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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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    \[\int_{0}^{1}\pi(\sqrt[4]{x})^2dx-\int_{0}^{1}\pi(x^4)^2dx\]

  3. ta123
    • 4 years ago
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    how come but side are the same

  4. across
    • 4 years ago
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    What do you mean?

  5. ta123
    • 4 years ago
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    oh, wait my eyes are playing tricks on me I thought I saw a square root in the right integral

  6. across
    • 4 years ago
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    Yeah, integrals will do that to your eyes. :P

  7. ta123
    • 4 years ago
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    I got 7pi/9 is that what you got?

  8. across
    • 4 years ago
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    They're subtracting.

  9. across
    • 4 years ago
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    \[\pi\int_{0}^{1}\sqrt{x}dx=\frac{2}{3}\pi\left[x^{\frac{3}{2}}\right]_{0}^{1}=\frac{2}{3}\pi,\]\[\pi\int_{0}^{1}x^8dx=\pi\left[\frac{1}{9}x^9\right]_{0}^{1}=\frac{1}{5}=\frac{1}{9}\pi,\]\[\frac{2}{3}\pi-\frac{1}{9}\pi=\frac{5}{9}\pi.\]

  10. ta123
    • 4 years ago
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    dang I integrated wrong by different number number to the exponents beside the number 1

  11. ta123
    • 4 years ago
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    oops! I forgot "adding" in between by and different

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