Homework 5 Correction Model

1. (30 pts) Find the derivatives of the following functions.

a) (10 pts)
Z 3 Z x
t2 t2
F (x) = e dt = e dt (2 points)
x 3
FTC part 1:
Z
d ⇣ ⌘ d ⇣ x
t2
⌘
x2
F (x) = e dt = e (8 points)
dx dx 3

b) (10 pts)
Z 5x
2
t2
G(x) = ex e dt
4

Recognizing that we need to employ product rule and correctly doing it (3 points) and
then correctly using FTC to di↵erentiate the integral:
Z
d ⇣ 5x t2 ⌘ 2 d(5x) 2
e dt = e (5x) = 5e 25x (6 points)
d(5x) 4 dx
)
Z 5x
0 x2 t2 24x2
G (x) = 2xe e dt + 5e (1 point)
4

c) (10 pts)
Z x3
t2
H(x) = e dt
e x2

We start by splitting the interval and thus the also the integral (additive property):
Z c Z x3
t2 t2
H(x) = e dt + e dt (2 points)
e x2 c
Pick c=0 (arbitrary) )
Z 0 Z x3
t2 t2
H(x) = e dt + e dt
e x2 0

1
 Recognizing that FTC1 can be used (rewritting it into the form):
Z e x2 Z x3
t2 t2
H(x) = e dt + e dt (1 point)
0
| {z } | 0 {z }
I1 I2

Correctly di↵erentiating the first integral:

d ⇣Z e x2
⌘ x2 2 d(e x )
2
x2 2
t2 x2
I10 = x2 )
e dt = e (e )
= e (e )
( 2x)e
d(e 0 dx
2x2
(x2 +e )
= 2xe (4 points)

Correctly di↵erentiating the second integral:

Z
d ⇣ x3
t2
⌘
(x3 )2 d(x3 ) x6
I20 = e dt = e = 3x2 e (2 points)
d(x3 ) 0 dx
Writing down the complete correct answer:
2x2
(x2 +e x6
H 0 (x) = 2xe )
+ 3x2 e (1 point)

Award full points if FTC2 was correctly used instead.

2. (30 pts) We consider the integral

Z 1
I= xn e ax
dx
0

where a > 0 and n = 0, 1, 2, 3, ...

a) (5 pts) For n=0 we obtain:
Z 1 h 1 i1 h it
1 1 1 1
e ax dx = e ax
= lim e ax
= lim e ax
+ e0 =
0 a 0 t!1 a 0 t!1 a a a

b) (10 pts) We apply integration by parts:

Z 1 h 1 it Z 1
n
xn e ax dx = lim xn e ax
+ xn 1
e ax
dx
0 t!1 a 0 a 0

Looking at the limit, we see that we have a ” 1 1 ” situation and hence we can apply
l’Hopital. After the first, second, and so on di↵erentiation we still have the same problem.
Therefore, we apply l’Hopital n-times, and the limit becomes:
⇣ 1 tn ⌘ ⇣ 1 n! ⌘
lim = lim =0
t!1 a eat t!1 an+1 eat

2
 Hence we find that:
Z 1 Z 1
n
xn e ax
dx = xn 1
e ax
dx
0 a 0

Award:
• 3 points for correct integration by parts
• 3 points for spotting and correctly applying l’Hopital
• 3 points for providing a feasible argument as to why the limit goes to zero after nth
application of l’Hopital
• 1 point for the correct final answer
c) (5 pts) Here, we could either apply integration by parts, just like in (b), or use the result
of (b) as follows:
Z 1 Z Z
n 1 n 1 ax (b) n ⇣ n 1 1 n 2 ax ⌘
xn e ax dx = x e dx = x e dx
0 a 0 a a 0

d) (10 pts) First, we check if the statement is true for n = 1:

Z 1 Z 1 ⇣h 1 it ⌘ 1 ⇣
ax (b) 1 1 1 1⌘ 1
xe dx = e ax dx = lim e ax = lim e ax
+ = 2
0 a 0 a t!1 a 0 a t!1 a a a
which indeed holds for n = 1. Next we assume that the statement holds for n = k, i.e.
Z 1
k(k 1)(k 2) . . . 1
xk e ax dx =
0 ak+1
So we now need to show that it holds for n = k + 1, i.e.
Z 1
(k + 1)(k)(k 1)(k 2) . . . 1
xk+1 e ax dx =
0 ak+2

