Q.330. Area enclosed by a curve whose parametric equations are given.

Question 330.
Find the area enclosed by the x-axis and the curve whose parametric equations are
x = 2 + e^t and y = t - t^2.

Answer 330.
To find the points of intersection of the curve with the x-axis, putting y = 0
=> t - t^2 = 0
=> t (1 - t) = 0
=> t = 0 or t = 1.

=> required area
= ∫ (t=0 to 1) y dx
= ∫ (t=0 to 1) (t - t^2) e^t dt ... [x = 2 + e^t => dx = e^t dt]
= [(t - t^2 - (1 - 2t) + (-2)] e^t ... [t=0 to 1]

[Note: Here, the formula of integration,
∫ f (t) e^t dt = [f (t) - f '(t) + f "(t) - f "' (t) + .....] e^t + c is used.]

= (- t^2 + 3t - 3) e^t ... [t=0 to 1]
= (- 1 + 3 - 3) e - (- 3)
= 3 - e.

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