What is $f (x) = \int {(x - 1)}^{2} + 3 x d x$ $f(x) = int (x-1)^2+3x dx$ if $f (3) = 2$ $f(3)=2$ ?

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What is $f (x) = \int {(x - 1)}^{2} + 3 x d x$ $f(x) = int (x-1)^2+3x dx$ if $f (3) = 2$ $f(3)=2$ ?

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Answer 1 · 2016-11-19T23:37:08

Simply expand ${(x - 1)}^{2}$ $(x- 1)^2$ and then integrate using the rule $\int x^{n} (d x) = \frac{x^{n + 1}}{n + 1} + C$ $intx^n(dx) = x^(n + 1)/(n + 1) + C$ .

$f (x) = \int x^{2} - 2 x + 1 + 3 x d x$ $f(x) = intx^2 - 2x + 1 + 3xdx$

$f (x) = \int x^{2} - x + 1 d x$ $f(x) = intx^2 - x + 1dx$

$f (x) = \frac{2}{3} x^{3} - \frac{1}{2} x^{2} + x + C$ $f(x) = 2/3x^3 - 1/2x^2 + x + C$

We can solve for $C$ $C$ now.

When $x = 3$ $x = 3$ , $y = 2$ $y = 2$ :

$2 = \frac{2}{3} {(3)}^{3} - \frac{1}{2} {(3)}^{2} + 3 + C$ $2 = 2/3(3)^3 - 1/2(3)^2 + 3 + C$

$2 = 18 - 9/2 + 3 + C$

$-19 + 9/2= C$

$-29/2 = C$

Hence, $f(x) = 2/3x^3 - 1/2x^2 + x - 29/2$ .

Hopefully this helps!

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What is $f (x) = \int {(x - 1)}^{2} + 3 x d x$ $f(x) = int (x-1)^2+3x dx$ if $f (3) = 2$ $f(3)=2$ ?

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