Problem 1: Find these integrals
(a) \int (e-4x+1) dx= -1/4*e-4x+x+c
(b) \int13 x*(x2-2)7 dx, put u=x2-2, 1/2*\int-17 u7 du=1/16*u8 from -1 to 7 is 1/16*(78-(-1)8).
(c) \int((sec(x))2tan(x) dx, put u=tan(x), so du=sec(x)2dx, \int u du= u2/2+c, so tan(x)2/2+c.
Problem 2: (a) Find \limx->Pi/2 sin(4x)/cos(x). Since sin(2Pi)=cos(Pi/2)=0, this is a 0/0 so we use L'Hopitals rule. We get
\limx->Pi/2 4cos(4x)/(-sin(x))=-4.
(b) limn->\infty 1/n*\sumi=1 ((i/n)2+(i/n)). This is the limit of the Riemann sum approximations to the integral \int01 (x2+x) dx= 1/3+1/2=5/6.
Problem 3: Find the area of the region bounded below by the curve y=3x and bounded above by the curve y=4-x2.
These curves intersect when 3x=4-x2, or x=1, x=-4. So the area is \int-41 (4-x2-3x) dx= 4x-x3/3-3x2/2 evaluated from -4 to 1, which is 125/6.
Problem 4: True or False? (Please give your reason in complete sentences. An answer alone is worth zero.) If we define for x>0, F(x)=\intx 2x 1/t dt, then F(x) is a constant independent of x.
Solution #1: True. F(x)=ln(2x)-ln(x)=ln(x)+ln(2)-ln(x)=ln(2) is independent of x.
Solution #2: True. F '(x)=1/(2x)*2-1/x=0, so F(x) must be a constant, independent of x.
Problem 5: A rectangular box with square top and bottom is to be built with fixed volume V. The top and bottom cost $3 per square inch and the sides cost $1 per square inch. Find the dimensions of the cheapest box.
Let the square bottom and top have side lengths x, let the height be h. Then x2*h=V. The cost of the box is
cost=2*x2*3+4*x*h=6x2+4*x*V/x2=6x2+4V/x.
If we take d/dx we get d/dx(cost)= 12x-4V/x2=0 when x3=V/3, so x=cube root of V/3, and h= V/(cube root (V/3))2.
This is a local min, from either the first or second derivative tests. The cost goes to infinity as x approaches the endpoints of x=0 and x=infinity, so it is the global min too.