Download Notes for Test 1.

Notes for Test 1.
Math 142, Calculus II, Fall 2016
1
Definite Integrals
Example 1)
The integral of a piecewise function
Find the following integral:
Z
4
f (x) dx,
−3

2

 x if − 3 ≤ x ≤ −1,
where f (x) =
2x if − 1 < x ≤ 0,

 √x
if 0 ≤ x ≤ 4
Solution:
We use the property of splitting the integral over the interval of integration
Z 4
Z −1
Z 0
Z 4
f (x) dx =
f (x) dx +
f (x) dx +
f (x) dx
−3
−3
−1
0
Z 4
Z 0
Z −1
√
2
2x dx +
x dx
x dx +
=
−1
−3
3 −1
0
0
4
x 2
= + x2 + x3/2 −1
0
3 −3
3
−1 −27
2 √ 3
−
( 4) − 0
=
+ (0 − 1) +
3
3
3
2
39
26
− 1 + (8) =
= 13.
=
3
3
3
J
Example 2)
The integral of an absolute value function (Still piecewise function)
Find the following integral:
Z 3π/2
|sin x| dx.
0
Solution:
The sin x function is non-negative for x in the first and second quadrants, i.e., x ∈ [0, π], and nonpositive in the third and fourth quadrants, i.e, x ∈ [π, 2π]. Thus, on the interval of integration
[0, 3π/2], we have
(
sin x
if 0 ≤ x ≤ π,
|sin x| =
− sin x if π ≤ x ≤ 3π/2.
Notes for Test 1.
2
Math 142, Calculus II, Fall 2016
Therefore,
Z
3π/2
Z
π
Z
3π/2
|sin x| dx =
sin x dx +
(− sin x) dx
0
π
π
= − cos x + cos x|π3π/2
0
0
3π
− cos π)
2
= −(−1 − 1) + (0 − (−1)) = 3.
= −(cos π − cos 0) + (cos
J
Fundamental Theorem of Calculus II
Recall:
FTC(II): Suppose f (t) is a continuous function on [a, b]. Then, for any x ∈ [a, b]
Z x
d
f (t) dt = f (x).
(Case 1)
dx a
If instead of x you have a function g(x), then by the chain rule of derivative
Z g(x)
d
(Case 2)
f (t) dt = f (g(x)) · g 0 (x).
dx a
If instead of a you have a function h(x), then by splitting the integral into two integrals and then
using the chain rule of derivative
!
Z g(x)
Z g(x)
Z a
d
d
f (t) dt
(Case 3)
f (t) dt =
f (t) dt +
dx h(x)
dx
a
h(x)
!
Z g(x)
Z h(x)
d
f (t) dt
=
f (t) dt +
−
dx
a
a
Z h(x)
Z g(x)
d
d
=−
f (t) dt +
f (t) dt
dx a
dx a
= −f (h(x)) · h0 (x) + f (g(x)) · g 0 (x)
= f (g(x)) · g 0 (x) − f (h(x)) · h0 (x).
Example 1)
d
dy
Z
y
tan2 (x2 − 1) dx = tan2 (y 2 − 1).
0
Example 2)
d
dz
Z
csc(z/3)
e
0
x3 −x
csc3 (z/3)−csc(z/3)
dx = e
1
· − csc(z/3) cot(z/3) ·
3
1
3
= − csc(z/3) cot(z/3)ecsc (z/3)−csc(z/3) .
3
Notes for Test 1.
Math 142, Calculus II, Fall 2016
Note: The derivative of csc(z/3) by using chain rule is
d z 1
d
csc(z/3) = − csc(z/3) cot(z/3) ·
= − csc(z/3) cot(z/3).
dz
dz 3
3
Example 3)
d
dx
Z
x3
sec(t + 5) dt = sec(x3 + 5) · (x3 )0 − sec(sin x + 5) · (sin x)0
sin x
= 3x2 sec(x3 + 5) − cos x sec(sin x + 5).
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