9>Area of region
Area of square-area under graph
(2*1)-(Integration of (3x+1)1/2 by 0,1)
2-[2/9*(3X+1)3/2],0,1
=4/9
ii> Volume of region
volume of cylinder-volume under graph
Pi*2*2-Pi integration of [(3x+1)1/2]2 to 0,1
4Pi-Pi[1/6*(3x+1)2] to 0,1
4Pi-Pi[8/3-1/6]
4Pi-2.5Pi
1.5Pi
iii> Differentiate da eqn & substitute point of x to find da gradient
for x=1 m=3/4 for x=0 m=3/2
tanQ=3/4 tanQ=3/2
Q=36.87 56.31
Thare4 Acute angel =56.31-36.87
=19.4
[tanQ because gradient is also =y/x]