Answer:
a. 
b. ![\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Cdfrac%7B1%7D%7B108%7D%20%5B%20145%20%5Csqrt%7B145%7D%20-%201%5D%7D%7D)
Step-by-step explanation:
Evaluate integral _C x ds where C is
a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)
i . e

where;
x = t , y = t/2
the derivative of x with respect to t is:

the derivative of y with respect to t is:

and t varies from 0 to 12.
we all know that:

∴


![= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B%5Csqrt%7B5%7D%7D%7B2%7D%20%5C%20%5C%20%5B%5Cdfrac%7Bt%5E2%7D%7B2%7D%5D%5E%7B12%7D_0)

= 
b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)
Given that:
x = t ; y = 3t²
the derivative of x with respect to t is:

the derivative of y with respect to t is:


Hence; the integral _C x ds is:

Let consider u to be equal to 1 + 36t²
1 + 36t² = u
Then, the differential of t with respect to u is :
76 tdt = du

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145
Thus;



![\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}](https://tex.z-dn.net/?f=%5Cmathtt%7B%3D%20%5Cdfrac%7B2%7D%7B216%7D%20%5B%20145%20%5Csqrt%7B145%7D%20-%201%5D%7D)
![\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20%5Cdfrac%7B1%7D%7B108%7D%20%5B%20145%20%5Csqrt%7B145%7D%20-%201%5D%7D%7D)