What is $f (x) = \int {(x - 3)}^{2} - 3 x + 4 d x$ $f(x) = int (x-3)^2-3x+4 dx$ if $f (2) = 1$ $f(2) = 1$ ?

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Answer 1 · 2016-01-24T22:17:25

$f (x) = \frac{x^{3}}{3} - \frac{9 x^{2}}{2} + 13 x - \frac{29}{3}$ $f(x) = x^3/3 -(9x^2)/2 +13x -29/3$

First, expand the integrant as follow

$\int ({(x - 3)}^{2} - 3 x + 4) d x = \int (x^{2} - 6 x + 9 - 3 x + 4) d x$ $int((x-3)^2 -3x +4)dx = int(x^2-6x+9-3x+4)dx$

$= \int (x^{2} - 9 x + 13) d x$ $=int(x^2-9x+13)dx$

Then we can integrate this using the power rule like this

$f (x) = \frac{x^{3}}{3} - \frac{9 x^{2}}{2} + 13 x + C$ $f(x) = x^3/3-(9x^2)/2+13x+C$ #

We are given $f (2) = 1$ $f(2) = 1$ , substitute this into the $f (x)$ $f(x)$ to solve for C

$1 = \frac{2^{3}}{3} - \frac{9 {(2)}^{2}}{2} + 13 (2) + C$ $1= (2^3)/3-(9(2)^2)/2+13(2) +C$

$1 = \frac{8}{3} - \frac{36}{2} + 26 + C$ $1= 8/3 -36/2 +26+C$

$1 = \frac{8}{3} - 18 + 26 + C$ $1= 8/3 -18 +26+C$

$C = - \frac{29}{3}$ $C= -29/3$

So $f (x) = \frac{x^{3}}{3} - \frac{9 x^{2}}{2} + 13 x - \frac{29}{3}$ $f(x) = x^3/3 -(9x^2)/2 +13x -29/3$

What is f(x) = int (x-3)^2-3x+4 dxf(x)=∫(x−3)2−3x+4dxf(x) = int (x-3)^2-3x+4 dx if f(2) = 1 f(2)=1f(2) = 1 ?