EvaluateJcF. dr.F(x, y) = xi + yjC: r(t) = (5t + 8)i + tj,0 ≤t≤1

A line integral is used to find the integral of a function over the vector function. Line integral has two types of paths. The line integral of the open path and the line integral in a closed path. A line integral is denoted by ∫C where c represents the curve. In line integral, we are dealing with integration with the limit of a curve.The given function is, F(x,y)=xi+yj........(1) r(t)=(5t+8)i+tj......(2) where r(t) in the form of xi+yj.That is, r(t)=xi+yj.....(3) Comparing equations (2) and (3), x=5t+8y=t Now differentiate equation (2) on both sides, r(t)=(5t+8)i+tj dr=(5×1+0)idt+1×jdt dr=5idt+jdt dr=(5i+j)dt........(4) The given integral is, ∫CF.dr Put equation (4),(1) in the integral, ∫CF.dr=∫C(xi+yj)(5i+j)dt ∫CF.dr=∫C(5x+y)dt Put value of x=5t+8,y=t in the integral, ∫CF.dr=∫C(5×5t+8+t)dt =∫C5×5t+8×5+tdt =∫C25t+40+tdt ∫CF.dr=∫C26t+40dt The value of t very from 0 to 1 given.Applying an integral, ∫CF.dr=∫0126t+40dt =26t22+40t01 By using ∫xndx=xn+1n+1 ∫CF.dr=13t2+40t01 Now Upper Limit-Lower Limit applying, ∫CF.dr=13(1)2+40(1)-13(0)2+40(0) =(13×1+40)-0 =13+40∫CF.dr=53sq.unit The value of the integral is 53 sq. unit.