Graphics Programs Reference
In-Depth Information
we get
I
=
0
.
75 [0
.
555 556(0
.
098 08)
+
0
.
888 889(0
.
628 16)
+
0
.
555 556(0
.
952 16)]
=
0
.
856 37
−
1
.
5
cos
x dx
Comparisonwith
=
0
.
856 38 showsthat the discrepancyis within the
roundoff error.
PROBLEM SET 6.2
1. Evaluate
π
ln
x
2
dx
x
2
−
2
x
+
1
with Gauss-Legendrequadrature. Use (a)two nodes and (b)four nodes.
Use Gauss-Laguerrequadraturetoevaluate
0
x
2
)
3
e
−
x
dx
.
2.
(1
−
3.
Use Gauss-Chebyshev quadrature with six nodes to evaluate
π/
2
dx
√
sin
x
0
t
2
.
4. The integral
0
sin
x dx
isevaluatedwith Gauss-Legendrequadrature using four
nodes.What are the boundson the truncation errorresulting fromthe quadrature?
5. How many nodes are requiredinGauss-Laguerrequadraturetoevaluate
0
e
−
x
sin
x dx
to six decimal places?
6. Evaluate as accuratelyas possible
1
Compare the result with the “exact”value 2
.
62206.
Hint:
substitute sin
x
=
2
x
+
1
√
x
(1
x
)
dx
−
0
2.
7. Compute
0
sin
x
ln
x dx
to four decimal places.
8. Calculate the boundson the truncation errorif
0
x
sin
x dx
isevaluatedwith
Gauss-Legendrequadrature using three nodes.What is the actualerror?
9. Evaluate
0
sinh
x
Hint:
substitute
x
=
(1
+
t
)
/
x
dx
to four decimal places.
10.Evaluate the integral
/
∞
x dx
e
x
+
1
0
to six decimal places.
Hint:
substitute
e
x
=
1
/
t
.
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