Q8 of 123 Page 22

The area enclosed between the curves y =log_e (x + e), x = log_eand the x-axis is

y =log_e (x + e) and x = log_e 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,

(Ans)

If A_n be the area bounded by the curve y = (tan x)ⁿ and the lines x = 0, y = 0 and x = π/4, then for x > 2

The area of the region formed by x² + y² – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is

The area of the region bounded by the parabola (y – 2)² =x – 1, the tangent to it at the point with the ordinate 3 and the x-axis is

The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the x-axis is

The area enclosed between the curves y =loge (x + e), x = logeand the x-axis is