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anonymous
 5 years ago
evaluate the integral to 3 decimal places
anonymous
 5 years ago
evaluate the integral to 3 decimal places

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{3}\sqrt{9x ^{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral 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
 5 years ago
Best ResponseYou've already chosen the best response.0can u use \[\int\limits_{}^{}(ax+b)^{n}dx= (ax+b)^{n+1}/a(n+1) +c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i haven't used integration with trigonometric identities yet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the problem is that i get zero when i integrate using that method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i know its not possible to get zero area in that problem lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ax +b will not work because you have the form b  ax^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see do you no what form i should use exlcuding the use of trig forms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt learn integration using trigonometric identities

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0i would use trig substitution x=3sin(u)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0im not sure how else to do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there isnt another method?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0well because its a definite integral you could approximate area using graphing software or sum of rectangles

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it asks to evaluate the integral does that mean i cant use the area formulae of a circle

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i didnt notice that before yeah this is just the area of the upper part of a circle of radius 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it asks to evaluate it doesn't necessarily mean through the use of integration does it

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0it can but in this case there is a more practical and simple method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see well thankyou for your help :)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \sqrt{a^2  x^2} dx = \frac{1}{2} x \sqrt{a^2x^2} +\frac{1}{2}a^2\tan^{1}\frac{x}{\sqrt{a^2x^2}}+C\] http://integraltable.com/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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:)
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