Z 1 Z
k+1 ax (b) k + 1 1 k ax assumption k + 1 k(k 1)(k 2) . . . 1
x e dx = x e dx =
0 a 0 a ak+1
(k + 1)(k)(k 1)(k 2) . . . 1
=
ak+2
Since the statement holds for n = 1 and after assuming it holds for n = k it held for
n = k + 1, th statement is true for all integers n 1.
Award:
• 2 points for base step
• 1 point for the assumption and 1 point for saying what is needed to show
• 4 points for proving the statement is true for n = k+1, with the use of the assumption
• 2 points for a conclusion

3
Problem 3 (20 pts)
We are given the integral

∫ sin−1 𝑥 𝑑𝑥 = ∫ arcsin 𝑥 𝑑𝑥
But as arcsines are trouble, we simplify by substitution:

𝑡 = arcsin 𝑥 → 𝑥 = sin 𝑡 3 pts

𝑑𝑥 2 pts
𝑑𝑡
= cos 𝑡 → 𝑑𝑥 = cos 𝑡 𝑑𝑡

Substituting these definitions yields

∫ arcsin 𝑥 𝑑𝑥 = ∫ 𝑡 cos 𝑡 𝑑𝑡

This integral must be solved by parts, i.e.

∫ 𝑓𝑔′ = 𝑓𝑔 − ∫ 𝑓′𝑔
We pick our parts as such:
4 pts
𝑓 = 𝑡 → 𝑓′ = 1
𝑔′ = cos 𝑡 → 𝑔 = sin 𝑡
To obtain
2 pts
∫ 𝑡 cos 𝑡 𝑑𝑡 = 𝑡 sin 𝑡 − ∫ sin 𝑡 𝑑𝑡
However, we know that
4 pts
∫ sin 𝑡 𝑑𝑡 = − cos 𝑡 + 𝐶
So this can be solved as forgetting C: -2 pts

forgetting minus: -2 pts

∫ 𝑡 cos 𝑡 𝑑𝑡 = 𝑡 sin 𝑡 + cos 𝑡 + 𝐶

We then resubstitute 𝑡 = arcsin 𝑥 to obtain

arcsin 𝑥 sin(arcsin 𝑥) + cos(arcsin 𝑥) + 𝐶

But as

sin(arcsin 𝑥) = 𝑥 1 pt

and
2 pts
cos(arcsin 𝑥) = √1 − 𝑥 2
This finally becomes
2 pts
∫ arcsin 𝑥 𝑑𝑥 = 𝑥 arcsin 𝑥 + √1 − 𝑥 2 + 𝐶
Problem 4 (20 pts)
We are given the integral
𝑥+4
∫ 𝑥 2 −5𝑥+6 𝑑𝑥

Factoring the denominator gives

𝑥+4 2 pts
∫ (𝑥−2)(𝑥−3) 𝑑𝑥

To solve, we wish to express this as a partial fraction, i.e.

𝑥+4 𝐴 𝐵
= + 3 pts
(𝑥−2)(𝑥−3) 𝑥−2 𝑥−3

Multiplying both sides by the denominator yields

𝐴(𝑥 − 3) + 𝐵(𝑥 − 2) = 𝑥 + 4
𝐴𝑥 − 3𝐴 + 𝐵𝑥 − 2𝐵 = 𝑥 + 4 5 pts

(𝐴 + 𝐵)𝑥 + (−3𝐴 − 2𝐵) = 𝑥 + 4

To satisfy this, the following system of two equations emerges:

𝐴 + 𝐵 = 1 ∧ −3𝐴 − 2𝐵 = 4
𝐴 = 1 − 𝐵 ∧ 3𝐴 + 2𝐵 = −4 2 pts

Some substituting and rewriting yields a value for 𝐵:

3(1 − 𝐵) + 2𝐵 = −4
3 − 𝐵 = −4 → 𝐵 = 7 2 pts
Subbing this back into the first equation gives a value for 𝐴:

→ 𝐴 = 1 − 7 = −6 1 pt

Hence, we have found for partial fractions

𝑥+4 6 7
(𝑥−2)(𝑥−3)
= − 𝑥−2 + 𝑥−3

And so, for the integral

𝑥+4 1 1 1 pt
∫ (𝑥−2)(𝑥−3) 𝑑𝑥 = 7 ∫ 𝑥−3 𝑑𝑥 − 6 ∫ 𝑥−2 𝑑𝑥

Which finally yields

𝑥+4
∫ (𝑥−2)(𝑥−3) 𝑑𝑥 = 7 ln|𝑥 − 3| − 6 ln|𝑥 − 2| + 𝐶 4 pts