00:01
Hi here for the given question.
00:03
We are given that there is a line integral over c, which is x square plus y square dx plus 2 xy dy by c is a constant arc for the parabola c is a arc of the parabola y is equal to x square from 0 0 2 1 1 followed by the line segment 1 1 2 0 0.
00:48
So here using the direct method, we need to compute the value of the parameterization curve.
00:55
So here the parameterization will be for y is equal to x square.
00:59
So here let x is equal to t.
01:01
So y will be equal to t square.
01:03
So here in our case, we have dx is equal to dt which further implies dy is equal to 2t dt.
01:10
So here further using this value of a line integral can be written as here.
01:16
We have integration over 0 to 1 for x square plus y square dx plus 2 xy dy.
01:27
So here by substituting the above value of x and y in terms of t this can be further reduced to integration over 0 to 1 t square plus t to the power 4 multiplied with dt plus 4 times t to the power 4 dt.
01:43
So simplifying this again here we can say that our equation will be equal to t square plus 5 t to the power 4 integration over 0 to 1 dt.
01:58
So here solving this further and simplifying this here.
02:02
We have value as t cube upon 3 plus 5 times t to the power 5 upon 5 and the limit is from 0 to 1.
02:10
So again, this can be written as 1 by 3 plus 1.
02:13
So this is equal to 4 by 3.
02:15
So here further the parameterization is from 1 to 1 to 0 0 for the next step.
02:21
So here let x of u is equal to 1 minus u and y of u is equal to 1 minus u.
02:29
So further we have dx is equal to minus du and further dy is equal to minus du.
02:36
So using this value in our above line integral and simplifying further we can have the value of integration as integration over here in our case the integration will be for the line segment.
02:51
So here it will be from 1 to 0 and here in our case further this can be written as 1 minus u whole square plus 1 minus u whole square multiplied with minus du plus 2 multiplied with 1 minus u multiplied with 1 minus u minus du...