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

Answer 1

Ezra Adams

#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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Answer 2

Emilia Aguilar

To find the function ( f(x) = \int_{-2}^{x} -2x^2 + \frac{1}{x} dx ) given that ( f(2) = -1 ), we need to first integrate the given function and then use the provided information to solve for the constant of integration.

First, integrate the function:

( \int -2x^2 + \frac{1}{x} dx = -\frac{2}{3}x^3 + \ln|x| + C )

Next, substitute the limits of integration (-2 and x) into the integrated function:

( f(x) = -\frac{2}{3}x^3 + \ln|x| + C )

Using the given condition ( f(2) = -1 ):

( -1 = -\frac{2}{3}(2)^3 + \ln|2| + C )

Solve for ( C ):

( C = -\frac{16}{3} - \ln(2) )

Therefore, the function ( f(x) ) is:

( f(x) = -\frac{2}{3}x^3 + \ln|x| -\frac{16}{3} - \ln(2) )

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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.