- [1] Use the substitution u = 4 - 2x to convert the
integral, from x = 0 to x = 1 of 2x(4 - 2x)
^{1/2}with respect to x, into an integral in terms of u only. You do not need to evaluate the integral.

Step 1: u = 4 when x = 0, and u = 2 when x = 1, so the new integral is from u = 4 to u = 2.

Step 2: du = -2 dx

Step 3: The new integrand is (-1/2)(4 - u)(u)^{1/2}(but don't forget to put the du in the actual integral) - [2] Let R be the region bounded by the curves
y = x
^{1/2}, y = 2 - x, and y = 0. Sketch the region R, and express its area in terms of one or more integrals. You do not need to evaluate the integrals.

Step 1: The left edge of R is given by y = x^{1/2}, the right edge of R is given by y = 2 - x, they cross at the point (1,1), and the bottom edge of R is the x-axis. Step 2: This is easiest as an integral in y. Solve for x in terms of y, to get x = y^{2}from y = x^{1/2}, and x = 2 - y from y = 2 - x. Now subtract the left edge from the right edge (so the integrand is (2 - y) - y^{2}) and integrate from y = 0 to y = 1. - [3] Let
**R**be the region bounded by the curves y = 1, x = 1, and y = 100/x^{2}. Sketch the region**R**, and write down an integral for the volume of the solid formed by revolving**R**about the y-axis. You do not need to evaluate the integral.

The answer depends on whether you use washers or cylinders.

WASHERS: The integral is in terms of y, since for washers you use y as your variable if the axis of revolution is parallel to the y-axis (and you use x as your variable if the axis of revolution is parallel to the x-axis). From the sketch we see the bottom washer occurs at y = 1 and the top one at that y for which x = 1 (i.e., for y = 100 since this is what we get when we plug x = 1 into y = 100/x^{2}). The inner radius of every washer is the same in this case, r = 1. The outer radius R is the x-coordinate of points on the curve y = 100/x^{2}, so we need to solve for x. We get x = 10/y^{1/2}. So the answer is the integral of (Pi(10/y^{1/2})^{2}- Pi(1)^{2})dy from y = 1 to y = 100 with respect to y.

CYLINDERS: The integral is in terms of x, since for cylinders you use x as your variable if the axis of revolution is parallel to the y-axis (and you use y as your variable if the axis of revolution is parallel to the x-axis; i.e., it is opposite to what you do for washers). From the sketch we see the innermost cylinder occurs at x = 1 and the outermost one at x = 10 (i.e., the x value at which the line y = 1 and the curve y = 100/x^{2}cross). The height of a cylinder is 100/x^{2}, and the radius r is x, so the answer is the integral of 2Pi(x)(100/x^{2})dx from x = 1 to x = 10 with respect to x. - [4] A force of 10 pounds stretches a spring 2 inches. Set up an
integral for the work done in stretching the spring 3 inches
beyond its natural length. You do not need to evaluate the
integral, but indicate what units of work the integral will
evaluate to.

If we work in lbs and feet, then the spring constant is k = 10lbs/(1/6)ft = 60lbs/ft. Thus the work, in ft.lbs, is the integral from x = 0 to x = 1/4 of 60x dx. - [5] Baby bear's porridge is 150F when Mother Bear sets it out
at 9:00 am. The den is a comfy (for bears) 50F. At 9:10 am,
the porridge is 120F, still too hot for Baby Bear, who likes
porridge to be 99F. What time will it be when Baby Bear's
porridge is just right? (Assume the porridge cools according
to Newton's Law of cooling.)

The formula for temperature T(t) as a function of time t (in minutes) is such that the temperature difference T(t) - T_{a}with the ambient temperature T_{a}decays exponentially. Thus T(t) - T_{a}= (T(0) - T_{a}) e^{-kt}. Let t = 0 be 9am. Then T(0) = 150F, and T_{a}= 50, so T(t) = 50 + 100e^{-kt}. We'll need to find k, using the other data we're given: 120F = T(10) = 50 + 100e^{-k10}, which tells us that 0.70 = e^{-k10}, or -ln(0.70)/10 = k. Now we can find which t gives T(t) = 99F: 99 = T(t) = 50 + 100e^{-kt}implies 0.49 = e^{-kt}, hence t = -ln(0.49)/k = 10 ln(0.49)/ln(0.70) = 20.

