A curve is given by `y={(sqrt(4-x^2)),0lt=x<1and sqrt((3x)),1lt=xlt=3.` Find the area lying between the curve and x-axis.

Find the area bounded by the curves y=sqrt(x),2y+3=x and x -axis.

The area enclosed between the curves y = x^(3) and y = sqrt(x) is

(i) Draw a rough sketch of y = sin 2x and determine the area enclosed by the curve, x-axis and the lines x = (pi)/(4) and x = (3pi)/(4). (ii) Draw the graph of y = cos 3x, 0 lt x le (pi)/(6) and find the area between the curve and the axes.

Area under the curve y=sqrt(3x+4) between x=0 and x=4 is

Make a rough sketch of the graph of the function y=4-x^2,\ 0lt=xlt=2 and determine the area enclosed by the curve, the x-axis and the lines x=0\ a n d\ x=2.

If A is the area lying between the curve y=sin x and x-axis between x=0 and x=pi//2 . Area of the region between the curve y=sin 2x and x -axis in the same interval is given by

The area of the region lying between the line x-y+2=0 and the curve x=sqrt(y)and y-axis , is

y = sin ^(-1)(2xsqrt(1 - x^(2))),-(1)/sqrt(2) lt x lt (1)/sqrt(2)

y = sec^(-1)((1)/(2x^(2) -1 )), 0 lt x lt (1)/(sqrt(2))