Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5).

Answer:[tex]\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}[/tex]Step-by-step explanation:The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately. Let  [tex]C_1,C_2[/tex]   be the two paths. Recall that if we parametrize a path C as [tex](r_1(t),r_2(t),r_3(t))[/tex] with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is [tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt[/tex] Given any two points P, Q we can parametrize the line segment from P to Q as r(t) = tQ + (1-t)P with 0≤ t≤ 1 The parametrization of the line segment from (1,1,1) to (2,2,2) is r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t) r'(t) = (1,1,1) and  [tex]\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}[/tex] The parametrization of the line segment from (2,2,2) to  (-9,6,5) is r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  r'(t) = (-11,4,3) and  [tex]\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}[/tex] Hence [tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]