- 1.
- (a) Integrate x(1 + 4x
^{2})^{ -1/2}with respect to x.

Use the substitution u = 1 + 4x^{2}. Then du = 8x dx, and we need to integrate (1/8) u^{ -1/2}du. We get (1/4) u^{1/2}+ C, or (1/4) (1 + 4x^{2})^{1/2}+ C. - (c) Integrate x(4 + x)
^{ -1/2}with respect to x, from x = 0 to x = 5.

Use the substitution u = 4 + x, so du = dx and we need to integrate (u - 4)(u)^{ -1/2}du = (u^{1/2}- 4u^{ -1/2}) du from u = 4 to u = 9, which gives ((2/3)u^{3/2}- 8u^{1/2})|_{4}^{9}= (18 - 24) - (16/3 - 16) = 14/3.

- (a) Integrate x(1 + 4x
- 2. Let R be the region enclosed by x = 1 + y
^{2}, y = x - 1.- (a) Find (but don't evaluate) an integral whose value gives
the exact area of the region R.

These curves cross at the points (1,0) and (2,1). We can integrate either in terms of x or y.

In terms of y we get: the integral from y = 0 to y = 1 of ((y+1) - (1 + y^{2})) dy.

In terms of x we get: the integral from x = 1 to x = 2 of ((x - 1)^{1/2}- (x - 1)) dx.

- (b) Find (but don't evaluate) an integral whose value gives
the volume of the solid obtained by revolving the region R
about the y-axis, using the method of shells.

We integrate with respect to y from y = 0 to y = 1. What we integrate is (Pi(y + 1)^{2}- Pi(1 + y^{2})^{2}) dy. - (c) Find (but don't evaluate) an integral whose value gives
the volume of the solid obtained by revolving the region
R about the vertical line x = 0, using the method of
cylindrical washers.

We integrate with respect to x from x = 1 to x = 2. What we integrate is 2Pi(x)((x - 1)^{1/2}- (x - 1)) dx.

- (a) Find (but don't evaluate) an integral whose value gives
the exact area of the region R.
- 3. A water tank is in the shape of a right circular cone
of altitude 10 feet and base radius
5 feet, with its base at the ground. If the tank is
full, find an integral but do not evaluate for the work
done in pumping the top 5 feet of the water out of the top.

Let h be the height off the ground of a horizontal slice of water in the tank. Let r be the radius of that slice (which is circular). Using similar triangles, we get (10 - h)/r = (total height)/(base radius) = 10/5 = 2, so r = (10 - h)/2:| ___|___ radius = r, at height h | total height = 10 | | _________|_________ base radius = 5, at height h = 0

Each slice of water of thickness dh weighs 62.4(Pi)r^{2}dh pounds, and must be lifted 10 - h feet, for 62.4(Pi)((10 - h)/2)^{2}(10 - h)dh ft.lbs. We must do this for h = 5 to h = 10 (i.e., for the upper 5 feet of water). So the answer is the integral from h = 5 to h = 10 of 62.4(Pi)((10 - h)/2)^{2}(10 - h)dh. - 5. An underwater observatory has a circular hatch
of 1 foot radius on its (vertical) wall
with the hatch's top 20 feet below the surface. Write down but
do not evaluate an integral whose value is the force that is
required to hold the hatch in place against the water.

Choose as coordinate system x where x = 0 is the bottom of the window, x = 1 is the top and x = 22 is at the surface of the water. The force on a horizontal slice of window at some position x is 62.4(Depth)(width)(thickness), where Depth = 22 - x feet underwater, width = horizontal width of window at position x = (1 - (1 - x)^{2})^{1/2}by the Pythagorean theorem, and thickness is dx. So now we take the integral from x = 0 to x = 2 of 62.4(22 - x)(1 - (1 - x)^{2})^{1/2}dx.

- 1.
- a. Integrate (sin x)/(3 + cos x) with respect to x.

Substitute u = 3 + cos x, so du = -sin x dx, which gives the integral of (-1/u) du, or -ln|u| + C = -ln|3 + cos x| + C. - b. Integrate (ln x)
^{7}/x with respect to x from x = 1 to x = e.

Substitute u = ln x, so du = dx/x, which gives the integral of u^{7}du, or u^{8}/8 + C = (ln x)^{8}/8 + C. - c. Integrate x
^{ -1/2}sec^{2}(x^{1/2}) with respect to x.

