First, integrate the velocity function to get the indefinite position function:

\(\displaystyle \int 4t-1dt=4(\frac{t^2}{2})-t+C\)

Now, plug in your initial conditions to solve for C:

\(\displaystyle 2(2^2)-2+C=2; C=-4\)

Now, plug back into the position function:

\(\displaystyle x(t)=2t^2-t-4\).

First, write the integral expression for this problem:

\(\displaystyle \int 2t^2-1dt\)

Now, integrate. Remember to add one to the exponent and then put that result on the denominator:

\(\displaystyle 2(\frac{t^3}{3})-t+C\)

Now, plug in your initial conditions to find C:

\(\displaystyle \frac{2(1)}{3}-1+C=2; C=\frac{7}{3}\)

Now, plug your C into the position function:

\(\displaystyle x(t)=\frac{2}{3}t^3-t+\frac{7}{3}\).

First, set up the integral expression:

\(\displaystyle \int 2t-1dt\)

Now, integrate. Remember to raise the exponent by 1 and put that result on the denominator. Also remember to add a C because it is an indefinite integral:

\(\displaystyle \frac{2t^2}{2}-t+C=t^2-t+C\)

Plug in your initial conditions to solve for C: 

\(\displaystyle (3^2)-3+C=4; C=-2\)

Plug back in for C and get your position function:

\(\displaystyle x(t)=t^2-t-2\).

Remember that the integral of the velocity function is the position function. First, write the integral expression:

\(\displaystyle \int 4t+2dt\)

Integrate. Remember to raise the exponent by 1 and then put that result on the denominator:

\(\displaystyle \frac{4t^2}{2}+2t=2t^2+2t+C\)

Now, plug in your initial conditions to solve for C:

\(\displaystyle 2(2^2)+2(2)+C=1; C=-11\)

Plug back in to get your position function:

\(\displaystyle x(t)=2t^2+2t-11\)

Recall that the integral of the velocity function is the position function. Therefore, set up the integral:
\(\displaystyle \int 2t-1dt\)

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

\(\displaystyle \frac{2t^2}{2}-t=t^2-t+C\)

Now, plug in your initial conditions to find C:
\(\displaystyle 1^2-1+C=2; C=2\)

Plug back in to get your position function:

\(\displaystyle x(t)=t^2-t+2\)

Recall that integrating the velocity function will yield you the position function:

\(\displaystyle \int 6t-1dt\)

Now, integrate:

\(\displaystyle \frac{6t^2}{2}-t=3t^2-t+C\)

Now, to find your C, plug in your initial conditions:

\(\displaystyle 3(1^2)-1+C=2; C=0\)

Since, C is 0, your position function is:

\(\displaystyle x(t)=3t^2-t\).

Recall that integrating the velocity function will yield the position function:

\(\displaystyle \int 4t+5dt\)

Now, integrate.

\(\displaystyle \frac{4t^2}{2}+5t=2t^2+5t+C\)

To find your C, plug in your initial conditions:

\(\displaystyle 2(0^2)+0+C=3; C=3\)

Plug your C back in to get your position function:

\(\displaystyle x(t)=2t^2+5t+3\)

First, set up the integral expression:

\(\displaystyle \int 4x^2-3xdx\)

Next, integrate:

\(\displaystyle \frac{4x^3}{3}-\frac{3x^2}{2}+C\)

Now, plug in your initial conditions:

\(\displaystyle \frac{4}{3}(1^3)-\frac{3}{2}(1^2)+C=2; C=\frac{13}{6}\)

Now, plug your C back into the position function:

\(\displaystyle x(t)=\frac{4}{3}x^3-\frac{3}{2}x^2+\frac{13}{6}\)

Recall that the integral of the velocity function is the position function:

\(\displaystyle \int 10t-2dt\)

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

\(\displaystyle \frac{10t^2}{2}-2t=5t^2-2t+C\)

Now, plug in your initial conditions:

\(\displaystyle 5(1^2)-2(1)+C=2; C=-1\)

Now, plug in your C to your position function:

\(\displaystyle x(t)=5t^2-2t-1\)

Recall that the integral of the velocity function is the position function. Write the integral expression:

\(\displaystyle \int 2t-1dt\)

Now, integrate:

\(\displaystyle \frac{2t^2}{2}-t=t^2-t+C\)

Plug in your initial conditions to get your C:
\(\displaystyle 1^2-1+C=2; C=2\)

Plug your C back into the position function:

\(\displaystyle x(t)=t^2-t+2\)