SOLUTION OF HW4 October 09, 2011 Z |(x − 1)(3x − 1)| dx.

advertisement
SOLUTION OF HW4
MINGFENG ZHAO
October 09, 2011
Z
2
|(x − 1)(3x − 1)| dx.
1. [Page 83, Problem 16] Compute
0
Proof. First, we look at (x − 1)(3x − 1) ≥ 0, then we can get
x ≥ 1,
or x ≤
1
.
3
Hence, we can get
2
Z
|(x − 1)(3x − 1)| dx
0
1
3
Z
1
3
1
3
Z
[3x2 − 4x + 1] dx −
Z
2
1
1
[3x2 − 4x + 1] dx +
1
3
0
|(x − 1)(3x − 1)| dx
(x − 1)(3x − 1) dx
1
3
1
3
Z
Z
(x − 1)(3x − 1) dx +
0
2
1
1
Z
(x − 1)(3x − 1) dx −
=
=
Z
|(x − 1)(3x − 1)| dx +
0
=
1
Z
|(x − 1)(3x − 1)| dx +
=
Z
2
[3x2 − 4x + 1] dx
1
1
1
2
[x3 − 2x2 + x]03 − [x3 − 2x2 + x] 1 + [x3 − 2x2 + x]1
3
=
1
2 1
1
2 1
− +
− 0 − [1 − 2 + 1] +
− +
+ [8 − 8 + 2] − [1 − 2 + 1]
27 9 3
27 9 3
=
2
4 2
− + +2
27 9 3
=
2 − 12 + 18 + 54
27
=
62
27
=
2
8
.
27
1
2
MINGFENG ZHAO
Z
−4
2. [Page 83, Problem 20] Compute
(x + 4)10 dx.
−2
Proof.
Z
=
−4
1
· [x + 4]10+1 1 + 10
−2
=
1 · [−4 + 4]11 − [−2 + 4]11
11
−4
(x + 4)10 dx
−2
= −
211
11
= −
2048
11
= −186
2
.
11
3. [Page 83, Problem 21] Find all values of c for which
Z
c
x(1 − x) dx = 0.
a.
Z0 c
|x(1 − x)| dx = 0.
b.
0
Proof. a. Since we have
Z
0
c
x(1 − x) dx
=
0
Z
=
c
[x − x2 ] dx
0
c
x3 x2
−
2
3 0
2
c
c3
=
−
−0
2
3
=
=
c2
c3
−
2
3
=
3c2 − 2c3
6
=
c2 [3 − 2c]
.
6
SOLUTION OF HW4
3
So we get c2 [3 − 2c] = 0, therefore, we have
c = 0,
3
.
2
b. Firstly, we look at x(1 − x) ≥ 0, then we get
0 ≤ x ≤ 1.
So if c ≤ 0, then
c
Z
0
|x(1 − x)| dx
=
0
c
Z
=
−
x(1 − x) dx
0
c
Z
=
[x − x2 ] dx
−
0
c
x3 x2
−
2
3 0
2
c
c3
−
−
−0
2
3
=
=
−
=
−
c2
c3
+
2
3
=
−
3c2 − 2c3
6
=
−
c2 [3 − 2c]
.
6
Then we get c2 [3 − 2c] = 0, then c = 0 or c = 32 , but we know that c ≤ 0, then in this case, we have
c = 0.
If 0 < c ≤ 1, then we have
Z
0
c
|x(1 − x)| dx
=
0
Z
c
x(1 − x) dx
=
0
Z
=
0
c
[x − x2 ] dx
4
MINGFENG ZHAO
c
x2
x3 −
2
3 0
2
c
c3
−
−0
2
3
=
=
=
c3
c2
+
2
3
=
3c2 − 2c3
6
=
c2 [3 − 2c]
.
6
Then we get c2 [3 − 2c] = 0, then c = 0 or c = 23 , but we know that 0 < c ≤ 1, then in this case, we
do not have solution.
If c > 1, then we know that
Z
c
Z
|x(1 − x)| dx
1
≥
0
|x(1 − x)| dx
0
Z
1
x(1 − x) dx
=
0
Z
=
1
[x − x2 ] dx
0
1
x2
x3 −
=
2
3 0
1 1
=
−
−0
2 3
=
1
6
> 0.
So in this case, we do not have solution.
Z c
In a summary, if
|x(1 − x)| dx = 0, then
0
c = 0.
SOLUTION OF HW4
5
Z
4. [Page 83, Problem 23] Find a quadratic polynomial P for which P (0) = P (1) = 0, and
1
P (x) dx =
0
1.
Proof. Since P is a quadratic polynomial, then we can assume that there exists some a, b, c ∈ R such
that
Q(x) = ax2 + bx + c.
Since 0 = P (0) = c, then c = 0. And by 0 = P (1) = a + b + c = a + b, then b = −a. On the other
hand,
1
Z
1
=
P (x) dx
0
1
Z
[ax2 + bx + c] dx
=
0
1
Z
[ax2 − ax] dx
=
0
=
=
=
1
ax2 ax3
−
3
2 0
ha
3
−
ai
−0
2
2a − 3a
6
a
= − .
6
That is, we get 1 = − a6 , so a = −6, which implies that b = −a = 6, and
P (x) = −6x2 + 6x.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: [email protected]
Download