Substitute u = x^{1/2}, so du = (1/2)x^{ -1/2}dx, which gives the integral of 2 sec^{2}(u) du, or 2 tan(u) + C = 2 tan(x^{1/2}) + C.

- a. Integrate (sin x)/(3 + cos x) with respect to x.
- 2. Let R be the region enclosed by y = x
^{4}, y = x^{1/2}from x = 0 to x = 1.- a. Find (but don't evaluate) an integral whose value gives the
exact area of the region R.

The two curves cross at (0,0) and (1,1); we can integrate using either x or y. Using x we get the integral from x = 0 to x = 1 of (x^{1/2}- x^{4}) dx. Using y we get the integral from y = 0 to y = 1 of (y^{1/4}- y^{2}) dy. - b. Find (but don't evaluate) an integral whose value gives the
volume of the solid obtained by revolving the region R about
the x-axis, by using the method of slicing.

Washer method, revolved about x-axis, so we use the x variable. We take the integral from x = 0 to x = 1 of (Pi(x^{1/2})^{2}- Pi(x^{4})^{2}) dx. - c. Find (but don't evaluate) an integral whose value gives the
volume of the solid obtained by revolving the region R about
the y-axis, by using the method of cylindrical shells.

Cylinder method, revolved about y-axis so we use the other variable (i.e., not y, hence x again). We get the integral from x = 0 to x = 1 of 2Pi(x)(x^{1/2}- x^{4}) dx.

- a. Find (but don't evaluate) an integral whose value gives the
exact area of the region R.
- 3. A tank has a square base whose length is 5 feet and
rectangular sides of height 3 feet. Assume that the tank is
filled with water weighing 62.5 lb/ft
^{3}.- a. Find a Riemann sum whose value approximates the work
required to pump all of the water over the top of the tank.

Let h be height off the ground, so h = 0 is the bottom of the tank and h = 3 the top. This looks just like the integral but with a sum instead of an integral sign, and Delta h instead of dh. - b. Write down but do not evaluate an integral whose value is
exactly the work required to pump all of the water over the
top of the tank.

We have the integral from h = 0 to h = 3 of 62.5*5^{2}(3 - h) dh. - c. Write down but do not evaluate an integral whose value is
exactly the force exerted by the water on one side of the tank.

We have the integral from h = 0 to h = 3 of 62.5*Depth*width*thickness = 62.5*(3 - h)5 dh.

- a. Find a Riemann sum whose value approximates the work
required to pump all of the water over the top of the tank.
- 4. Find f
^{-1}(x) if y = f(x) = 4 + 3e^{2x}.

Solve x in terms of y: y = 4 + 3e^{2x}means (y - 4)/3 = e^{2x}, so ln((y - 4)/3) = 2x hence x = (1/2)ln((y - 4)/3). Thus f^{-1}(x) = (1/2)ln((x - 4)/3). - 5. This question deals with the function f(x) = 2x
^{3}+ 5x - 1.- a. Show that f
^{-1}exists.

First, f'(x) = 6x^{2}+ 5; since 6x^{2}is never negative, we see f'(x) is always at least 5 and hence positive. Therefore, f(x) is always increasing, so f(x) must be one to one (i.e., it passes the horizontal line test, since the graph can never reverse direction to hit the same horizontal line twice). This means that f^{-1}(x) exists. - b. Find the equation of the tangent line to the function
y = f
^{-1}(x) at the point (6, 1).

The tangent line is y = m(x - 6) + 1, where m is the derivative of f^{-1}(x) at x = 6. But f(1) = 6, so m = 1/f'(1) = 1/(6*1^{2}+ 5) = 1/11, hence the tangent line is y = (x - 6)/11 + 1.

- a. Show that f
- 6. Let y(t) be the amount of radioactive element present
at time t >= 0, and assume
that y(t) satisfies the equation: dy/dt = -0.3y.
Write down the exact form of y(t) and find the half-life
of the radioactive element.

We know that y(t) = y(0)e^{-kt}, where k = 0.3. The half-life is that time t = T such that y(T) = (1/2)y(0); i.e., such that (1/2)y(0) = y(0)e^{-kT}, or 1/2 = e^{-kT}, hence T = (ln(1/2))/(-k) = (ln 2)/k = (ln 2)/0.3.