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

Answer 1

Isabella Ackerman

#f(x)=2 sqrt(x+3) + 3#

I believe that the given must be like #f(x)= int (1/sqrt(x+3)) dx# and #f(1)=7#

Also #f(x)=int (x+3)^(-1/2)*dx# The integral becomes like this after integration

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

which simplifies to

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

and

#f(x)=2 sqrt(x+3)+C#

but #f(1)=7#

#f(1)=7=2 sqrt(1+3)+C#

#7=2*sqrt(4)+C# #7=2*2+C# #C=3#

Finally

#f(x)=2 sqrt(x+3)+3#

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Answer 2

Emma Aldridge

To find ( f(x) = \int \frac{1}{\sqrt{x+3}} , dx ) when ( f(1) = 7 ), you need to evaluate the definite integral of ( \frac{1}{\sqrt{x+3}} ) and then use the given condition to find the value of the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.