Consider the function $f(x)$ whose second derivative is $f^{\prime \prime }(x)=10x+3\sin (x)$. If $f(0)=2$ and $f^{\prime }(0)=3$, what is $f(4)$ ?

Step 1 ：Given that the second derivative of the function $f(x)$ is $f^{″}(x)=10x+3\sin (x)$, we can find the first derivative $f^{\prime }(x)$ by integrating $f^{″}(x)$ with respect to $x$.

Step 2 ：The integral of $10x$ with respect to $x$ is $5x^2$ and the integral of $3\sin (x)$ with respect to $x$ is $-3\cos (x)$. Therefore, $f^{\prime }(x)=5x^2-3\cos (x)+C$, where $C$ is the constant of integration.

Step 3 ：Given that $f^{\prime }(0)=3$, we can substitute $x=0$ into the equation $f^{\prime }(x)=5x^2-3\cos (x)+C$ to find the value of $C$. This gives $3=5(0{)}^2-3\cos (0)+C$, which simplifies to $C=3+3=6$. Therefore, $f^{\prime }(x)=5x^2-3\cos (x)+6$.

Step 4 ：We can find the function $f(x)$ by integrating $f^{\prime }(x)$ with respect to $x$. The integral of $5x^2$ with respect to $x$ is $\frac{5}{3}{x}^3$, the integral of $-3\cos (x)$ with respect to $x$ is $-3\sin (x)$, and the integral of $6$ with respect to $x$ is $6x$. Therefore, $f(x)=\frac{5}{3}{x}^3-3\sin (x)+6x+D$, where $D$ is the constant of integration.

Step 5 ：Given that $f(0)=2$, we can substitute $x=0$ into the equation $f(x)=\frac{5}{3}{x}^3-3\sin (x)+6x+D$ to find the value of $D$. This gives $2=\frac{5}{3}(0{)}^3-3\sin (0)+6(0)+D$, which simplifies to $D=2$. Therefore, $f(x)=\frac{5}{3}{x}^3-3\sin (x)+6x+2$.

Step 6 ：Finally, we can find the value of $f(4)$ by substituting $x=4$ into the equation $f(x)=\frac{5}{3}{x}^3-3\sin (x)+6x+2$. This gives $f(4)=\frac{5}{3}(4{)}^3-3\sin (4)+6(4)+2$.

Step 7 ：Calculating the above expression gives $f(4)=\frac{5}{3}(64)-3\sin (4)+24+2$.

Step 8 ：Simplifying the above expression gives $f(4)=\frac{320}{3}-3\sin (4)+26$.

Step 9 ：Therefore, the value of $f(4)$ is $\boxed{{\displaystyle \frac{320}{3}-3\sin (4)+26}}$.