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ta123

  • 4 years ago

need help on the attachment @Calculus1

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

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

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

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

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

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

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

  9. across
    • 4 years ago
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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

  10. across
    • 4 years ago
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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 :/

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

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

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

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

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