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

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Answer 1 · 2016-01-07T11:25:01

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

Integrating f(x): $\frac{x^{3}}{3} - \frac{3}{2} x^{2} + c$ $x^3/3 - 3/2x^2 + c$

f(2 ) = 1 enables the constant of integration ( c ) to be found by evaluating for x = 2 , y = 1

$\Rightarrow \frac{2^{3}}{3} - 3 \times \frac{2^{2}}{2} + c = 1$ $rArr 2^3/3 -3 xx 2^2/2 + c = 1$

$\Rightarrow \frac{8}{3} - 6 + c = 1$ $rArr 8/3 - 6 + c = 1$

$\Rightarrow c = 1 + 6 - \frac{8}{3} = \frac{13}{3}$ $rArr c = 1 + 6 - 8/3 = 13/3$

$\Rightarrow f (x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + \frac{13}{3}$ $rArr f(x) = 1/3 x^3 - 3/2 x^2 + 13/3$

What is f(x) = int x^2 - 3x f(x)=∫x2−3xf(x) = int x^2 - 3x if f(2)=1 f(2)=1f(2)=1 ?