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Solutions to Quiz 1 in Math 106a: Integration using Substitution, Exercises of Calculus

Aliah UniversityCalculus

The solutions to quiz 1 in math 106a, focusing on the integration technique using substitution. It includes step-by-step calculations for two examples.

Typology: Exercises

2012/2013

Uploaded on 03/16/2013

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Math 106a Solutions
Quiz 1
9/17/10
Evaluate both integrals using substitution.
1. Zex
1 + e2xdx
Rewrite the integrand as follows: Zex
1+(ex)2dx
Let u=ex, then du =exdx.
Therefore, Zex
1 + e2xdx =Zexdx
1+(ex)2=Zdu
1 + u2= arctan u+C= arctan(ex) + C.
2. Zโˆš3
0
2xp1 + x2dx
Let u= 1 + x2, then du = 2x dx.
Note, if x= 0 then u= 1 + 02= 1. Likewise, if x=โˆš3, then u= 1 + (โˆš3)2= 4. This gives us
our new limits of integration.
Therefore, Zโˆš3
0
2xp1 + x2dx =Z4
1
โˆšu du =Z4
1
u1/2du =๎˜”2
3u3/2๎˜•4
1
=2
3๎˜(โˆš4)3
โˆ’(โˆš1)3๎˜‘=14
3
1

Partial preview of the text

Download Solutions to Quiz 1 in Math 106a: Integration using Substitution and more Exercises Calculus in PDF only on Docsity!

Math 106a Solutions

Quiz 1

9/17/

Evaluate both integrals using substitution.

ex

1 + e^2 x^

dx

Rewrite the integrand as follows:

ex

1 + (ex)^2

dx

Let u = ex, then du = exdx.

Therefore,

ex 1 + e^2 x^

dx =

exdx 1 + (ex)^2

du 1 + u^2

= arctan u + C = arctan(ex) + C.

0

2 x

1 + x^2 dx

Let u = 1 + x^2 , then du = 2x dx.

Note, if x = 0 then u = 1 + 0^2 = 1. Likewise, if x =

3, then u = 1 + (

3)^2 = 4. This gives us our new limits of integration.

Therefore,

0

2 x

1 + x^2 dx =

1

u du =

1

u^1 /^2 du =

[

u^3 /^2

] 4

1

4)^3 โˆ’ (

1)^3