The area enclosed between the curves y = loge (x + e), x = loge(1/y)

Question

The area enclosed between the curves y = log_e (x + e), x = log_e\(\left(\frac{1}{y}\right)\) and the x-axis is

A. 2

B. 1

C. 4

D. None of these

Answer 1

Correct answer is A.

y =log_e (x + e) and x = log_e\(\left(\frac{1}{y}\right)\) look like –

The curves intersect at (0, 1)

(Putting x = 0 in the 2 curves, y = log_e (e) = 1 and 0 = log_e (1/y),

i.e., y = 1/e⁰ = 1)

So, bounds are x = 1 – e to x = 0 for the first curve and then x = 0 to apparently x = ∞ for the second curve.

Therefore,