Consider the function $f(x)$ whose second derivative is $f^{\prime \prime}(x)=10 x+3 \sin (x)$. 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 $f'(x)$ 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, $f'(x) = 5x^2 - 3\cos(x) + C$, where $C$ is the constant of integration.

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

Step 4 ：We can find the function $f(x)$ by integrating $f'(x)$ 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 $\boxed{\frac{320}{3} - 3\sin(4) + 26}$.