anonymous
  • anonymous
evaluate the integral to 3 decimal places
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{-3}^{3}\sqrt{9-x ^{2}}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
can u use \[\int\limits_{}^{}(ax+b)^{n}dx= (ax+b)^{n+1}/a(n+1) +c\]

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anonymous
  • anonymous
i haven't used integration with trigonometric identities yet
anonymous
  • anonymous
the problem is that i get zero when i integrate using that method
anonymous
  • anonymous
and i know its not possible to get zero area in that problem lol
anonymous
  • anonymous
ax +b will not work because you have the form b - ax^2.
anonymous
  • anonymous
oh i see do you no what form i should use exlcuding the use of trig forms
anonymous
  • anonymous
i didnt learn integration using trigonometric identities
dumbcow
  • dumbcow
i would use trig substitution x=3sin(u)
dumbcow
  • dumbcow
im not sure how else to do it
anonymous
  • anonymous
so there isnt another method?
anonymous
  • anonymous
oh i see
dumbcow
  • dumbcow
well because its a definite integral you could approximate area using graphing software or sum of rectangles
anonymous
  • anonymous
it asks to evaluate the integral does that mean i cant use the area formulae of a circle
dumbcow
  • dumbcow
hmm i didnt notice that before yeah this is just the area of the upper part of a circle of radius 3
dumbcow
  • dumbcow
so A= pi*r^2/2
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
if it asks to evaluate it doesn't necessarily mean through the use of integration does it
dumbcow
  • dumbcow
it can but in this case there is a more practical and simple method
anonymous
  • anonymous
i see well thankyou for your help :)
dumbcow
  • dumbcow
\[\int\limits \sqrt{a^2 - x^2} dx = \frac{1}{2} x \sqrt{a^2-x^2} +\frac{1}{2}a^2\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}+C\] http://integral-table.com/
anonymous
  • anonymous
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:)
dumbcow
  • dumbcow
i agree

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