MATH 107-253 Recitation 3
JD Nir
Avery 230 • Office Hours: TBD
[email protected]
www.math.unl.edu/∼jnir2/107-253.html
August 31, 2015
p. 304 #14: Assume f 0 is given by the graph in Figure 6.13. Suppose f is continuous
and that f (3) = 0.
(a) Sketch a graph of f .
(b) Find f (0) and f (7).
R7
(c) Find 0 f 0 (x) dx in two different ways.
f 0 (x)
2
1
1
−1
−2
4
3
2
6 7
5
(a) Because f 0 (x) is constant on each interval, f (x) will be linear with varying slopes. We start at
f (3) because we know f (3) = 0 and then we can draw:
2
1
−1
1
2
4
3
5
6
7
−2
(b) We can check our graph which indicates f (0) = −1 and f (7) = −1.
1
MATH 107-253 Recitation 3
(c) One way is to manually calculate the area:
AreaRec 1 + AreaRec 2 + AreaRec 3 + AreaRec 4 + AreaRec 5 = (2 × 1) + (1 × −1) + (1 × 2) + (2 × −2) + (1 × 1)
= 2 + (−1) + 2 + (−4) + 1
=0
The
R 7 0 second way the book wants you to use is the Fundamental Theorem of Calculus to say
0 f (x) dx = f (7) − f (0). But that is actually wrong! Remember the statement of the FTC:
If f is continuous on the interval [a, b] and f (t) = F 0 (t), then
Z
a
b
f (t) dt = F (b) − F (a).
Becuase f 0 is not continuous, we can’t use the FTC!
p. 130 #35: Find an antiderivative F (x) with F 0 (x) = f (x) = 2 + 4x + 5x2 and F (0) = 0.
Is there only one possible solution?
First let’s find the general antiderivative:
Z
Z
Z
Z
2
2 + 4x + 5x dx = 2 dx + 4x dx + 5x2 dx
x2
x3
+5·
+C
2
3
5
= 2x + 2x2 + x3 + C
3
= 2x + 4 ·
Now to make sure F (0) = 0, we can plug in and sovle for C:
5
F (0) = 0 = 2(0) + 2(0)2 + (0)3 + C
3
So C = 0 and F (x) = 2x + 2x2 + 35 x3 . There can’t be any other solutions because the restraint
that F (0) = 0 forces C = 0 and all antiderivatives must be of the form 2x + 2x2 + 53 x3 + C.
Z √
1
p. 130 #51: Find the indefinite integral of
t t+ √
dt.
t t
First, because the integral is the sum of two other integrals, let’s split it up:
Z √
Z √
Z
1
1
√ dx
t t+ √
dt = t t dt +
t t
t t
Now let’s get the integral into a form we’re more comfortable with. Remember that
xa × xb = xa+b :
Z √
Z
Z
Z
3
1
1
√ dx = t 2 dt +
t t dt +
3 dx
t t
t2
2
√
1
x = x 2 and
MATH 107-253 Recitation 3
Also remember that
1
x
= x−1 so
Z
Z
Z
Z
3
3
3
1
t 2 dt +
t 2 dt + t− 2 dx
3 dx =
t2
Now both of these are in a form we recognize!
Z
Z
1
3
3
1 5
1
t 2 dt + t− 2 dx = 5 t 2 + C1 + 1 t− 2 + C2
−2
2
Each integral has a separate constant, but since we are adding them to solve our original integral
we can combine them into one constant. We can also simplfy our fractions and get
1
5
2
5
t 2 + C1 +
or if you prefer
1
1 −1
2 5
2 + C =
t 2 − 2t− 2 + C
2
1t
5
−2
1
2 5
2 5
2
t 2 − 2t− 2 + C = t 2 − √ + C.
5
5
t
2
1 + y2
dy exactly [as in ln(3π)], using the Fundamental They
1
orem, and numerically [ln(3π) ≈ 2.243]
Z
p. 130 #59: Evaluate
If we’re going to use the Fundamental Theorem, we’ll need an antiderivative.
At first, this integral doesn’t appear to be in the form we’re used to working with, but note that:
Z
Z
1 y2
1 + y2
dy =
+
dy when y 6= 0.
y
y
y
That note about y 6= 0 is important; this function is not defined at y = 0. Fortunately we will only
be focusing on the interval [1, 2] so we can use this equality.
Now we can split up our integral into two parts and write each part in a way that we know how to
find the antiderivative:
Z
Z
Z
1 y2
1
+
dy =
dy + y dy
y
y
y
Then we use our antiderivative rules:
Z
Z
1
1
dy + y dy = ln(y) + y 2 + C
y
2
We can choose any antiderivative for the Fundamental Theorem, so let’s choose C = 0. Then we
can say
Z 2
1 + y2
1
1
dy = F (2) − F (1) = (ln(2) + 22 ) − (ln(1) + (1)2 )
y
2
2
1
Noting that ln(1) = 0 we get
Z
1
2
1 + y2
3
dy = ln(2) + ≈ 2.193
y
2
3