c is the arc of the curve\n7. $int_{c}(x + 2y)dx+x^{2}dy$

Explanation:

Step1: Parameterize the line - segments

Let (C = C_1 + C_2). For (C_1): The line - segment from ((0,0)) to ((2,1)). The parametric equations are (x=t), (y=\frac{1}{2}t), (0\leq t\leq2). Then (dx = dt), (dy=\frac{1}{2}dt). [ \begin{align*} \int_{C_1}(x + 2y)dx+x^{2}dy&=\int_{0}^{2}(t + 2\times\frac{1}{2}t)dt+t^{2}\times\frac{1}{2}dt\ &=\int_{0}^{2}(t + t)dt+\frac{1}{2}t^{2}dt\ &=\int_{0}^{2}(2t+\frac{1}{2}t^{2})dt \end{align*} ] [ \begin{align*} \int_{0}^{2}(2t+\frac{1}{2}t^{2})dt&=\left[t^{2}+\frac{1}{6}t^{3}\right]0^2\ &=4+\frac{8}{6}\ &=4+\frac{4}{3}=\frac{12 + 4}{3}=\frac{16}{3} \end{align*} ] For (C_2): The line - segment from ((2,1)) to ((3,0)). The parametric equations are (x=2+(3 - 2)t=2 + t), (y=1+(0 - 1)t=1 - t), (0\leq t\leq1). Then (dx = dt), (dy=-dt). [ \begin{align*} \int{C_2}(x + 2y)dx+x^{2}dy&=\int_{0}^{1}((2 + t)+2(1 - t))dt+(2 + t)^{2}(-dt)\ &=\int_{0}^{1}(2 + t+2 - 2t)dt-(4 + 4t+t^{2})dt\ &=\int_{0}^{1}(4 - t-(4 + 4t+t^{2}))dt\ &=\int_{0}^{1}(-5t - t^{2})dt \end{align*} ] [ \begin{align*} \int_{0}^{1}(-5t - t^{2})dt&=\left[-\frac{5}{2}t^{2}-\frac{1}{3}t^{3}\right]_0^1\ &=-\frac{5}{2}-\frac{1}{3}\ &=-\frac{15 + 2}{6}=-\frac{17}{6} \end{align*} ]

Step2: Calculate the total line - integral

[ \begin{align*} \int_{C}(x + 2y)dx+x^{2}dy&=\int_{C_1}(x + 2y)dx+x^{2}dy+\int_{C_2}(x + 2y)dx+x^{2}dy\ &=\frac{16}{3}-\frac{17}{6}\ &=\frac{32-17}{6}=\frac{15}{6}=\frac{5}{2} \end{align*} ]

Answer:

(\frac{5}{2})