Question

Evaluate the line integral, where C is the given curve. (x 6y) dx x2 dy, C C consists of line segments from (0, 0) to (6, 1) and from (6, 1) to (7, 0)

Answer 1

Answer:

$\mathbf{\int _c (x+6y)\ dx + x^2 \ dy = \dfrac{91}{6}}$

Step-by-step explanation:

The integral of the curve is given as:

$\int (x +6y) \ dx +x^2 \ dy$

Where the line segment from (0,0) to (6,1) is:

0 < x < 6 and $y = \dfrac{x}{6} \implies dy = \dfrac{dx}{6}$

Thus;

$\int ^{6,1}_{0,0} (x+ 6 * \dfrac{x}{6}) \ dx + \dfrac{x^2*dx}{6 } = \int^6_0 (x+x+\dfrac{x^2}{6}) \ dx$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx = \int^6_0 (2x +\dfrac{x^2}{6}) \ dx$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx = \begin {bmatrix} \dfrac{2x^2}{2}+\dfrac{x^3}{18} \end {bmatrix}^6_0$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx = \begin {bmatrix} x^2+\dfrac{x^3}{18} \end {bmatrix}^6_0$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx = \begin {bmatrix} (6)^2+\dfrac{(6)^3}{18} \end {bmatrix}$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx = \begin {bmatrix} 36+12 \end {bmatrix}$

$\int^6_0 (x+x+\dfrac{x^2}{6}) \ dx =48 --- (1)$

Similarly; at (6,1) to (7,0)

$\implies 6 \leq x \leq 7 \ and \ y = -x +7$

Thus; the integral can be computed as :

$\int ^{(7,0)}_{(6,1)} \ (x+6(-x+7)) \ dx + x^2 (-dx) = \int^7_6 (-5x +42 -x^2) \ dx$

$\int^7_6 (-5x +42 -x^2) \ dx= \begin {bmatrix}- \dfrac{5x^2}{2}+ 42x -\dfrac{x^3}{3} \end {bmatrix}^7_6$

$\int^7_6 (-5x +42 -x^2) \ dx= \begin {bmatrix}- \dfrac{5(7^2-6^2)}{2}+ 42(7-6) -\dfrac{(7^3-6^3)}{3} \end {bmatrix}$

$\int^7_6 (-5x +42 -x^2) \ dx= \begin {bmatrix}- \dfrac{5(49-36)}{2}+ 42(1) -\dfrac{(343-216)}{3} \end {bmatrix}$

$\int^7_6 (-5x +42 -x^2) \ dx= \begin {bmatrix}- \dfrac{5(13)}{2}+ 42 -\dfrac{(127)}{3} \end {bmatrix}$

$\int^7_6 (-5x +42 -x^2) \ dx= \begin {bmatrix}- \dfrac{-195+252-254}{6}\end {bmatrix}$

$\int^7_6 (-5x +42 -x^2) \ dx= - \dfrac{-197}{6} --- (2)$

By adding (1) and (2) ; we have:

$\int _c (x+6y)\ dx + x^2 \ dy = 48 + (-\dfrac{197}{6})$

$\int _c (x+6y)\ dx + x^2 \ dy = (\dfrac{288-197}{6})$

$\mathbf{\int _c (x+6y)\ dx + x^2 \ dy = \dfrac{91}{6}}$