Question 1099289
The derivative of a sum of functions is the sum of the derivatives,
so the same applies to antiderivatives,
So to find a function with {{{x^3-2sqrt(x)}}} for a derivative,
we need to find what function has for derivative {{{x^3}}} ,
and what function has for derivative {{{sqrt(x)=x^"1 / 2"}}} .
We also know that if we find a function with {{{x^3-2sqrt(x)}}} for a derivative,
adding any constant {{{C}}} , we get another function with the same derivative.
We will use {{{f(1)=4}}} to figure out what constant to add.
 
We know about derivatives of polynomials.
For any exponent {{{r}}} (integer or not, as long as {{{r<>-1}}} ), 
if {{{g(x)=x^(r+1)/(r+1)}}} , {{{"g ' ( x )"=x^r}}} ,
and that lets us figure out the antiderivative of any {{{x^r}}} ,
except {{{x^(-1)=1/x}}} .
 
So, the antiderivative of {{{x^3}}} is {{{x^4/4}}} ;
and the antiderivative of {{{sqrt(x)=x^"1 / 2"}}} is {{{x^"3 / 2"/"3 / 2"=(2/3)x^"3 / 2"=2x^"3 / 2"/3}}} .
 
So, we can see that {{{f(x)=x^4/4-2(2x^"3 / 2"/3)+C=x^4/4-4x^"3 / 2"/3+C}}}
for some constant {{{C}}} .
Then, {{{f(1)=1/4-4/3+C}}} .
If {{{f(1)=4}}} ,
{{{1/4-4/3+C=4}}}
{{{C=4+4/3-1/4}}}
{{{C=61/12}}} ,
and {{{highlight(f(x)=x^4/4-4x^"3/2"/3+61/12)}}}