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## Homework Statement

Evaluate the line integral [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along

(a) a straight line from (0,1) to (1,2);

(b) the parabola x=t, y=t

^{2}+ 1;

(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

## Homework Equations

Equation of line: y = mx + c

## The Attempt at a Solution

(a) [/B]To convert given integral into one variable I used Equation of line

y=mx + c

where slope m = [itex]\frac {y_2 - y_1}{x_2 - x_1} = \frac{2-1}{1-0}=1[/itex]

and y intercept c = 1

which gives us y = x+1

dy = dx

Thus given integral

[itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]

becomes one variable x dependent

[itex] = ∫^{1}_{0} (x^2-x-1)dx + (({x+1})^2+x)dx[/itex]

[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+2x+1+x)dx[/itex]

[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+3x+1)dx[/itex]

[itex] = ∫^{1}_{0} (x^2-x-1+x^2+3x+1)dx[/itex]

[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex] ------------------ (1)

solving

[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]

applying limits gives

[itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

**(b)**need some hints for this part

**(c)**straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

from equation (1) above

[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex]

solving

[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]

from (0,1) to (1,1)

we use only x: 0 to 1

we get [itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

from (1,1) to (1,2)

we use only x: 1 to 1

we get [itex]\frac{2}{3} - \frac{2}{3} + 1 - 1 = 0 [/itex]

hence adding both parts gives [itex] \frac{5}{3} [/itex]

Please tell me am I on the right track?

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