What is $f (x) = \int - x^{2} + x - 4 d x$ $f(x) = int -x^2+x-4 dx$ if $f (2) = - 3$ $f(2) = -3$ ?

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What is $f (x) = \int - x^{2} + x - 4 d x$ $f(x) = int -x^2+x-4 dx$ if $f (2) = - 3$ $f(2) = -3$ ?

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Answer 1 · 2016-10-08T18:46:40

$y = - \frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 4 x + \frac{17}{3}$ $y=-1/3x^3+1/2x^2-4x+17/3$

Explanation:

Integrate each term using the $power rule$ $color(blue)"power rule"$

$color(orange)"Reminder "color(red)(bar(ul(|color(white)(a/a)color(black)(int(ax^n)dx=a/(n+1)x^(n+1))|)))$

$rArrint(-x^2+x-4)dx$

$y=-1/3x^3+1/2x^2-4x+c$

To find c, the constant of integration, substitute x = 2 , y = - 3

$rArr-1/3(2)^3+1/2(2)^2-4(2)+c=-3$

$rArrc=-3+8/3-2+8=5 2/3=17/3$

$rArry=-1/3x^3+1/2x^2-4x+17/3$

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What is $f (x) = \int - x^{2} + x - 4 d x$ $f(x) = int -x^2+x-4 dx$ if $f (2) = - 3$ $f(2) = -3$ ?

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What is f(x) = int -x^2+x-4 dxf(x)=∫−x2+x−4dxf(x) = int -x^2+x-4 dx if f(2) = -3 f(2)=−3f(2) = -3 ?

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What is $f (x) = \int - x^{2} + x - 4 d x$ $f(x) = int -x^2+x-4 dx$ if $f (2) = - 3$ $f(2) = -3$ ?