Finding original functions of second derivatives expressions I have an antiderivative problem.
Given some function:
$$f''(t)=t+\sqrt t $$
And the conditions when the $t = 1$:
$$ f(1)=1; $$
$$ f'(1)=2 $$
How do I go about deriving the original function given that I must do this with antiderivatives? My logic has been to just take the antiderivative of the original $f''(t)$ and then take the antiderivative of $f'(t)$ but I am wrong in this approach?
 A: HINT: When you integrate, you add an arbitrary constant. Since you have to integrate twice to find $f'(t)$ and $f(t)$, at each step substitute in the given values and solve for the constants.
Try that. If you are still stuck then keep reading.
First we integrate to get $f'(t)$:
$$f'(t)=\frac{t^2}{2}+\frac{2}{3}t\sqrt{t}+C$$
And since we have that$$ f'(1)=2,$$
$$\frac{1}{2}+\frac{2}{3}+C=2$$
and so
$$C=\frac{5}{6}$$
Just use this process again to find $f$.
A: Since $f''(t)=t+\sqrt{t}$ you get $f'(t)=\frac12t^2+\frac23t^{\frac32}+c$. With $2=f'(1)=\frac12+\frac23+c$ you get $c=2-\frac12-\frac23=\frac56$. So $f'(t)=\frac12t+\frac23t^{\frac32}+\frac56$. Now you get $f(t)=\frac16t^3+\frac4{15}t^{\frac52}+\frac56t+d$. With $1=f(1)=\frac16+\frac{4}{15}+\frac56+d$ you get $d=-\frac4{15}$.
Finally you have $f(t)=\frac16t^3+\frac4{15}t^{\frac52}+\frac56t-\frac4{15}$. 
A: $$f''(t)=t+\sqrt t=t+t^{\frac{1}{2}}$$
$$f'(t)=\frac{t^2}{2}+\frac{2}{3}t^{\frac{3}{2}}+c_1$$
$$f(t)=\frac{t^3}{6}+\frac{4}{15}t^{\frac{5}{2}}+c_1t+c_2$$
$$f'(1)=\frac{1}{2}+\frac{2}{3}+c_1=2$$
$$c_1=\frac{5}{6}$$
$$f(1)=\frac{1}{6}+\frac{4}{15}+c_1+c_2$$
$$f(1)=\frac{1}{6}+\frac{4}{15}+1+c_2=1$$
$$c_2=-\frac{13}{30}$$
