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Calculus II: Midterm I Solutions - Integration and Improper Integrals, Exams of Calculus

Solutions to the calculus midterm i exam, focusing on integration and improper integrals. Topics covered include finding derivatives, evaluating integrals using substitution and l'hopital's rule, and calculating volumes of solids obtained by rotating regions about an axis.

Typology: Exams

2009/2010

Uploaded on 02/25/2010

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HONORS CALC II F 08, MIDTERM I
(1) Find the derivative of the function Zx
x
et
tdt.
(Suggestion: Write the integral as a difference between two integrals, one from 0 to xand
another from 0 to x.
Write f(t) = et
t. Set
I(x) = Zx
x
f(t)dt =I1(x)I2(x),where I1(x) = Zx
0
f(t)dt, I2(t) = Zx
0
f(t)dt.
Hence by FTC and the chain rule,
I0(x) = f(x)
f(x)
2x=ex
x
ex
2x=2exex
2x.
(2) Evaluate the integral
Zdx
x2x2
16.
Change x= 4 sec t. Then dx = 4 tan tsec tdt and x2
16 = 16 tan2t. So
I=Ztan tsec t
16 sec2ttan tdt =1
16 Zcos tdt =sin t
16 +C,
Now t= arccos(4/x), so sin t=1cos2t, and we have
I=p116/x2
16 +C=x2
16
16x+C.
(3) Evaluate the integral
Zdx
1 + x+x.
(You don’t need to change variables, you can do algebraic manipulation using the identity
(a+b)(ab) = a2
b2).
Following the suggestion, we note that
1
1 + x+x=1
1 + x+x
1 + xx
1 + xx=1 + xx.
Therefore
I=2
3(1 + x)3/2
x3/2+C.
(4) Evaluate the improper integral
Z100
0
tln tdt.
Integrate by parts
Z100
0
tln tdt =2
3t3/2ln t
100
0
2
3Z100
0
t1/2dt =2000
3ln 100
4000
9=4000
3ln 10
1
3.
1
pf2

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HONORS CALC II F 08, MIDTERM I

(1) Find the derivative of the function

∫ (^) x √x

et t dt. (Suggestion: Write the integral as a difference between two integrals, one from 0 to x and another from 0 to √x. Write f (t) = e tt. Set

I(x) =

∫ (^) x √x^ f^ (t)dt^ =^ I^1 (x)^ −^ I^2 (x),^ where^ I^1 (x) =

∫ (^) x

0

f (t)dt, I 2 (t) =

∫ √x

0

f (t)dt.

Hence by FTC and the chain rule,

I′(x) = f (x) − f^ (

x) 2

x

= e

x x

− e

√x

2 x

=^2 e

x (^) − e√x 2 x

(2) Evaluate the integral (^) ∫ dx x^2

x^2 − 16

Change x = 4 sec t. Then dx = 4 tan t sec tdt and x^2 − 16 = 16 tan^2 t. So

I =

tan t sec t 16 sec^2 t tan t

dt = 1 16

cos tdt = sin^ t 16

+ C,

Now t = arccos(4/x), so sin t =

1 − cos^2 t, and we have

I =

1 − 16 /x^2 16

+ C =

x^2 − 16 16 x

+ C.

(3) Evaluate the integral (^) ∫ √ dx 1 + x +

x

(You don’t need to change variables, you can do algebraic manipulation using the identity (a + b)(a − b) = a^2 − b^2 ). Following the suggestion, we note that

√^1 1 + x +

x

√^1

1 + x +

x

1 + x −

√ x 1 + x −

x

1 + x −

x.

Therefore I =^2 3

(1 + x)^3 /^2 − x^3 /^2

+ C.

(4) Evaluate the improper integral ∫ (^100)

0

t ln tdt.

Integrate by parts ∫ (^100)

0

t ln tdt =^2 3

t^3 /^2 ln t

100

0

0

t^1 /^2 dt =^2000 3

ln 100 − 4000 9

=^4000

ln 10 − 1 3

1

Note that we have used L’Hospital’s rule to compute

lim t→ 0 t^3 /^2 ln t = lim t→ 0 ln^ t t−^3 /^2

= lim t→ 0

1 t − 32 t−^5 /^2

Alternatively, you can substitute u = ln

t = 12 ln t and then t = e^2 u, so dt = 2e^2 udu ∫ (^100)

0

t ln tdt =

∫ (^) ln 10

−∞

eu 2 u 2 e^2 udu = 4

∫ (^) ln 10

−∞

e^3 uudu =

3 e

3 u(u − 1 3 )

ln 10

−∞

3 (ln 10^ −^

(5) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis: y = 2, y = x^2 ; about the y axis. We’ll use y as the independent variable, and so

V = π

0

x^2 (y)dy = π

0

ydy = 2π

2