18.01 Problem Set 7 Solutions
Part II: 15 points
1a) Using an idea from section PI of the supplementary notes, evaluate the definite integral
Z
0
1
dx
.
+1
x2
The idea appearing on page 2 of PI is to use the substitution x = tan(θ). As x = tan(θ) varies from 0
to 1, the angle θ is varying from 0 to π/4. We have
dx = sec2 (θ)dθ = (1 + tan2 (θ))dθ.
Plugging all this back into the original integral gives
Z
1
0
dx
=
x2 + 1
Z
π/4
0
1 + tan2 (θ)
dθ =
1 + tan2 (θ)
Z
π/4
dθ = π/4.
0
b) Calculate the upper and lower sums for the integral in (a) with four equal subintervals.
Since the interval has length 1, the endpoints for the subintervals are
x0 = 0,
x1 = 1/4,
x2 = 1/2,
x3 = 3/4,
x4 = 1.
The function x2 + 1 is positive and increasing for x ≥ 0, so the integrand 1/(x2 + 1) is positive and
decreasing. So the maximum value on each subinterval is at the left endpoint, and the minimum value at
the right endpoint. The upper Riemann sum (using maximum values) is therefore
1
1
1
1
1
+
+
+
4 1 + 02
1 + (1/4)2
1 + (1/2)2
1 + (3/4)2
1
1
1
1 1
+
+
+
=
4 1 17/16 5/4 25/16
1 25 · 17 16 · 25 4 · 5 · 17 16 · 17
,
+
+
+
=
4 25 · 17 25 · 17
25 · 17
25 · 17
which is 1437/1700. Similarly, the lower Riemann sum (using minimum values) is
1
1
1
1
1
+
+
+
4 1 + (1/4)2
1 + (1/2)2 1 + (3/4)2
1 + 12
1
1
1
1
1
=
+
+
+
4 17/16 5/4 25/16 2
1 16 · 50 4 · 10 · 17 32 · 17 25 · 17
,
+
+
+
=
4 50 · 17
50 · 17
50 · 17 50 · 17
which is 2449/3400.
c) Explain why π is between 2449/850 and 2874/850. (The average of these two numbers is
5323/1700 = 3.131 . . ., which is not terrible as an estimate of π.)
Any definite integral must be between the upper and lower Riemann sums for any subdivision of the
interval. In our case, for the subdivision into four equal parts and the Riemann sums we just calculated,
this says that
2449
π
≤ ≤ 2874/3400.
3400
4
Multiplying these inequalities by 4 gives
2449
≤ π ≤ 2874/850,
850
as you were asked to show.
√
2a) Calculate the area under the graph of y = 3x + 5 from x = 0 to x = 2. (This is exercise
3C-2a.)
R2√
This is 0 3x + 5dx. Since what makes the integral tricky is the 3x + 5, it’s reasonable to try the
substitution u = 3x + 5. As x increases from 0 to 2, u increases from 5 to 11; and du = 3dx, or dx = du
3 .
The integral therefore becomes
Z
5
11
√ du
1 u3/2
u
=
3
3/2 3
11
5
=
√
√
2
(11 11 − 5 5).
9
√
√
b) Calculate the area under the graph of x = (y 2 − 5)/3 from y = 5 to y = 11. (I’ve called the
independent variable y instead of x, just to confuse you; the area is an integral of something
times dy.)
This is
Z
√
√
5
11
√11
√
√
√
√
√
√
y2 − 5
dy = y 3 /9 − 5y/3 √ = [11 11/9 − 5 11/3 − 5 5/9 + 5 5/3] = [−4 11/9 + 10 5/9].
3
5
√
c) If both of us did the arithmetic right, your answers to (a) and (b) should add up to 2 11.
Explain how you could have predicted that before doing the calculation.
√
The equation y = 3x + 5 (for non-negative values of y) is exactly the same thing as x = (y 2 − 5)/3.
The graph of this equation (in either form) divides the rectangle
√
0 ≤ x ≤ 2,
0 ≤ y ≤ 11
into two pieces. The lower right piece is the area we computed in (a), and the upper left
√ piece is the area we
computed in (b). Adding these two pieces gives the whole rectangle, whose area is 2 11. Here is a rough
picture; notice that the scales on the x and y axes are different, which is confusing.
3a) Calculate the area under the graph of x = ey from y = 0 to y = ln(a).
This is
Z
ln(a)
ey dy = ey
0
ln(a)
0
= eln(a) − e0 = a − 1.
b) Using the idea from problem 2c and the answer to (a), calculate the area under the graph
of y = ln(x) from x = 1 to x = a. (You’re not supposed to know yet how to find an antiderivative
of ln(x).)
This area and the one in (a) together fill up the rectangle
0 ≤ x ≤ a,
0 ≤ y ≤ ln(a),
which has area a ln(a). Since we already know the area in (a), the conclusion is
Z
1
a
ln(x)dx = a ln(a) − (a − 1) = a ln(a) − a + 1.
c) Using the answer to (b), write down an antiderivative of ln(x).
The second FTC says that the function of a written in (b) must have derivative ln(a) (with respect to
a). Changing the name of the variable to x gives
Z
ln(x)dx = x ln(x) − x + 1.