What is #f(x) = int -2x^2+1/xdx# if #f(2)=-1 #?

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Answer 1 · 2017-02-17T16:06:34

#f(x)=(-2x^3)/3+ln|x|+13/3-ln2, or,#

#f(x)=1/3(13-2x^3)+ln|x/2|.#

#f(x)=int(-2x^2+1/x)dx=int-2x^2dx+int1/xdx#

#=-2intx^2dx+ln|x|=-2(x^(2+1)/(2+1))+ln|x|#

#:. f(x)=(-2x^3)/3+ln|x|+C........(star)#

But, #f(2)=-1 rArr ((-2)(2)^3)/3+ln|2|+C=-1#

#rArr C=16/3-1-ln2=13/3-ln2.#

Hence, by #(star)#,

#f(x)=(-2x^3)/3+ln|x|+13/3-ln2, or,#

#f(x)=1/3(13-2x^3)+ln|x/2|.#

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