What is # int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx #?
1 Answer
# int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx = 11/405 sqrt(3) #
Explanation:
We seek:
# I = int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx #
We can perform a substitution of the form:
# u = 4-9x^2 iff 9x^2=4-u# , and#(du)/dx = -18x #
This substitution will require an associated change of limits from
When
#x= { (1/3), (2/3) :} => u= { (3), (0) :}#
And we manipulate the integral as follows:
# I = int_(1/3)^(2/3) \ (9x^2)/9(-18x)/(-18)sqrt(4-9x^2) \ dx #
# \ \ = -1/162 \ int_(1/3)^(2/3) \ (9x^2)(-18x)sqrt(4-9x^2) \ dx #
# \ \ = -1/162 \ int_(3)^(0) \ (4-u)sqrt(u) \ du #
# \ \ = 1/162 \ int_(0)^(3) \ (4-u)u^(1/2) \ du #
# \ \ = 1/162 \ int_(0)^(3) \ 4u^(1/2) - u^(3/2) \ du #
# \ \ = 1/162 \ [ (4u^(3/2))/(3/2) - (u^(5/2))/(5/2) ]_(0)^(3) #
# \ \ = 2/162 [ (4u^(3/2))/(3) - (u^(5/2))/(5) ]_(0)^(3) #
# \ \ = 1/81 (4/3 (3)^(3/2) - 1/5 (3)^(5/2) ) #
# \ \ = 1/81 (4/3 (3sqrt(3)) - 1/5 (9sqrt(3)) ) #
# \ \ = 1/81 (12/3-9/5) sqrt(3) #
# \ \ = 1/81 (11/5) sqrt(3) #
# \ \ = 11/405 sqrt(3) #
So the last answer is correct.
