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Parametric Integration Simplified Revision Notes

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9.2.2 Parametric Integration

Integration of Parametric Equations

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📝Problem:

Find the area between the curve and the xx-axis between the given limits.

Given:

x=1+t,y=11tx = 1 + t, \quad y = 1 - \frac{1}{t}

Solution:

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  1. General Formula:
  • The area under a curve for a parametric equation is given by:
ydx\int y \, dx

For the given curve, the area is:

26(11t)dx\int_{2}^{6} \left(1 - \frac{1}{t}\right) \, dx
  1. Substituting:
  • Since yy is given in terms of t, we substitute:
y=11ty = 1 - \frac{1}{t}

Now we need to replace dxdx with an expression in terms of tt.

  1. Finding dxdt\frac{dx}{dt}:
  • Given x=1+tx = 1 + t:
dxdt=1\frac{dx}{dt} = 1

Therefore, dx=dtdx = dt.

  1. Adjusting the Integral:
  • Replace dxdx with dt in the integral:
t1t2(11t)dt\int_{t_1}^{t_2} \left(1 - \frac{1}{t}\right) \, dt
  1. Finding the Limits:
  • The limits were initially for dxdx. We now convert them to limits for tt:
  • Lower limit: When x=2x = 2, 2=1+t2 = 1 + t gives t=1t = 1.
  • Upper limit: When x=6x = 6, 6=1+t6 = 1 + t gives t=5t = 5.
  • Therefore, the integral becomes:
I=15(11t)dtI = \int_{1}^{5} \left(1 - \frac{1}{t}\right) \, dt
  1. Evaluating the Integral:
  • Solve the integral:
I=[tlnt]15=[(5ln5)(1ln1)]I = \left[t - \ln|t|\right]_{1}^{5} = \left[(5 - \ln 5) - (1 - \cancel{\ln 1})\right]
  • Simplify the expression:
I=:highlight[4ln5]I = :highlight[4 - \ln 5]

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📑Example: Find the Shaded Area

Given:

x=3t,y=sin(t),tRx = 3t, \quad y = \sin(t), \quad t \in \mathbb{R}

Explanation:

  1. Identifying the Limits:
  • In this example, no explicit limits are provided. However, it is clear that the limits lie on the xx-axis where y=0y = 0.
  • The sine function, sin(t)=0\sin(t)=0, when t=0t = 0 and t=πt = \pi. image
  1. Setting Up the Integral:
  • To find the area, we integrate ydxy \, dx:
ydx=sin(t)dx\int y \, dx = \int \sin(t) \, dx
  • We find dxdt=3\frac{dx}{dt} = 3, which gives us dx=3dtdx = 3 \, dt.
  1. Adjusting the Integral:
  • Substitute dx=3dtdx = 3 \, dt into the integral:
I=0πsin(t)×3dt=0π3sin(t)dtI = \int_{0}^{\pi} \sin(t) \times 3 \, dt = \int_{0}^{\pi} 3 \sin(t) \, dt
  1. Evaluating the Integral:
  • Calculate the integral:
I=[3cos(t)]0πI = \left[ -3 \cos(t) \right]_{0}^{\pi}
  • Substitute the limits:
=(3cos(π))(3cos(0))= (-3 \cos(\pi)) - (-3 \cos(0))
  • Simplify:
=3+3=:highlight[6]= 3 + 3 = :highlight[6]
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