If the area bounded by the x-axis, the curve `y=f(x), (f(x)gt0)" and the lines "x=1, x=b " is equal to "sqrt(b^(2)+1)-sqrt(2)" for all "bgt1,` then find f(x).

The area bounded by the x-axis,the curve y=f(x), and the lines x=1,x=b is equal to sqrt(b^(2)+1)-sqrt(2) for all b>1, then f(x) is sqrt(x-1)(b)sqrt(x+1)sqrt(x^(2)+1)(d)(x)/(sqrt(1+x^(2)))

The area bounded by the x-axis the curve y = f(x) and the lines x = a and x = b is equal to sqrt(b^(2) - a^(2)) AA b > a (a is constant). Find one such f(x).

Area bounded by the line y=x, curve y=f(x),(f(x)gtx,AA xgt1) and the lines x=1,x=t is (t-sqrt(1+t^2)-(1+sqrt2)) for all tgt1 . Find f(x) .

If the area bounded by the curve y=f(x), x-axis and the ordinates x=1 and x=b is (b-1) sin (3b+4), then-

The area bounded by the curve x=f(y) , the Y-axis and the two lines y=a and y=b is equal to

Area bounded by the curve y = f (x) and the lines x =a, =b and the x axis is :

Find the area bounded by the curve 2x^2-y=0 and the lines x=3, y=1 and the x-axis all in first quadrant.