Question

evaluate the integral to 3 decimal places

Answer 1

\[\int\limits_{-3}^{3}\sqrt{9-x ^{2}}\]

Answer 2

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

Answer 3

can u use \[\int\limits_{}^{}(ax+b)^{n}dx= (ax+b)^{n+1}/a(n+1) +c\]

Answer 4

i haven't used integration with trigonometric identities yet

Answer 5

the problem is that i get zero when i integrate using that method

Answer 6

and i know its not possible to get zero area in that problem lol

Answer 7

ax +b will not work because you have the form b - ax^2.

Answer 8

oh i see do you no what form i should use exlcuding the use of trig forms

Answer 9

i didnt learn integration using trigonometric identities

Answer 10

i would use trig substitution
x=3sin(u)

Answer 11

im not sure how else to do it

Answer 12

so there isnt another method?

Answer 13

oh i see

Answer 14

well because its a definite integral you could approximate area using graphing software or sum of rectangles

Answer 15

it asks to evaluate the integral does that mean i cant use the area formulae of a circle

Answer 16

hmm i didnt notice that before yeah this is just the area of the upper part of a circle of radius 3

Answer 17

so A= pi*r^2/2

Answer 18

yes it is

Answer 19

if it asks to evaluate it doesn't necessarily mean through the use of integration does it

Answer 20

it can but in this case there is a more practical and simple method

Answer 21

i see well thankyou for your help :)

Answer 22

\[\int\limits \sqrt{a^2 - x^2} dx = \frac{1}{2} x \sqrt{a^2-x^2} +\frac{1}{2}a^2\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}+C\]

http://integral-table.com/

Answer 23

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:)

Answer 24

i agree