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.

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

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

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

Solve sin 3x = - (1)/(sqrt2) , 0 lt x lt 2pi.

A function y =f(x) is defined as follow: " " y= f(x) ={ (x^(2) ,when 0 lt xlt 1),( sqrt x ,when xgt 1):} Find the area above the x-axis included between the curve y=f(x) and the line x=4

(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.

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.

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.

y = sin^(-1) x + sin^(-1)sqrt(1 - x^2), 0 lt x lt 1 find (dy)/(dx) .

Prove that pi/6 lt int_0^1(dx) (sqrt(4-x^2-x^3)) lt pi/(4sqrt(2))