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anonymous

  • 5 years ago

evaluate the integral to 3 decimal places

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{-3}^{3}\sqrt{9-x ^{2}}\]

  2. anonymous
    • 5 years ago
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    integral of the form sqrt(a^2 - x^2) = (x/2)*sqrt(a^2 - x^2) + ((a^2)/2)*arcsin(u/a) + constant, evaluated at from -3 to 3 = (9*pi)/2 to 3 decimal places = 14.137

  3. anonymous
    • 5 years ago
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    can u use \[\int\limits_{}^{}(ax+b)^{n}dx= (ax+b)^{n+1}/a(n+1) +c\]

  4. anonymous
    • 5 years ago
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    i haven't used integration with trigonometric identities yet

  5. anonymous
    • 5 years ago
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    the problem is that i get zero when i integrate using that method

  6. anonymous
    • 5 years ago
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    and i know its not possible to get zero area in that problem lol

  7. anonymous
    • 5 years ago
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    ax +b will not work because you have the form b - ax^2.

  8. anonymous
    • 5 years ago
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    oh i see do you no what form i should use exlcuding the use of trig forms

  9. anonymous
    • 5 years ago
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    i didnt learn integration using trigonometric identities

  10. dumbcow
    • 5 years ago
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    i would use trig substitution x=3sin(u)

  11. dumbcow
    • 5 years ago
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    im not sure how else to do it

  12. anonymous
    • 5 years ago
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    so there isnt another method?

  13. anonymous
    • 5 years ago
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    oh i see

  14. dumbcow
    • 5 years ago
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    well because its a definite integral you could approximate area using graphing software or sum of rectangles

  15. anonymous
    • 5 years ago
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    it asks to evaluate the integral does that mean i cant use the area formulae of a circle

  16. dumbcow
    • 5 years ago
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    hmm i didnt notice that before yeah this is just the area of the upper part of a circle of radius 3

  17. dumbcow
    • 5 years ago
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    so A= pi*r^2/2

  18. anonymous
    • 5 years ago
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    yes it is

  19. anonymous
    • 5 years ago
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    if it asks to evaluate it doesn't necessarily mean through the use of integration does it

  20. dumbcow
    • 5 years ago
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    it can but in this case there is a more practical and simple method

  21. anonymous
    • 5 years ago
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    i see well thankyou for your help :)

  22. dumbcow
    • 5 years ago
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    \[\int\limits \sqrt{a^2 - x^2} dx = \frac{1}{2} x \sqrt{a^2-x^2} +\frac{1}{2}a^2\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}+C\] http://integral-table.com/

  23. anonymous
    • 5 years ago
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    so that is the form used to integrate that question. I havent learnt it yet so i will assume that using area formulae and a graphy is the way to go lol:)

  24. dumbcow
    • 5 years ago
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    i agree

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