74 Higher Engineering Mathematics
=
[
θ−
θ^3
18
+
θ^5
600
−
θ^7
7 ( 5040 )
+···
] 1
0
= 1 −
1
18
+
1
600
−
1
7 ( 5040 )
+···
=0.946,correct to 3 significant figures.
Problem 15. Evaluate
∫ 0. 4
0 xln(^1 +x)dxusing
Maclaurin’s theorem, correct to 3 decimal places.
From Problem 6,
ln( 1 +x)=x−
x^2
2
+
x^3
3
−
x^4
4
+
x^5
5
−···
Hence
∫ 0. 4
0
xln( 1 +x)dx
=
∫ 0. 4
0
x
(
x−
x^2
2
+
x^3
3
−
x^4
4
+
x^5
5
−···
)
dx
=
∫ 0. 4
0
(
x^2 −
x^3
2
+
x^4
3
−
x^5
4
+
x^6
5
−···
)
dx
=
[
x^3
3
−
x^4
8
+
x^5
15
−
x^6
24
+
x^7
35
−···
] 0. 4
0
=
(
( 0. 4 )^3
3
−
( 0. 4 )^4
8
+
( 0. 4 )^5
15
−
( 0. 4 )^6
24
+
( 0. 4 )^7
35
−···
)
−( 0 )
= 0. 02133 − 0. 0032 + 0. 0006827 −···
=0.019,correct to 3 decimal places.
Now try the following exercise
Exercise 33 Further problems on
numerical integration using Maclaurin’s
series
- Evaluate
∫ 0. 6
0. 2 3e
sinθdθ, correct to 3 decimal
places, using Maclaurin’s series. [1.784]
- Use Maclaurin’s theorem to expand cos2θand
hence evaluate, correct to 2 decimal places,
∫ 1
0
cos2θ
θ
1
3
dθ. [0.88]
- Determine the value of
∫ 1
0
√
θcosθdθ,cor-
rect to 2 significant figures, using Maclaurin’s
series. [0.53]
- Use Maclaurin’s theorem to expand√
xln(x+ 1 ) as a power series. Hence
evaluate, correct to 3 decimal places,∫ - 5
0
√
xln(x+ 1 )dx. [0.061]
8.6 Limiting values
It is sometimes necessary to find limits of the form
lim
x→a
{
f(x)
g(x)
}
,wheref(a)=0andg(a)=0.
For example,
lim
x→ 1
{
x^2 + 3 x− 4
x^2 − 7 x+ 6
}
=
1 + 3 − 4
1 − 7 + 6
=
0
0
and^00 is generally referred to as indeterminate.
For certain limits a knowledge of series can sometimes
help.
For example,
lim
x→ 0
{
tanx−x
x^3
}
≡lim
x→ 0
⎧
⎪⎨
⎪⎩
x+
1
3
x^3 +···−x
x^3
⎫
⎪⎬
⎪⎭
from Problem 5
=lim
x→ 0
⎧
⎪⎨
⎪⎩
1
3
x^3 +···
x^3
⎫
⎪⎬
⎪⎭
=lim
x→ 0
{
1
3
}
=
1
3
Similarly,
lim
x→ 0
{
sinhx
x
}
≡lim
x→ 0
⎧
⎪⎪
⎨
⎪⎪
⎩
x+
x^3
3!
+
x^5
5!
+
x
⎫
⎪⎪
⎬
⎪⎪
⎭
from Problem 11
=lim
x→ 0
{
1 +
x^2
3!
+
x^4
5!
+···
}
= 1