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Math  /  Calculus

QuestionEvaluate the integral I1=∫L1(x+y)dx+(y−x)dyI_{1}=\int_{L_{1}}(x+y) d x+(y-x) d y where L1L_{1} is the path from P(1,1)P(1,1) to Q(2,2)Q(2,2) via S(1,2)S(1,2). I1=I_{1}=

Studdy Solution

STEP 1

1. The integral is a line integral over a piecewise linear path.
2. The path L1 L_1 consists of two segments: from P(1,1) P(1,1) to S(1,2) S(1,2) and from S(1,2) S(1,2) to Q(2,2) Q(2,2) .
3. We will evaluate the line integral separately over each segment and then sum the results.

STEP 2

1. Parameterize the path from P(1,1) P(1,1) to S(1,2) S(1,2) and evaluate the integral over this segment.
2. Parameterize the path from S(1,2) S(1,2) to Q(2,2) Q(2,2) and evaluate the integral over this segment.
3. Sum the integrals from both segments to find the total integral I1 I_1 .

STEP 3

Parameterize the path from P(1,1) P(1,1) to S(1,2) S(1,2) . This path is vertical, so x=1 x = 1 and y y varies from 1 to 2. The parameterization can be x(t)=1 x(t) = 1 and y(t)=t y(t) = t for t t in [1,2][1, 2].
The differential changes are dx=0 dx = 0 and dy=dt dy = dt .
Substitute into the integral:
∫12((1+t)⋅0+(t−1)⋅dt) \int_{1}^{2} \left((1 + t) \cdot 0 + (t - 1) \cdot dt \right)
Simplify and evaluate:
∫12(t−1) dt \int_{1}^{2} (t - 1) \, dt
Calculate the integral:
[t22−t]12=(42−2)−(12−1)=(2−2)−(12−1)=0+12=12 \left[ \frac{t^2}{2} - t \right]_{1}^{2} = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) = (2 - 2) - \left( \frac{1}{2} - 1 \right) = 0 + \frac{1}{2} = \frac{1}{2}

STEP 4

Parameterize the path from S(1,2) S(1,2) to Q(2,2) Q(2,2) . This path is horizontal, so y=2 y = 2 and x x varies from 1 to 2. The parameterization can be x(t)=t x(t) = t and y(t)=2 y(t) = 2 for t t in [1,2][1, 2].
The differential changes are dx=dt dx = dt and dy=0 dy = 0 .
Substitute into the integral:
∫12((t+2)⋅dt+(2−t)⋅0) \int_{1}^{2} \left((t + 2) \cdot dt + (2 - t) \cdot 0 \right)
Simplify and evaluate:
∫12(t+2) dt \int_{1}^{2} (t + 2) \, dt
Calculate the integral:
[t22+2t]12=(42+4)−(12+2)=(2+4)−(12+2)=6−52=122−52=72 \left[ \frac{t^2}{2} + 2t \right]_{1}^{2} = \left( \frac{4}{2} + 4 \right) - \left( \frac{1}{2} + 2 \right) = (2 + 4) - \left( \frac{1}{2} + 2 \right) = 6 - \frac{5}{2} = \frac{12}{2} - \frac{5}{2} = \frac{7}{2}

STEP 5

Sum the integrals from both segments to find the total integral I1 I_1 :
I1=12+72=82=4 I_1 = \frac{1}{2} + \frac{7}{2} = \frac{8}{2} = 4
The value of the integral I1 I_1 is:
4 \boxed{4}

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