2 years ago
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  2. across
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    \[f(x)=2x^2\]

    • 2 years ago
  3. across
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    \[f'(x)=4x\implies f(1)=4,\]\[f(1)=2.\]

    • 2 years ago
  4. across
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    \[y=4x-2.\]

    • 2 years ago
  5. across
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    \[f_{1}^{-1}(x)=\sqrt{\frac{1}{2}x}\]\[f_{2}^{-1}(x)=\frac{x+2}{4}\]

    • 2 years ago
  6. across
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    \[\int_{0}^{2}\int_{\sqrt{\frac{y}{2}}}^{\frac{y+2}{4}}dxdy\]

    • 2 years ago
  7. across
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    There it is. Can you evaluate that integral?

    • 2 years ago
  8. ta123
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    I believe so, you just have to separate both them right?

    • 2 years ago
  9. across
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    \[\frac{1}{4}\int_{0}^{2}ydy+\frac{1}{2}\int_{0}^{2}dy-\sqrt{\frac{1}{2}}\int_{0}^{2}\sqrt{y}dy\]yep

    • 2 years ago
  10. across
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    \[\frac{1}{8}\left[y^2\right]_{0}^{2}+\frac{1}{2}\left[y\right]_{0}^{2}-\frac{1}{3}\sqrt{2}\left[y^{\frac{3}{2}}\right]_{0}^{2}\]seems a bit tedious though, but it gives the right answer :/

    • 2 years ago
  11. across
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    There has got to be an easier method.

    • 2 years ago
  12. ta123
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    thanks again across, like the new pic of you

    • 2 years ago
  13. across
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    Thought it'd look more "professional." n_O Thank you.

    • 2 years ago
  14. ta123
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    "professional",lol

    • 2 years ago
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