What is #f(x) = int (3x+5)^2-x dx# if #f(1)=2 #?

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Answer 1 · 2016-01-28T17:14:51

#f(x)=3x^3+29/2x^2+25x-81/2#

The easiest way to integrate this is by expanding #(3x+5)^2# and subtracting #x# to get

#int9x^2+29x+25dx#

We can now integrate each term individually using the rule

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

This gives us the indefinite integral of

#=(9x^(2+1))/(2+1)+(29x^(1+1))/(1+1)+(25x^(0+1))/(0+1)+C#

#=3x^3+29/2x^2+25x+C#

We can now find #C#, the constant of integration for this particular function, since we know that #f(1)=2#.

#f(1)=3(1^3)+29/2(1^2)+25(1)+C=2#

#85/2+C=2#

#C=-81/2#

Thus,

#f(x)=3x^3+29/2x^2+25x-81/2#