# Wed, Feb 5 ```Clicker Question 1
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What is the volume of the solid formed when
area enclosed by the line y = x and the curve
y = x2 is revolved around the x- axis?
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A.  / 6
B. 1 / 6
C. 2 / 15
D.  / 15
E. 2 / 6
Clicker Question 2
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What is the volume of the solid formed when
the curve y = x on the interval [0, 1] is
revolved around the line y = -1?
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A. 11  / 6
B. 3 / 2
C. 5 
D. 17 / 6
E. 13  / 6
Techniques of Integration (2/5/14)
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Should be called “techniques of anti-differentiation”.
Finding derivatives involves “facts” and “rules”.
It is a mechanical process.
Finding anti-derivatives is not mechanical. The only
rules are Sum &amp; Difference and Constant Multiplier.
There are no Product, Quotient, or Chain Rules.
We need “techniques” rather than just rules.
The first two techniques are algebraic manipulation
and substitution (which tries to reverse the Chain
Rule).
Integration By Parts
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Whereas substitution techniques tries (if
possible) to reverse the chain rule,
“integration by parts” tries to reverse the
product rule.
Example: x ex dx ??
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Substitution? No!
Question: Can the integrand be split into a
product of one part with a nice derivative and
another part whose anti-derivative isn’t bad?
Reversing the product rule
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If u and v are functions of x, then by the
product rule: d/dx (u v) = u v  + u  v
Rewrite: u v  = d/dx (u v) - u  v
Antidifferentiate both sides, obtaining the
Integration by Parts Formula:
 u v  dx = u v -  u  v dx
The hope, of course, is that u  v is easier to
integrate than u v  was!
Back to the Example
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x ex dx ??
Note x gets simpler when you take its derivative and
ex’s anti-derivative is no worse, so we try letting
u = x and v  = ex
Then u  = 1 and v = ex , so rebuild, using the
Parts Formula:
x ex dx = x ex - ex dx = x ex – ex + C
A quick check, which of course involves the product
rule, shows this is right.
Clicker Question 3
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x cos(x) dx ?
– A. &frac12; x 2 sin(x) + C
– B. -&frac12; x 2 sin(x) + C
– C. x cos(x) – sin(x) + C
– D. x sin(x) – cos(x) + C
– E. x sin(x) + cos(x) + C
Antiderivative of the natural log
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We now can figure out the antiderivative of
the natural log function.
Try letting u = ln(x) and v ' = 1.
Assignment
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Hand-in #1 is due at class time on Friday.
For Friday, read Section 7.1 and do
Exercises 3, 5, 9, 11,15, 23, 27 and 64b
and 65.
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