Question

if f(x) is a polynomial of degree three f(0) = -1 and f(1) = 0. Also,0 is a stationary point of f(x) does not have an extremum at x=0, then the value of integral (given above).​

Answer 1

Here f(x) is a third degree polynomial. Let,

$\displaystyle\small\longrightarrow f(x)=ax^3+bx^2+cx+d$

Its first derivative is,

$\displaystyle\small\longrightarrow f'(x)=3ax^2+2bx+c$

and second derivative is,

$\displaystyle\small\longrightarrow f''(x)=6ax+2b$

Given that,

$\displaystyle\small\longrightarrow f(0)=-1$

$\displaystyle\small\longrightarrow a(0)^3+b(0)^2+c(0)+d=-1$

$\displaystyle\small\longrightarrow\underline{d=-1}$

And,

$\displaystyle\small\longrightarrow f(1)=0$

$\displaystyle\small\longrightarrow a(1)^3+b(1)^2+c(1)-1=0$

$\displaystyle\small\longrightarrow a+b+c=1\quad\quad\dots(1)$

Given that f(x) has a stationary point, but does not have any extremum at x = 0, i.e., f(x) has a point of inflection at x = 0.

That means both $\displaystyle\smallf'(0)=0$ being a stationary point and $\displaystyle\smallf''(0)=0$ being a point of inflection.

So,

$\displaystyle\small\longrightarrow f'(0)=0$

$\displaystyle\small\longrightarrow 3a(0)^2+2b(0)+c=0$

$\displaystyle\small\longrightarrow\underline{c=0}$

and,

$\displaystyle\small\longrightarrow f''(0)=0$

$\displaystyle\small\longrightarrow 6a(0)+2b=0$

$\displaystyle\small\longrightarrow\underline{b=0}$

Then from (1),

$\displaystyle\small\longrightarrow\underline{a=1}$

Therefore,

$\displaystyle\small\longrightarrow f(x)=x^3-1$

Hence,

$\displaystyle\small\longrightarrow\int\dfrac{f(x)}{x^3-1}\ dx=\int\dfrac{x^3-1}{x^3-1}\ dx$

$\displaystyle\small\longrightarrow\int\dfrac{f(x)}{x^3-1}\ dx=\int dx$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{\int\dfrac{f(x)}{x^3-1}\ dx=x+C}} }$