Evaluate the integral between (1,1) and (4,2): [ ( x + y ) d x + ( y − x ) d y ] [(x+y)dx + (y-x)dy] [( x + y ) d x + ( y − x ) d y ]
a) a straight line.
b) straight lines from (1,1) to (1,2) and then to (4,2).
c) the curve x = 2 t 2 + t + 1 x = 2t^2 + t + 1 x = 2 t 2 + t + 1 y = t 2 + 1 y = t^2 + 1 y = t 2 + 1
∫ L ( x + y ) d x + ( y − x ) d y \int_L (x + y) \, dx + (y - x) \, dy ∫ L ( x + y ) d x + ( y − x ) d y
a) The equation of the straight line passing through the points (1,1) and (4,2) is x − 1 4 − 1 = y − 1 2 − 1 \frac{x-1}{4-1} = \frac{y-1}{2-1} 4 − 1 x − 1 = 2 − 1 y − 1 y = 1 3 x + 2 3 y = \frac{1}{3}x + \frac{2}{3} y = 3 1 x + 3 2 d y = 1 3 d x dy = \frac{1}{3}dx d y = 3 1 d x
∫ L ( x + y ) d x + ( y − x ) d y = ∫ 1 4 ( ( x + 1 3 x + 2 3 ) + ( 1 3 x + 2 3 − x ) × 1 3 ) d x = = ∫ 1 4 ( 10 9 x + 8 9 ) d x = ( 5 9 x 2 + 8 9 x ) ∣ 1 4 = ( 80 9 + 32 9 − 5 9 − 8 9 ) = 11 \begin{aligned}
\int_L (x + y) \, dx + (y - x) \, dy &= \int_1^4 \left( \left(x + \frac{1}{3}x + \frac{2}{3}\right) + \left( \frac{1}{3}x + \frac{2}{3} - x \right) \times \frac{1}{3} \right) dx = \\
&= \int_1^4 \left( \frac{10}{9}x + \frac{8}{9} \right) dx = \left( \frac{5}{9}x^2 + \frac{8}{9}x \right) \Bigg|_{1}^{4} = \left( \frac{80}{9} + \frac{32}{9} - \frac{5}{9} - \frac{8}{9} \right) = 11
\end{aligned} ∫ L ( x + y ) d x + ( y − x ) d y = ∫ 1 4 ( ( x + 3 1 x + 3 2 ) + ( 3 1 x + 3 2 − x ) × 3 1 ) d x = = ∫ 1 4 ( 9 10 x + 9 8 ) d x = ( 9 5 x 2 + 9 8 x ) ∣ ∣ 1 4 = ( 9 80 + 9 32 − 9 5 − 9 8 ) = 11
b) The equation of the straight line passing through the points (1,1) and (1,2) is x = 1 x = 1 x = 1 d x = 0 dx = 0 d x = 0 y = 2 y = 2 y = 2 d y = 0 dy = 0 d y = 0
Hence
∫ L ( x + y ) d x + ( y − x ) d y = ∫ 1 2 ( y − 1 ) d y + ∫ 1 4 ( x + 2 ) d x = ( y 2 2 − y ) ∣ 1 2 + ( x 2 2 + 2 x ) ∣ 1 4 = = ( 4 2 − 2 − 1 2 + 1 ) + ( 16 2 + 8 − 1 2 − 2 ) = 14 \begin{aligned}
\int_L (x + y) \, dx + (y - x) \, dy &= \int_1^2 (y - 1) \, dy + \int_1^4 (x + 2) \, dx = \left( \frac{y^2}{2} - y \right) \Bigg|_{1}^{2} + \left( \frac{x^2}{2} + 2x \right) \Bigg|_{1}^{4} = \\
&= \left( \frac{4}{2} - 2 - \frac{1}{2} + 1 \right) + \left( \frac{16}{2} + 8 - \frac{1}{2} - 2 \right) = 14
\end{aligned} ∫ L ( x + y ) d x + ( y − x ) d y = ∫ 1 2 ( y − 1 ) d y + ∫ 1 4 ( x + 2 ) d x = ( 2 y 2 − y ) ∣ ∣ 1 2 + ( 2 x 2 + 2 x ) ∣ ∣ 1 4 = = ( 2 4 − 2 − 2 1 + 1 ) + ( 2 16 + 8 − 2 1 − 2 ) = 14
c) x = 2 t 2 + t + 1 x = 2t^2 + t + 1 x = 2 t 2 + t + 1 y = t 2 + 1 y = t^2 + 1 y = t 2 + 1
d x = ( 4 t + 1 ) d t and d y = 2 t d t dx = (4t + 1)dt \text{ and } dy = 2t \, dt d x = ( 4 t + 1 ) d t and d y = 2 t d t
When (1,1), then 2 t 2 + t + 1 = 1 2t^2 + t + 1 = 1 2 t 2 + t + 1 = 1 t 2 + 1 = 1 t^2 + 1 = 1 t 2 + 1 = 1 t = 0 t = 0 t = 0
When (4,2), then 2 t 2 + t + 1 = 4 2t^2 + t + 1 = 4 2 t 2 + t + 1 = 4 t 2 + 1 = 2 t^2 + 1 = 2 t 2 + 1 = 2 t = 1 t = 1 t = 1
∫ L ( x + y ) d x + ( y − x ) d y = ∫ 0 1 ( ( 2 t 2 + t + 1 + t 2 + 1 ) ( 4 t + 1 ) + ( t 2 + 1 − 2 t 2 − t − 1 ) 2 t ) d t = = ∫ 0 1 ( 10 t 3 + 5 t 2 + 9 t + 2 ) d t = ( 5 t 4 2 + 5 t 3 3 + 9 t 2 2 + 2 t ) ∣ 0 1 = ( 5 2 + 5 3 + 9 2 + 2 ) = 10 2 3 \begin{aligned}
\int_L (x + y) \, dx + (y - x) \, dy &= \int_0^1 \left( (2t^2 + t + 1 + t^2 + 1)(4t + 1) + (t^2 + 1 - 2t^2 - t - 1)2t \right) dt = \\
&= \int_0^1 (10t^3 + 5t^2 + 9t + 2) dt = \left( \frac{5t^4}{2} + \frac{5t^3}{3} + \frac{9t^2}{2} + 2t \right) \Bigg|_{0}^{1} = \left( \frac{5}{2} + \frac{5}{3} + \frac{9}{2} + 2 \right) = 10\frac{2}{3}
\end{aligned} ∫ L ( x + y ) d x + ( y − x ) d y = ∫ 0 1 ( ( 2 t 2 + t + 1 + t 2 + 1 ) ( 4 t + 1 ) + ( t 2 + 1 − 2 t 2 − t − 1 ) 2 t ) d t = = ∫ 0 1 ( 10 t 3 + 5 t 2 + 9 t + 2 ) d t = ( 2 5 t 4 + 3 5 t 3 + 2 9 t 2 + 2 t ) ∣ ∣ 0 1 = ( 2 5 + 3 5 + 2 9 + 2 ) = 10 3 2 
