Recitation 13: Antiderivatives & Integrals Galore
Joseph Wells
Arizona State University
May 3, 2014
Example (§5.1, #1).
a. Does the right Riemann sum underestimate or overestimate the area of the
region under the graph of a positive decreasing function? Explain.
b. Does the left Riemann sum underestimate or overestimate the area of the region
under the graph of a positive increasing function?
y
y
4
4
3
2
2
1
x
x
1
2
3
4
5
1
2
3
4
5
Example (§5.1, #3). Use the tabulated values of f to evaluate the left and right Riemann
x
1 1.5 2 2.5 3 3.5 4 4.5 5
sums for n = 8 in the interval [1, 5].
f (x) 0 2 3 2 2 1 0 2 3
Solution.
So L = (0.5)(0+2+3+2+2+1+0+2) = 6 and R = (0.5)(2+3+2+2+1+0+2+3) =
15
.
2
Example (§5.2, #4).
Z
a
x dx in terms of a.
a. Use geometry to find a formula for
0
Z
b
|f (x)| dx = 0, what can you conclude about f ?
b. If f is integrable and
a
Solution.
a. Notice that, for 0 ≤ x ≤ a, the area under f (x) is a triangle with base a and
height a. The area of the triangle is thus 12 a2 .
b. We can conclude that f (x) = 0.
Example (§5.3, #1). Evaluate the following integral using the Fundamental Theorem of
Z 7π/4
(sin x + cos x) dx
Calculus. Discuss whether your result is consistent with the figure.
−π/4
Solution.
We solve this in the usual fashion:
7π/4
Z 7π/4
(sin x + cos x) dx = − cos x + sin x
−π/4
−π/4
h
π i
π
7π
7π
= − cos
+ sin −
+ sin
− − cos −
4
4
4
4
" √
√ # " √
√ #
2
2
2
2
= −
−
− −
−
2
2
2
2
= 0.
d
Example (§5.3, #3). Simplify the expression:
dx
Z
10
x2
dz
.
+1
z2
Solution.
By the Fundamental Theorem of Calculus,
Z 10
d
d f (z) dz =
F (10) − F (x2 )
dx x2
dx
= −2xf (x2 ),
(chain rule)
so in particular,
d
dx
Z
10
x2
dz
2x
=
−
.
z2 + 1
x4 + 1
As justification that this is correct, we can brute force our way through it (although in
general, it’s rare that we could compute the antiderivative so easily, if even at all.)
d
dx
Z
10
x2
"
10 #
dz
d
arctan(z)
=
2
z2 + 1
dx
x
d =
arctan(10) − arctan(x2 )
dx
2x
=− 4
.
x +1
Example (§5.4, #3). Use symmetry to evaluate the following integrals. Draw a figure
to interpret your result.
Z 2π
a.
cos x dx
0
Z
b.
2π
sin x dx
0
Solution.
a. In
R πour interval, cos x is symmetric about x = π. So we can rewrite the integral
as 2 0 cos x dx. Looking at it again, we see that in our new interval, cos x is
R π/2
Rπ
antisymmetric about π2 meaning 0 cos x dx = − π/2 cos x dx. Thus
Z
2π
π
Z
cos x dx
cos x dx = 2
0
0
π/2
Z
=2
π
Z
cos x dx + 2
0
π/2
Z
cos x dx
π/2
Z π/2
cos x dx − 2
=2
cos x dx
0
0
= 0.
b. In our interval, sin x is antisymmetric about x = π, so
Thus
Z
2π
Z
π
sin x dx =
0
sin x dx +
0
= 0.
sin x dx = −
sin x dx
Zπ π
sin x dx −
=
0
2π
Z
Z0 π
Rπ
sin x dx
0
R 2π
π
cos x dx.
Assignment
MAT270 Recitation Notebook
§5.1, Problems 2
§5.2, Problems 1
§5.3, Problems 2,4
§5.4, Problems 2