The acceleration function is $\tag{1}a(t)=4+t;\ \ 0\le t\le10.$ So, for example, the acceleration at $t=0$ is $a(0)=1+0=1$.
To find the velocity function, recall that the derivative of the velocity function is the acceleration function. So to find the velocity function, we can first find an antiderivative of the acceleration function $(1)$. An antiderivative of $a(t)=4+t$ is $4t+{1\over2}t^2+C$. So, for $0\le t\le10$ $\tag{2} v(t)=4t+\textstyle{1\over2}t^2+C $ for some constant $C$. We need to find the value of $C$. To do this, we use the information that $v(0)=5$. Using equation ${2}$, we have $ 5=v(0)=4\cdot0 +\textstyle{1\over 2} \cdot0^2+C\ \ \Rightarrow\ \ C=5. $ Thus, the velocity function is $\tag{3}v(t)=4t+\textstyle{1\over2}t^2+5;\ \ 0\le t\le10.$
Alternatively, you could just compute the velocity function via, for $0\le t\le 10$ $ v(t)-v(0) =\int_0^t 4+s\,ds= (4s+\textstyle{1\over2}s^2)|_0^t=4t+\textstyle{1\over2}t^2, $ which gives $(3)$ by moving $-v(0)=-5$ in the equation above to the other side.
Towards finding the distance traveled, we could first find the displacement function. The derivative of the displacement function is the velocity function. So the displacement function, $d$, can be obtained by finding an antiderivative of $(3)$: $ d(t)=2t^2+\textstyle{1\over 6}t^3+5t+D, \ \ 0\le t\le10, $ for some constant $D$. We can't find the value of $D$ here, but it's not needed to find the distance traveled from $t=0$ to $t=10$. Since $v(t)$ is nonnegative for $0\le t\le10$, the object is always moving to the right; so the total distance traveled from $t=0$ to $t=10$ is $d(10)-d(0) =[ 200+(1000/6) +50+D] -[ 0+0+0-D] = 200+(1000/6) +50.$
Alternatively, to find the distance traveled, you could compute $ \int_0^{10} v(t)\,dt $; which would lead to exactly the same expression as above.
Here it is important to realize that $v(t)\ge0$ for $0\le t\le 10$. For a general velocity function $v$, the expression $\int_a^b v(t)\, dt$ gives the displacement from $t=a$ to $t=b$. The total distance traveled is $\int_a^b |v(t)|\, dt$. The two can of course be different...