Test 5 Solutions

Note: due to the limitations of HTML, the following changes in notation will be used throughout this web page:

• The integral of f(x) from a to b will be written here as: Int(f(x), dx, x=a..b).
• Since no Greek letters are available, the constant 3.14159265... will be written here as Pi.

I.    For each of the following, find U_f(P) and L_f(P).  (Recall that these stand for the upper sum and the lower sum of the function f with respect to the partition P.)

1. f(x) = 1/x, 1 < x < 4, P={1, 2, 3, 4}

This is a decreasing function, so in any interval, the maximum of f(x) will be at the left endpoint, and the minimum of f(x) will be at the right endpoint.  (This is clear from looking at a graph of y=1/x.) Therefore,

U_f(P) = (1)(1) + (1)(1/2) + (1)(1/3) = 11/6.
L_f(P) = (1)(1/2) + (1)(1/3) + (1)(1/4) = 13/12

This is an increasing function, so in any interval, the maximum of f(x) will be at the right endpoint, and the minimum will be at the left endpoint.  (In other words, this situation is exactly the opposite of what we had in problem #1.)

U_f(P) = (1)(2³) + (1)(3³) = 8+27 = 35
L_f(P) = (1)(1³) + (1)(2³) = 1+8 = 9.

II.    Calculate each of the following.

1. Int(2x, dx, x=2..6)

An antiderivative of 2x is x².  Evaluating at x=6 and x=2, then subtracting, gives us Int(2x, x=2..6) = 6² - 2² = 32.

An antiderivative of 2x is x².  Evaluating at x=2 and x=6, then subtracting, gives us Int(2x, x=6..2) = 2² - 6² = -32.
(Alternatively, simply make the observation that this integral must be the opposite of the integral in problem #1, since they are identical but for the order of the endpoints of the integral.)

An antiderivative of x³ is x⁴/4.  Evaluating at each endpoint and subtracting gives us 81/4 - 1/4 = 80/4 = 20.

This is an indefinite integral, so we must give an answer which accounts for ANY antiderivative.  This is done by allowing for a constant, C.  Therefore, the solution is x³ + x² + x + C, where C is any real number.

Rewrite the integrand as sec² x + sec x tan x.  These are antiderivatives of tan x and sec x, respectively.  Therefore, the solution is tan x + sec x + C,. where C is any real number.

1. F(1) = 0, since the integral of any function over the interval [1,1] (i.e., one point) is zero.

2. F'(1)

The important observation here is that F'(x) = x-1/x (since F(x) is the integral of the function x-1/x). Therefore, F'(1) = 1-1/1 = 0.

3. F''(1)

F'(x) = x-1/x, so F''(x) = 1 + x^-2. Therefore, F''(1) = 1 + 1 = 2.

IV.    Find the area of the region bounded by the given curves.

1. y=0, y=4-x²

The area bounded by these two curves is simply the area "under" the parabola y=4-x², between x=-2 and x=2. This is clear if you make a sketch. This area is found by evaluating the definite integral Int(4-x², dx, x=-2..2). Solution: 32/3.

2. y=x/2, y=x^1/2

These two curves intersect exactly twice: at x=0, and at x=4. (This is seen by setting x/2=x^1/2 and solving for x.) In this interval, x^1/2 > x/2 (again, this is clear from a sketch), and so the area is found by evaluating Int(x/2 - x^1/2, dx, x=0..4). Solution: 8/3.

3. y= cos x, y = sin x, x = 0, x = Pi/2

The area bounded by these curves is the area between (not beneath) the sine curve and the cosine curve, between x=0 and x=Pi/2. Note that cos(x) > sin(x) on [0, Pi/4], but sin(x) > cos(x) on [Pi/4, Pi/2]. Therefore, the total area is the sum of two definite integrals: Int(cos(x)-sin(x), dx, x=0..Pi/4) + Int(sin(x)-cos(x), dx, x=Pi/4..Pi/2). Each of these integrals is equal to 2^1/2-1. Therefore: Solution: 2(2^1/2 - 1)

V.   Solve each of the following.

1. A car is moving in a straight line at a speed of 84 ft per second. Then the driver brakes, resulting in a deceleration of 24 feet per second per second.

a. How long will it take for the car to come to a complete stop?
b. Once the brakes are applied, how far does the car travel before coming to a complete stop?

To solve these questions, we will need to find v(t) and x(t) (the velocity and position functions, respectively). Since a(t)=-24 (acceleration), and v(t) is an antiderivative of a(t), we know that v(t)=-24t + C, for some constant C. Furthermore, we know the initial velocity: v(0) = 84. (Remember: initial conditions correspond to t=0.) Therefore, C=84, and v(t) = 84 - 24t.

