The area enclosed between the curve y = 1 + x2, the y-axis, and the straight line y = 5 is given by
The line which is parallel to x-axis and crosses the curve y = $$\sqrt x$$ at an angle 45$$^\circ$$ is
The area included between the parabolas y2 = 4x and x2 = 4y is
The area enclosed by y = 3x $$-$$ 5, y = 0, x = 3 and x = 5 is
The area bounded by the parabolas y = 4x2, $$y = {{{x^2}} \over 9}$$ and the line y = 2 is
The area of the region bounded by y2 = x and y = | x | is
The area enclosed between y2 = x and y = x is
The area bounded by y2 = 4x and x2 = 4y is
The area bounded by the curves $$x=4-y^2$$ and the Y-axis is
Consider the function $$\mathrm{f}(x)=(x-2) \log _{\mathrm{e}} x$$. Then the equation $$x \log _{\mathrm{e}} x=2-x$$
Area of the figure bounded by the parabola $${y^2} + 8x = 16$$ and $${y^2} - 24x = 48$$ is
Let f be a non-negative function defined in $$[0,\pi /2]$$, f' exists and be continuous for all x and $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int\limits_0^x {f(t)dt} } $$ and f (0) = 0. Then
$$\{ (x,y):{x^2} + {y^2} \le 1 \le x + y\} $$ is
P = {(x, y) : x > 0, y > 0 and x2 + y2 = 1}
T = {(x, y) : x > 0, y > 0 and x8 + y8 < 1}
Then, P $$ \cap $$ T is