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If integrate f(x) dx = 1/3 * x ^ 3 - 2x ^ 2 + 3x - 5 then the value of f(-2) = ...

A. -9
B. -1
C. 7
D. 15
E. 23

please... T_T​

If integrate f(x) dx = 1/3 * x ^ 3 - 2x ^ 2 + 3x - 5 then the value of f(-2) = ... A-example-1
User LazyClown
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1 Answer

3 votes

Answer:

D. 15

Explanation:

Use the Fundamental Theorem of Calculus to solve this problem. Since f(x) is the derivative of 1/3 x^3 - 2x^2 + 3x - 5, you have:

f(x) = d/dx (1/3 x^3 - 2x^2 + 3x - 5) = x^2 - 4x + 3

To find f(-2), you substitute x = -2 into the expression for f(x):

f(-2) = (-2)^2 - 4(-2) + 3 = 4 + 8 + 3 = 15

User ATek
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