Calculus: Line Integrals

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Jeffrey Jordon · Accepted Answer

ExplanationLine integrals can be used to evaluate integrals of two- or three-dimensional curvesThe curve C is expressed by the following parametric equations given byx=t2 and y=2tThus, to evaluate the following integral given byI=∫cxyds we use the following line integral formula given by∫cf(x,y)ds=∫abf(x(t),y(t))|r′|dt=∫abf(x(t),y(t))|drdt|dt==∫abf(x(t),y(t))(dxxt)2+(dyxt)2dtSince x(t)=t2 and y(t)=2tWe want to evaluate ∫cxyds as followsFirstly, we must find the integrand f(x(t),y(t)) as followsf(x(t),y(t))=xy=t2(2t)=2t3Also, the derivatives of x(t) and y(t) must beddtx(t)=ddt(t2)=2t and ddty(t)=ddt(2t)=2Since 0≤t≤5, so that the upper and lower limits of the integral must be a=0 and b=5Thus, using the line integral formula, we get∫cxyds=∫abf(x(t),y(t))(dxxt)2+(dyxt)2dt=∫05(2t3)(2t)2+(2)2dt=2∫05t34t2+4dt=2∫05t34(t2+1)dt=2∫052t3(t2+1)dt=4∫05t3(t2+1)dtTo evaluate the following integral ∫05t3(t2+1)dt, we use the method of substitution as follows. Let u=t2+1, so thatDifferentiable both sides of the following equation u=t2+1, implies thatdu=2tdt and tdt=du2Also, we have t2=u−1Also, we find that the upper and lower limits of the integral must bet=0⇒u=02+1=1 and t=5⇒u=52+1=26Thus, the line integral becomes∫cxyds=4∫t=05t2t2+1tdt4∫t=026(u−1)u(du2)=42∫t=026(u−1)udu=2∫t=026(u32−12)duThus, the following line integral implies that∫cxyds=2[u525/2−u323/2]u=126=2[2u525−2u323]u=126=2[(2(26)525−2(26)323)−(2(1)525−2(1)323)]=815(94926+1)

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