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What is #f(x) = int -x^3+x-4 dx# if #f(2) = -3 #?

Answer 1

Christopher Agresta

#f(x) = int(-x^3+x-4)dx#

Use the power rule to integrate.

# = -x^4/4+x^2/2-4x+C# #" "# (item 1)

Find #C# by using the fact that #f(2) = -3#

#f(x) = -(2)^4/4+(2)^2/2-4(2) + C = -3#

#-4+2-8+C=-3#

# - 10 +C = -3#

# C = 7# #" "# (item 2)

Combining items (1) and (2), we finish with

#f(x) = -x^4/4+x^2/2-4x+7#

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Answer 2

Ezekiel Alexander

To find ( f(x) = \int_{-x^3}^{x-4} dx ) given that ( f(2) = -3 ), we integrate the function and then use the given value to solve for the constant of integration.

First, integrate the function:

[ f(x) = \int_{-x^3}^{x-4} dx ]

[ f(x) = \left[ x \right]_{-x^3}^{x-4} ]

[ f(x) = (x - (x-4)) - (-x^3) ]

[ f(x) = x - x + 4 + x^3 ]

[ f(x) = 4 + x^3 ]

Now, we use the given information that ( f(2) = -3 ) to solve for the constant of integration:

[ f(2) = 4 + (2)^3 = 4 + 8 = 12 ]

Since ( f(2) = -3 ), we have:

[ -3 = 12 + C ]

Solve for ( C ):

[ C = -3 - 12 = -15 ]

Therefore, ( f(x) = 4 + x^3 - 15 ).

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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.