# 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.