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

1. anonymous

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

2. 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

3. anonymous

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

4. anonymous

i haven't used integration with trigonometric identities yet

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

i didnt learn integration using trigonometric identities

10. anonymous

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

11. anonymous

im not sure how else to do it

12. anonymous

so there isnt another method?

13. anonymous

oh i see

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

so A= pi*r^2/2

18. anonymous

yes it is

19. anonymous

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

20. anonymous

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

21. anonymous

i see well thankyou for your help :)

22. anonymous

$\int\limits \sqrt{a^2 - x^2} dx = \frac{1}{2} x \sqrt{a^2-x^2} +\frac{1}{2}a^2\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}+C$ http://integral-table.com/

23. anonymous

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. anonymous

i agree