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Math  /  Calculus

QuestionEvaluate the integral I2=∫L2(x+y)dx+(y−x)dyI_{2}=\int_{L_{2}}(x+y) d x+(y-x) d y where L2L_{2} is the path from P(1,1)P(1,1) to Q(2,2)Q(2,2) via R(2,1)R(2,1). I2=I_{2}= □\square

Studdy Solution

STEP 1

1. The integral is a line integral over a piecewise linear path.
2. The path L2 L_2 consists of two segments: from P(1,1) P(1,1) to R(2,1) R(2,1) , and from R(2,1) R(2,1) to Q(2,2) Q(2,2) .
3. The line integral will be evaluated separately over each segment and then summed.

STEP 2

1. Parameterize the path from P(1,1) P(1,1) to R(2,1) R(2,1) .
2. Evaluate the line integral over the first segment.
3. Parameterize the path from R(2,1) R(2,1) to Q(2,2) Q(2,2) .
4. Evaluate the line integral over the second segment.
5. Sum the results from both segments.

STEP 3

Parameterize the path from P(1,1) P(1,1) to R(2,1) R(2,1) . This path is a horizontal line where y=1 y = 1 and x x varies from 1 to 2. We can parameterize it as:
x=t,y=1,where t∈[1,2] x = t, \quad y = 1, \quad \text{where } t \in [1, 2]

STEP 4

Substitute the parameterization into the integral and evaluate over the first segment:
∫12((t+1) dt+(1−t)⋅0) \int_{1}^{2} ((t + 1) \, dt + (1 - t) \cdot 0) =∫12(t+1) dt = \int_{1}^{2} (t + 1) \, dt
Calculate the integral:
=[t22+t]12 = \left[ \frac{t^2}{2} + t \right]_{1}^{2} =(222+2)−(122+1) = \left( \frac{2^2}{2} + 2 \right) - \left( \frac{1^2}{2} + 1 \right) =(2+2)−(12+1) = (2 + 2) - \left( \frac{1}{2} + 1 \right) =4−32 = 4 - \frac{3}{2} =82−32 = \frac{8}{2} - \frac{3}{2} =52 = \frac{5}{2}

STEP 5

Parameterize the path from R(2,1) R(2,1) to Q(2,2) Q(2,2) . This path is a vertical line where x=2 x = 2 and y y varies from 1 to 2. We can parameterize it as:
x=2,y=t,where t∈[1,2] x = 2, \quad y = t, \quad \text{where } t \in [1, 2]

STEP 6

Substitute the parameterization into the integral and evaluate over the second segment:
∫12((2+t)⋅0+(t−2) dt) \int_{1}^{2} ((2 + t) \cdot 0 + (t - 2) \, dt) =∫12(t−2) dt = \int_{1}^{2} (t - 2) \, dt
Calculate the integral:
=[t22−2t]12 = \left[ \frac{t^2}{2} - 2t \right]_{1}^{2} =(222−2×2)−(122−2×1) = \left( \frac{2^2}{2} - 2 \times 2 \right) - \left( \frac{1^2}{2} - 2 \times 1 \right) =(2−4)−(12−2) = (2 - 4) - \left( \frac{1}{2} - 2 \right) =−2−(12−2) = -2 - \left( \frac{1}{2} - 2 \right) =−2+32 = -2 + \frac{3}{2} =−42+32 = -\frac{4}{2} + \frac{3}{2} =−12 = -\frac{1}{2}

STEP 7

Sum the results from both segments to find the total integral:
I2=52+(−12) I_2 = \frac{5}{2} + \left(-\frac{1}{2}\right) =52−12 = \frac{5}{2} - \frac{1}{2} =42 = \frac{4}{2} =2 = 2
The value of the integral I2 I_2 is:
2 \boxed{2}

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