124. Find the function whose derivative is equal to $3 t^{2}-2 t+1$, given that at $t=2$ the function takes a value of 25.
Solution. $1^{0}$. From the condition, it follows that the desired function is an antiderivative of the function $3 t^{2}-2 t+1$; therefore, taking the indefinite integral of $3 t^{2}-2 t+1$, we find all antiderivatives of the given function:

$$
\int\left(3 t^{2}-2 t+1\right) d t=3 \int t^{2} d t-2 \int t d t+\int d t=t^{3}-t^{2}+t+C
$$

$2^{0}$. From the obtained expression $t^{3}-t^{2}+t+C$, which defines all antiderivatives of the function $3 t^{2}-2 t+1$, we will now find the desired antiderivative function. Using the additional condition (the value of the desired function is 25 when $t=2$), we will find the specific value of the integration constant $C$; we have $2^{3}-2^{2}+2+C=25$, from which $C=19$.

$3^{0}$. Thus, the desired function has the form $y=t^{3}-t^{2}+t+19$.