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...
Write $|v(t)|$ as a piecewise function:
$$
|v(t)|= \cases{5-3t, & $0\le t\le 5/3$ \cr 3t-5, & $5/3\le t\le 3$ }
$$
Then, split the integral $\int_0^3 |v(t)|\,dt $ into two pieces.
$$
\int_0^3 |v(t)|\,dt =\int_0^{5/3} |v(t)|\,dt +\int_{5/3}^3 |v(t)|\,dt =\int_0^{5/3} 5-3t\,dt +\int_{5/3}^3 3t-5\,dt,
$$
and evaluate.
Essentially, find the set over which $v$ is positive and the set over which $v$ negative; then integrate $|v|$ over these sets separately. On the set where $v$ is negative, you'd integrate $-v$ and over the set where $v$ is positive, you'd integrate $v$.
The difference between displacement and total distance should be clear: here the point travels to the left the first $5/3$ seconds then travels to the right from $t=5/3$ to $t=3$.
The displacement is the difference between the final and initial position of the point. Since the point traveled left and then right, this will be less than the total distance traveled (which is the sum of the distance traveled left with the distance traveled right).
For example if you move along a line left $4$ units then right $5$ units, the displacement is $1$ and the total distance traveled is $9$.
Best Answer
You have an expression for the velocity.
To calculate distance as opposed to displacement, you need to know the time(s) when $v=0$, since the particle reverses the direction of motion at these times, and integrate the velocity separately over these time intervals, taking the absolute value where necessary.
Clearly $v=0\Rightarrow t=1$
Therefore the distance travelled is $$\left|\int_0^1(t^2+4t-5)dt\right|+\left|\int_1^4(t^2+4t-5)dt\right|$$