74 Higher Engineering Mathematics
=[
θ−θ^3
18+θ^5
600−θ^7
7 ( 5040 )+···] 10= 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. 40xln( 1 +x)dx=∫ 0. 40x(
x−x^2
2+x^3
3−x^4
4+x^5
5−···)
dx=∫ 0. 40(
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. 40=(
( 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 exerciseExercise 33 Further problems on
numerical integration using Maclaurin’s
series- Evaluate
∫ 0. 6
0. 2 3esinθdθ, 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
0cos2θθ1
3dθ. [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
0and^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
3x^3 +···−x
x^3⎫
⎪⎬⎪⎭
from Problem 5=lim
x→ 0⎧
⎪⎨⎪⎩1
3x^3 +···
x^3⎫
⎪⎬⎪⎭=lim
x→ 0{
1
3}
=1
3Similarly,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