What is #f(x) = int xsqrt(3x) dx# if #f(3) = 0 #?

1 Answer
May 7, 2016

#=>f(x) = (2sqrt(3))/5x^(5/2) -54/5#

Explanation:

#f(x) = int xsqrt(3x) dx#
#=>f(x) = sqrt(3)int xsqrt(x) dx#
#=>f(x) = sqrt(3)int x^(3/2) dx#
#=>f(x) = sqrt(3)x^(3/2+1)/(3/2+1) +c# [where c = Integration constant]

#=>f(x) = (2sqrt(3))/5x^(5/2) +c.....(1)#

Again given condition is #f(3)=0#
So
#=>f(3) = (2sqrt(3))/5xx3^(5/2) +c#
#=>0= 2/5xx3^(5/2+1/2) +c#
#=>0= 2/5xx3^3 +c#
#=>0= 54/5 +c#
#=>c= -54/5 #
Hence we have , substituting the value of c in eq(1)

#=>f(x) = (2sqrt(3))/5x^(5/2) -54/5#