Question 126974
(3,f(3)) is at least a local minimum because the first derivitive goes to 0 at 3 and the second derivitive is positive at 3.  The picture proves that the minimum is absolute rather convincingly.


{{{graph(600,600,-5,5,-40,40,x^4-4x^3-5)}}}


2nd problem.  


{{{int(4-x^2,dx,0,2)}}}={{{(4(2)-(2^3/3))-(4(0)-(0^3/3))=8-8/3=16/3}}}


See the graph below for an illustration of the unreasonability of your answer.  The secant line {{{y=-2x+4}}} forms a right triangle with the axes with an area of 4.  By inspection you can see that the area of this triangle is less than the desired area under the function between the given limits, yet your answer of {{{8/3}}} is clearly less than 4.


{{{drawing(600,600,-5,5,-5,5,
grid(1),
graph(600,600,-5,5,-5,5,4-x^2,-2x+4))}}}