What is #f(x) = int 3x dx# if #f(2) = 7 #?

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Answer 1 · 2017-03-21T13:54:56

The answer is #f(x)=3/2x^2+1#

We need

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

So,

#f(x)=int3xdx=3/2x^2+C#

#f(2)=6+C=7#

#:. C=7-6=1#

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

Answer 2 · 2017-03-21T13:55:15

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

Evaluate the indefinite integral:

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

Use the initial condition to determine #C#:

#f(2) = 12/2+C = 7#

so:

#C = 7-6 = 1#

and:

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