For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 6 secs?

Question

For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 6 secs? 
 	
	-36
	36
	0
	55.596

For an object whose velocity in ft/sec is given by v(t) = -2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?
 	
	4.876
	2.667
	9.752
	2.438

Please help with these 2 questions.

M · Accepted Answer

The first one is -36, the second one is NOT 2.667. (FLVS)

bobpursley · Answer

d=int v(t)dt=int (- t^2+6)= = -1/3 t^3+6t over limits = -1/3(6^3)+36 = -1/3(216)+36=-72+36

Vikram · Answer

The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are squares. What is the volume, in cubic units, of the solid?

Can you please help me out on this one Bobby?

Also, for the second question I got 2.667 which I think is right.