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V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Section 5.3
Evaluating Definite Integrals Notes
V63.0121.006/016, Calculus I
New York University
April 20, 2010
Announcements
◮ April 16: Quiz 4 on §§4.1–4.4
◮ April 29: Movie Day!!
◮ April 30: Quiz 5 on §§5.1–5.4
◮ Monday, May 10, 12:00noon (not 10:00am as previously announced)
Final Exam
Image credit: docman
Announcements
Notes
◮ April 16: Quiz 4 on
§§4.1–4.4
◮ April 29: Movie Day!!
◮ April 30: Quiz 5 on
§§5.1–5.4
◮ Monday, May 10, 12:00noon
(not 10:00am as previously
announced) Final Exam
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 2 / 48
Homework: The Good
Notes
Most got problems 1 and 3 right.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 3 / 48
1
V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Homework: The Bad (steel pipe)
Notes
Problem
Asteel pipe is being carried down a hallway 9ft wide. At the end of the
hall there is a right-aangled turn into a narrower hallway 6ft wide. What is
the length of the longest pipe that can be carried horizontally around the
corner?
θ
θ 6 6
6sec
9
θ cscθ
9
9
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
Solution
Notes
Solution
The longest pipe that barely fits is the smallest pipe that almost doesn’t
fit. We want to find the minimum value of
f (θ) = asecθ +bcscθ
on the interval 0 < θ < π/2. (a = 9 and b = 6 in our problem.)
f ′(θ) = asecθtanθ −bcscθcotθ
sinθ cosθ asin3θ −bcos3θ
=acos2θ −b 2 = 2 2
sin θ sin θcos θ
So the critical point is when
asin3θ = bcos3θ =⇒ tan3θ = b
a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 5 / 48
Finding the minimum
If f ′(θ) = asecθtanθ − bcscθcotθ, then Notes
′′ 2 3 2 3
f (θ) = asecθtan θ +asec θ+bcscθcot θ+bcsc θ
which is positive on 0 < θ < π/2.
So the minimum value is
f (θ ) = asecθ +bcscθ
min min min
b b 1/3
where tan3θ = =⇒ tanθ = .
min a min a
Using
2 2 2 2
1+tan θ =sec θ 1+cot θ =csc θ
Weget the minimum value is
s r
b 2/3 a2/3
min = a 1+ a +b 1+ b
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 6 / 48
2
V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Simplifying
Notes
s r
2/3
b a 2/3
min = a 1+ a +b 1+ b
=bsb2/3 + a2/3 +asa2/3 + b2/3
b2/3 b2/3 a2/3 a2/3
b p 2/3 2/3 a p 2/3 2/3
=b1/3 b +a +a1/3 a +b
=b2/3pb2/3+a2/3+a2/3pa2/3+b2/3
2/3 2/3 p 2/3 2/3
=(b +a ) b +a
2/3 2/3 3/2
=(a +b )
If a = 9 and b = 6, then min ≈ 21.070.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 7 / 48
Homework: The Bad (Diving Board)
Notes
Problem
If a diver of mass m stands at the end of a diving board with length L and
linear density ρ, then the board takes on the shape of a curve y = f (x),
where
′′ 1 2
EIy =mg(L−x)+2ρg(L−x)
E and I are positive constants that depend of the material of the board
and g < 0 is the acceleration due to gravity.
(a) Find an expression for the shape of the curve.
(b) Use f(L) to estimate the distance below the horizontal at the end of
the board.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 8 / 48
Notes
Solution
Wehave
′′ 1 2
EIy (x) = mg(L−x)+ 2ρg(L−x)
Antidifferentiating once gives
′ 1 2 1 3
EIy (x) = −2mg(L−x) − 6ρg(L−x) +C
Once more:
1 3 1 4
EIy(x) = 6mg(L−x) + 24ρg(L−x) +Cx +D
where C and D are constants.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 9 / 48
3
V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Don’t stop there!
Notes
Plugging y(0) = 0 into
′ 1 3 1 4
EIy (x) = 6mg(L−x) + 24ρg(L−x) +Cx +D
gives
1 3 1 4 1 3 1 4
0 = 6mgL + 24ρgL +D =⇒ D =−6mgL − 24ρgL
Plugging y′(0) = 0 into
′ 1 2 1 3
EIy (x) = −2mg(L−x) − 6ρg(L−x) +C
gives
1 2 1 3 1 2 1 3
0 = −2mgL − 6ρgL +C =⇒ C = 2mgL + 6ρgL
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 10 / 48
Solution completed
So Notes
1 3 1 4
EIy(x) = 6mg(L−x) + 24ρg(L−x)
1 2 1 3 1 3 1 4
+ 2mgL +6L x−6mgL −24ρgL
which means
1 2 1 3 1 3 1 4
EIy(L) = 2mgL + 6L L−6mgL −24ρgL
1 3 1 4 1 3 1 4
=2mgL +6L −6mgL −24ρgL
1 3 1 4
=3mgL +8ρgL
3
=⇒ y(L)= gL m+ρL
EI 3 8
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 11 / 48
Homework: The Ugly
Notes
◮ Some students have gotten their hands on a solution manual and are
copying answers word for word.
◮ This is very easy to catch: the graders are following the same solution
manual.
◮ This is not very productive: the best you will do is ace 10% of your
course grade.
◮ This is a violation of academic integrity. I do not take it lightly.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 12 / 48
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