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Q1:

Given: $\int_{c} yd s$

$c: x=t^{2}$ (1)

$y=2 t$ (2)

$0 \leq t \leq 3$ (3)

Sol: (1) $d s=\sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t$

(2) $d s=\sqrt{(2 t)^{2}+(2)^{2}} d t$

$d s=\sqrt{4 t^{2}+4} d t$

$d s=2 \sqrt{t^{2}+1} d t$

(3) $\int_{c} y d s=\int_{0}^{3} 2 t(2 \sqrt{t^{2}+1} d t$

$=\int_{0}^{3} 4 t \sqrt{t^{2}+1} d t$

(4) $u=t^{2}+1 \Rightarrow 2 t d t=d u$

(5) $\int_{1}^{10} 2 \sqrt{u} d u=\int_{1}^{10} 2 \cdot u^{\frac{1}{2}} d u$

$=\left [2 \cdot \frac{3}{2} \cdot u^{\frac {3}{2}} \right]_{1}^{10}$

(6) $\int_{c} y d s = \frac{4}{3} \cdot 10 \sqrt{10}-\frac{4}{3} \cdot 1 \sqrt {1}$

$=\frac{40 \sqrt{10}-4}{3}$ #

Q2: Given:

$\int_{c}(x+2y) d x+x^{2} d y$

c: Line segmant

$(0,0) \rightarrow(2,1) \rightarrow(3,0)$

Req: Line Integral??

Sol: (1) graph c

(2) $C :\left\{\begin{array}{l}{x=t ; y=\frac {1}{2}t, \quad 0<t \leq 2} \\ {x=t ; y=3-t \quad 2<t \leq 3}\end{array}\right.$

(3) $=\int_{0}^{2} t+2\left(\frac{1}{2} t\right) d t+t^{2}\left(\frac{1}{2} d t\right)+\int_{2}^{3} t + 2(3-t)+t^{2}(-dt)$

(4) $=t^{2}+\left.\frac{1}{6} t^{3}\right|_{0} ^{2}+6 t-\frac{1}{2}t^{2}-\left.\frac{1}{3} t^{3}\right|_{2} ^{3}$

(5) $=4+\frac{8}{6}+18-\frac{9}{2}-9-12+2+\frac{8}{3}=\frac{5}{2}$