Application of Integration · Mathematics · WB JEE

The function f(x) which satisfies $$f(x) = f( - x) = {{f'(x)} \over x}$$ is given by

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

Select the correct statement from (a), (b), (c), (d). The function $$f(x) = x{e^{1 - x}}$$

The equation ex + x $$-$$ 1 = 0 has, apart from x = 0

The function $$f(x) = {e^{ax}} + {e^{ - ax}},a > 0$$ is monotonically increasing for

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

If $$x{{dy} \over {dx}} + y = x{{f(xy)} \over {f'(xy)}}$$, then $$|f(xy)|$$ is equal to

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 ...

Let f : R $$\to$$ R be such that f(0) = 0 and $$\left| {f'(x)} \right| \le 5$$ for all x. Then f(1) is in

The straight the through the origin which divides the area formed by the curves y = 2x $$-$$ x2, y = 0 and x = 1 into two equal halves is

The area bounded by the parabolas $$y = 4{x^2},y = {{{x^2}} \over 9}$$ and the straight line y = 2 is

If $${x^2} + {y^2} = {a^2}$$, then $$\int\limits_0^a {\sqrt {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} dx = } $$

Area in the first quadrant between the ellipses x2 + 2y2 = a2 and 2x2 + y2 = a2 is

The area of the region$$\{ (x,y):{x^2} + {y^2} \le 1 \le x + y\} $$ is

Let P and T be the subsets of k, y-plane defined byP = {(x, y) : x > 0, y > 0 and x2 + y2 = 1}T = {(x, y) : x > 0, y > 0 and x8 + y8 < ...

The area of the figure bounded by the parabolas x = $$-$$ 2y2 and x = 1 $$-$$ 3y2 is

Solve : $$({x^2} + 4{y^2} + 4xy)dy = (2x + 4y + 1)dx$$

The area of the figure bounded by the parabola $$x = - 2{y^2},\,x = 1 - 3{y^2}$$ is

The area bounded by y = x + 1 and y = cos x and the X-axis, is