Answer
$2\sqrt 3-4$
Work Step by Step
Re-write as:
$\int_C \sqrt {x+y} \ dx=\int_{C_{2}} \sqrt {x+y} \ dx+\int_{C_{2}} \sqrt {x+y} \ dx + \int_{C_{3}} \sqrt {x+y} \ dx....(1)$
$\int_{C_{1}} \sqrt {x+y} \ dx=\int_{0}^1 \sqrt {x+3x} \ dx = 2\int_0^1 \sqrt x d x \\= \dfrac{4}{3 } ..(2)$
$\int_{C_{2}} \sqrt {x+y} \ dx=\int_{1}^0 \sqrt {x+3} \ dx = \int_1^0 (x+3)^{1/2} dx \\=[\dfrac{2(x+3)^{3/2}}{3}]_0^1 \\ =2 \sqrt 3 - \dfrac{16}{3} ...(3) $
$\int_{C_{3}} \sqrt {x+y} \ dx=\int_{C_3} \sqrt {x+y} (0) =0 ...(4)$
Using equations (2), (3) and (4), equation (1) becomes:
$\int_C \sqrt {x+y} \ dx= \int_{C_{2}} \sqrt {x+y} \ dx+\int_{C_{2}} \sqrt {x+y} \ dx + \int_{C_{3}} \sqrt {x+y} \ dx \\=\dfrac{4}{3}+2\sqrt 3 - \dfrac{16}{3}+0 \\ =2\sqrt 3-4$
