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# NOTES ON MA114 HOMEWORK 5.1-5.3
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#  
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# p.12, no 7
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# The car's velocity at time t is 60-20*t. 
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> Int(60-20*t,t=0..4);\

> int(60-20*t,t=0..4);\

# Since this number is positive and east is the positive direction the
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# net effect of the car's motion is that it has traveled 80 miles to the
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# east.  The average value of  v(t) on the interval [0,4] is
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> Int(60-20*t,t=0..4)/(4-0);\

# or 20. We can see this graphically by:
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> plot({60-20*t,20},t=0..4);\

# Note that the area under the  average line  is exactly the integral of
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# the velocity over the interval [0,4].  That is, the distance covered
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# in the same interval by traveling at the average velocity is exactly
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# the distance traveled. The speed is the absolute value of the
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# velocity. The integral of the speed from 0 to 4 will be the total
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# distance covered rather than the net distance traveled.
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> Int(abs(60-20*t),t=0..4);\

> \

> int(abs(60-20*t),t=0..4);\

# The car's average speed will be
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> Int(abs(60-20*t),t=0..4)/4=int(abs(60-20*t),t=0..4)/4;\

# p 12 no 24.
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# We are given that 
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> Int(sin(x),x=0..Pi)=2;\

# The point of the problem is that this information and the periodicity
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# of the graph of sin and the fact that the graph of cos is just a shift
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# of that of sin is all we need to know to do most of these integrals.
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# This is a very important idea - using one integral which we know we
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# can then calculate many more using symmetry and the basic properties
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# of integrals.
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> plot({sin(x),cos(x)},x=0..4*Pi);\

> plot( abs(sin(x)) ,x=0..4*Pi);\

# From the graphs we see that the integral from 0 to 1 of  sin(x) must
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# be half that from 0 to Pi, or 1 and the integral from Pi to 2Pi must
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# be -2.
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# p. 15 no 34 Show  
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>  Int(exp(sin(x)),x=0..2*Pi)<=PI*(1+e);  \

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# Note from the graph of sin above that for Pi<=x<=2*Pi, sin(x) is
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# negative. Thus in that range we have sin(x)<=0 or 
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# exp(sin(x)<=exp(0)=1.  For x between 0 and Pi we have sin(x)<=1 and
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# exp(sin(x)<=exp(1)=e. Thus
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> Int(exp(sin(x)),x=0..2*Pi)<=Int(e,x=0..Pi)+Int(1,x=Pi..2*Pi);\

# We also want 
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> Pi*(1+1/e)<=Int(exp(sin(x)),x=0..2*Pi);\

> \

# Since for x from 0 to Pi we have 0<=sin(x) and for x from Pi to 2Pi we
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# have -1<=sin(x) we have
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> Int(exp(0),x=0.. Pi)+Int(exp(-1),x=Pi..\

> 2*Pi)<=Int(exp(sin(x)),x=0..2*Pi);\

# Putting these together complete the problem.
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# p. 25 no 14.
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# Note that  G and F differ by a constant, namely the integral of f 
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# from -3 to 2. The graph given is  that of the deriviative if G (and
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# H). A function is increasing if its derivative is positive so  F is
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# increasing whenever f is positive, that is when its graph is above the
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# x-axis, i.e from -5 to -3 and 1 to 5.  A functionhas a local minimum
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# as a point at which the deriviative is negative to the left and
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# positive to the right. Since f is negative to the left of x=1and
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# positive to the right it has a local minimum there. 
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# G achieves its minimum value at x=1 and maximum value at x=-3. At that
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# point its value is 3 while at 1 its
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# 3-2Pi. The area from 1 to 5 is 4 so the F(5)= 7-2Pi which is less than
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# 3. Thus the maximum value is at x=-3.              
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#  Since G and H differ by a constant their maximan and minima occur at
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# the same points. The values are easily calculated.
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# p 35, no1
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# The function f shown has a minimum between 1 and 2  where the sign of
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# its derivative changes from negative to positive. This means that the
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# second  derivative of F changes from negative to positive there - i.e.
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# that F has a flex point where its concavity changes from down to up.
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# Only Graph A has this property. 
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# For part (b) if F and g have the same derivative then they differ by a
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# constant. If  g(0)=2  then since F(0)=0, the graph of g is that of F
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# shifted up by 2 units.
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# p35  no 2
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# The critical points of a function are where its derivative does not
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# exist or is zero. This makes the critical poits of this f = -5 (end
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# point), -3.5, -2,2, 5 (end point). Points of inflection are where the
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# sign of the second derivative changes sign from positive to negative
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# or negative to positive.  The sign of the derivative of f (the second
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# derivative of F) changes sign at -3,-2,1, and 3. 
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# The G(x) given will have the function shown as its second derivative.
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# G will be concave up when its second derivative is positive, that is,
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# when the function f is positive.
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