# Integration - Trapazoidal Rule

• Feb 22nd 2008, 09:03 PM
KavX
Integration - Trapazoidal Rule
Hi im stuck on 3 questions from my maths txt book.

questions 7,11, and 12

http://img527.imageshack.us/img527/772/dsc00663zz8.jpg

7: 0.41
11: 0.94
12: 0.92

my answer for 7 comes up as 5.75, 11 and 12 are off aswell.
thx in advance all help is appreciated.
• Feb 23rd 2008, 01:01 AM
earboth
Quote:

Originally Posted by KavX
Hi im stuck on 3 questions from my maths txt book.

questions 7,11, and 12

7: 0.41
11: 0.94
12: 0.92

my answer for 7 comes up as 5.75, 11 and 12 are off aswell.
thx in advance all help is appreciated.

I'll show you how to do #7. The following questions can be done by the same method:

$\int_0^2 \left(\frac1{x+4}\right) dx$ is to be approximated by 2 stripes:

1. Divide the interval [0, 2] into 2 equal parts: [0, 1], (1, 2]. The length of each part is 1.

2. Calculate
$f(0) = \frac14$
$f(1) = \frac15$
$f(2) = \frac16$

3. Calculate the area of the 2 trapezoids:
$
\frac{\frac14 + \frac15}{2} \cdot 1 +
\frac{\frac15 + \frac16}{2} \cdot 1 = \frac18 + \frac15 + \frac1{12} = \frac{49}{120} \approx 0.4083...$
Since the upper border of the trapezoids is allways above the curve this value must be slightly too large.

4. The exact value is

$\int_0^2 \left(\frac1{x+4}\right) dx = \left. \ln(x+4) \right|_0^2 = \ln(6) - \ln(4) = \ln\left(\frac32\right) \approx 0.405...$
• Feb 23rd 2008, 01:06 PM
KavX
sweet thx , these are just a breeze now
i needed to check were im rushing figured out 11. working on 12 now

F(1)=1
F(2)=1/4
F(3)=1/9
F(4)=1/16
F(5)=1/25

h=5-1/4 = 1

1/2((1+1/25)+2(1/4 +1/9+1/16))

= 0.9436....

thx alot they dont seem much of a challenge now.