We now know enough to solve part (a), by setting v(t)=0 and solving for t:

Solution to (a): The car stops after 3.5 seconds.

Now, to find x(t). This is an antiderivative of v(t)=84-24t, and so x(t) = 84t - 12t² + C. We're given no information about the car's initial position (and thus, no information about C.) However, this won't matter, since all we need is the distance travelled in the first 3.5 seconds; that is, between t=0 and t=3.5 Therefore,

Distance	=	x(3.5) - x(0)
	=	[84(3.5) - 12(3.5)2 + C] - [0 + 0 + C]
	=	252 - 147 + C - C     (Note that the C's cancel out, as expected.)
	=	147

Solution to (b): The car travels 147 feet before coming to a complete stop.

Note: Equivalently, one could note that, since position is an antiderivative of velocity, the distance travelled is the definite integral: Int(84t - 12t², dt, t=0..3.5). As above, the result is 147 feet.

2. An object starts from the origin, and moves along the x-axis in such a way that its velocity is v(t) = sin(t) + 3^1/2cos(t), where t stands for time (measured in seconds).

a. At time t=Pi, what is the position of the object?
b. What is the total distance travelled by the object in the first Pi seconds? (That is, 0 < t < Pi.)

To find the position of the object at a specific time (as in (a)), we must find the function x(t) explicitly, including the constant term (if any). Since x(t) is an antiderivative of v(t), we must have x(t) = -cos(t) + 3^1/2sin(t) + C, where C is some constant to be determined. We are also told that the object starts from the origin; in other words, x(0) = 0. Plugging t=0 into the above equation for x(t), we get the equation 0 = -1 + 0 + C; therefore, C=1, and so x(t) = -cos(t) + 3^1/2sin(t) + 1. Now we can solve part (a):

x(Pi)	=	-cos(Pi) + 31/2sin(Pi) + 1
	=	-(-1) + 3(0) + 1
	=	2

Solution to (a): x(Pi) = 2

For part (b), we need to distinguish between distance travelled and change in position. The object's change in position is 2 units, since it starts at x=0 and ends up at x=2. However, this doesn't account for total distance travelled; in particular, movement to the left is counted as negative. To adjust for this, we need to break the movement up into two cases: movement to the left (i.e., v(t) < 0) and movement to the right (v(t) > 0).

Setting v(t)=0, we find that t=2Pi/3 is the only time in the interval [0, Pi] at which the object stops moving. Testing intervals, we find that v(t) > 0 (meaning the object moves to the right) when t < 2Pi/3, and v(t) < 0 (meaning the object moves to the left) when t > 2Pi/3. Thus, to find distance travelled, we integrate v(t) over two separate intervals:

Distance travelled = Int(sin(t) + 3^1/2cos(t), dt, t=0..2Pi/3) - Int(sin(t) + 3^1/2cos(t), dt, t=2Pi/3..Pi) Again: the integral from 2Pi/3 to Pi will turn out negative, since the object is moving to the left in this time interval. Since we want to count this movement as positive, we subtract, thus forcing the result to turn out positive (against its will, in a sense).

The two integrals above, when evaluated, give us 3 and -1, respectively. This means the object moves 3 units to the right, and then 1 unit to the left. The total distance travelled is 3 - (-1) = 4 units.

(Note: no units of any type were needed for this problem; answering simply "2" for part (a) and "4" for part (b) would be satisfactory.)

Extra Credit.    Each of the following problems is worth up to 5 points. I'll leave these problems open; the first person to give me a correct answer to any of these problems will get 5 extra credit points. (That's my way of thanking you for reading these solutions!)

1. Let F(x) = Int(1/t, dt, t=1..x). Show that for any a > 1 (a = some fixed constant), F(ax) = F(a) + F(x).

Hint: the trick to this problem is to first show that F(ax) - F(x) is a constant and then to show that this constant is specifically F(a).

2. Evaluate Int((4-x²)^1/2, dx, x=-2..2)

Hint: Sketch this region. Be precise, so that you get a good idea of the shape. What have you just drawn? Once you recognize it, you'll be able to give the area.

Do not try to find an antiderivative; most likely, you won't be able to. (It involves inverse trig functions, which we haven't discussed at all this semester!)

3. Evaluate:

Int(

cos(x) sin(x)+1

, dx,

x=0..Pi/2)

Hint: For this problem, it will be necessary to come up with an antiderivative. You'll need to be a bit clever; just play around with expressions involving sin(x) and/or cos(x), and see if you can find something whose derivative is the integrand in this problem.

Last modified 5/3/99 ludwick@math.temple.edu

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