MATH 104 Midterm Test Solutions
|Problem||Possible score||Average score|
1. (12 points) Find .
Let u = ln x. Then the integral becomes
Substitute t = cos u to get
2. (12 points) Find .
Do integration by parts with u = ln(x2 +x+1) and dv = dx/x2. Then du = (2x+1) dx/(x2 +x+1)
and v = −1/x. So
Now use partial fractions to compute the integral on the right.
This leads to 0 = A + B, 2 = A + C, 1 = A and we conclude that B = −1 and C = 1. So now we
In this integral we make the substitution u = x + 1/2 and then −u + 3/2 = −x + 1. So we get
Returning to the original variable x and combining all the pieces gives
3. (12 points) Find .
Start with a substitution w = ex. Then dw = ex dx. We get
Now do integration by parts, with u = arctanw and dv = w2 dw. So du = dw/(1+w2) and v = w3/3.
Thus our integral becomes
Long division gives w3/(1 + w2) = w − w/(1 + w2) so we have
4. (12 points) Find
Complete the square : x2 + 4x = (x + 2)2 − 4 and make the substitution u = x + 2 to get
Now make the substitution u = 2 sec θ . The integral becomes
since du = 2 sec θ tan θ dθ , u2 − 4 = 4 tan2θ. In addition u =
2 corresponds to sec θ = 1 or θ = 0 and
u = 4 corresponds to sec θ = 2 or cos θ = 1/2 or θ = π /3.
Continuing the integration
5. (12 points) Find .
Replace sin2θ by (1 − cos 2 θ)/2 to get
Putting all the pieces together our answer is
6. (12 points) Set up an integral for the area of the
region enclosed between the curve y = x3 − 2
and its tangent line at x = −1. JUST SET UP THE INTEGRAL. DO NOT COMPUTE A
NUMERICAL VALUE .
First we have to find the tangent line at the point (−1,−3) on the cubic curve . The slope is
dy/dx = 3x2 evaluated when x = −1. So the slope is 3. The line of slope 3 through the point
(−1,−3) turns out to be the line y = 3x.
If we sketch the graphs, it is clear that the tangent line lies above the cubic. So our top curve is
y = 3x and the bottom curve will be y = x3 − 2. We need to find the points where the tangent and
the cubic intersect. Of course one intersection is at x = −1. To find the other one we set the two
curves equal, and using the fact that x = −1 is a root we know that x + 1 should be a factor of the
cubic equation we get. Long division then lets us find the other root.
x3 − 2 = 3x => x3 − 3x − 2 = 0 => (x + 1)(x2 − x − 2) = 0 => (x + 1)(x + 1)(x − 2) = 0
So the other intersection is at x = 2. So the integral we want is
7. (12 points) The region R is bounded by the curves y =
ln x, y = 0 and x = e. The solid S is
obtained by revolving R around the y-axis.
(a) Set up an integral for the volume of S using the shell method .
We slice the region vertically as x runs from 1 where the graph of y = ln x crosses the x-axis
up to x = e, a given boundary curve for the region. The slice at x has length ln x. We rotate
about the y axis, so the radius is x. Thus
(b) Set up an integral for the volume of S using the disk
or washer method.
Now we have to slice horizontally. The slice at height y runs from the graph of y = ln x to the
vertical line x = e. This horizontal slice will generate a washer. The outer radius will be e and
the inner radius (the radius of the hole in the center) will be x = ey. We slice as y runs from 0
to 1. So our integral is
(c) Compute the volume of S. If we use the shell integral we compute using integration by parts.
Evaluating between 1 and e we eventually get
using the fact that ln e = 1 and ln 1 = 0.
8. (6 points) Sketch the curve given in polar coordinates by r = 1 + sin θ.
As runs from 0 to π /2 we move through the first quadrant in a counterclockwise direction and r
increases steadily from 1 to 2. As we move through the second quadrant, r decreases in a completely
symmetric fashion from 2 back to 1. As we swing through the third quadrant, the r values continue to
steadily decrease, reaching the value 0 as θ reaches 3π /2. So the curve curls into the origin, tangent
to the the θ = 3π /2 ray (the negative y -axis). In a symmetric way, as we swing through the fourth
quadrant, the r values increase from 0 to 1, so the curve comes out of the origin, again tangent to
the negative y-axis.
The resulting curve is a cardioid that crosses the positive x -axis at 1, the positive y-axis at 2, the
negative x-axis at −1 and has a cusp at the origin, centered on the negative y-axis.
9. (10 points) Find the length of the curve as x runs from 0 to 1.
Parametrize the curve as
So we need to compute 1 + (dy/dt)2:
Thus the speed is
So the arclength is