# hence expression

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```2GM
v 2 GM
is

 c , then show that v02 
R
2
y
constant and hence can be the value of the constant of integration in the
expression you come up with when you do the integration. Or, take the derivative
of the second expression and show that you get the first expression (without
integral signs).
2. Given an expression for position (which you are), velocity is its derivative and
acceleration is its second derivative. Hence,
a(t )  x ''(t )  6t  12
a. v(t )  x '(t )  3t 2  12t  9
b. Movement to the right means a positive velocity (velocity is a vector,
remember?), so we want to know when v(t) &gt; 0.
v(t )  0  t  1 or t  3 . Try specific values of v in the ranges (0,1),
(1,3), and (3,5):
v (0)  9
v (2)  3
v (4)  9
Hence, the particle is moving to the right on [0,1) and (3,5]
c. a(t )  0  t  2 and v(2) = -3
3. This is tricky and requires either two steps or looking at the concept of area from
a different perspective. If  f ( x)dx is the area between a curve and the x-axis,
1. Integrate the first equation to find
then you might suspect that we can integrate with respect to dy to get the area
between the curve and the y-axis. Draw in some rectangles perpendicular to the
y-axis if you don’t see this. If you do see it, then you need to realize that if the
base of each rectangle is dy the height of each rectangle is the distance between
the curve and the y-axis , and that distance is simply x . So, we want the sum of
all those rectangles of area x dy. But that’s just  xdy . Two keys here that you
should recall from substitution: the limits of integration and the function to be
integrated both have to be expressed in terms of the variable with respect to which
we’re integrating, in this case y. So, our limits of integration in terms of y are 0
and 4, and our integral becomes  xdy   ydy . In other words, the expression
we want is

4
0
1
y 2 dy . That expression equals 16 3 .
The other way to look at it is to draw the diagram and realize that the area we
want is the area of the rectangle formed by the interval on the x-axis [0,2] and the
interval on the y-axis [0.4] minus the area between the curve and the x-axis.
2
That expression is 8   x 2dx  8  8
3
0
which, conveniently, is also 16 . Isn’t math wonderful??
3
4. The definition of an odd function is that f(-x) = -f(x). Hence, the part of f for
which x &lt; 0 is simply the negative of the value of f when x&gt;0. We know in
general that the integral over a domain is the sum of the integrals over any two

contiguous subdomains—i.e., that
b
c
b
a
c
g ( x)dx   g ( x)dx   g ( x)dx provided that
a
a &lt; c &lt; b. (Draw the situation for some function g(x) if you don’t see why this is
true. So,

a
a
0
a
a
0
0
f ( x)dx   f ( x)dx   f ( x)dx . But from the definition of an odd
function, we know that

a
a
f ( x)dx   f ( x) .
0
a
a
a
a
0
0
Hence  f ( x)dx   f ( x)dx   f ( x)dx  0
5. Use u  1  x 2 as a substitution to get  1 1  x 2  2  C
3
6. First, show that the curve is positive over the given domain, perhaps by showing
that its derivative is never zero on that interval. Then,
3
area =


0


0
(2sin x  sin 2 x)dx
(2sin x  sin 2 x) dx 

1
2cos x  cos 2 x 
2
0
1
1 

2(1)  (1)   2 1  1 
2
2 

1
1
2 2  4
2
2
```