What is #f(x) = int sqrt(x+3) dx# if #f(1)=7 #?

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Answer 1 · 2018-05-26T13:52:27

#f(x)=2/3(1+3)^(3/2)+5/3#

#f(x)=intsqrt(x+3)dx#

#f(x)=int(x+3)^(1/2)dx#

now

#d/(dx)(x+3)^(3/2)=3/2(x+3)^(1/2)#

#f(x)=int(x+3)^(1/2)dx=2/3(x+3)^(3/2)+c#
now

#f(1)=7#

#:.2/3(1+3)^(3/2)+c=7#

#2/3xx8+c=7#

#16/3+c=7#

#=>c=7-16/3#

#c=5/3#

#:.f(x)=2/3(1+3)^(3/2)+5/3#