Trapezoid Rule is really ruling, need to be free, help!

**roseh** · March 23rd 2006, 04:09 AM

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I am sitting here stuck on these three problems. Please help me!

1. Find an approximation for the area in the first quadrant between the x-axis and the curve y=4-(x-2)^2 using 4 equally spaced intervals and a Left Hand Rieman Sum.

Im not sure about the left hand part. But I've tried and I come up with 10 and 10.666.

2. I have graph the function of y=1/x, where x goes from 0 to 5. Using the trapexoid rule with 4 equal intervals, I must approximate the area under the curve between x=1 and x=3.
Please help.

Okay, this is what I have for the second one so far. But for some reason I keep missing it from here. I know its just a matter of calculating but I just cant get it right.

f(x) = 1/x
a = 1
b = 3
n = 4

T = [(b - a)/(2n)]*[f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)]

T = [(3 - 1)/(2*4)]*[f(1) + 2f(3/2) + 2f(2) + 2f(5/2) + f(3)]

T = (1/4)[1 + 2(2/3) + 2(1/2) + 2(2/5) + (1/3)]

Now for the next trapezoid problem(ruler).

If the trapezoid rule is used with 5 interval,, what is the integral dx/1 + x^2 ,with limits 1,0.

Please, please help!

**CaptainBlack** · March 23rd 2006, 05:08 AM

Originally Posted by roseh

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I am sitting here stuck on these three problems. Please help me!

1. Find an approximation for the area in the first quadrant between the x-axis and the curve y=4-(x-2)^2 using 4 equally spaced intervals and a Left Hand Rieman Sum.

Im not sure about the left hand part. But I've tried and I come up with 10 and 10.666.

You are being asked to estimate:

$ \int_0^4 y(x) dx=\int_0^4 4-(x-2)^2\ dx $

using 4 equal sub-intervals of $[0,4]$ . These sub-intervals are:

$[0,1], [1,2], [2,3]$ and $[3,4]$ .

As we are instructed to use left hand Riemann sums we will approximate the
function on each of these subintervals by its value at the left hand point of
the sub-interval, so:

$ \int_0^4 y(x) dx\approx (y(0)\times \delta x) +(y(1) \times \delta x)$ $+ (y(2) \times \delta x) + (y(3) \times \delta x) $ ,

where $\delta x$ is the width of the intervals in this case $\delta x=1$

Hence:

$ \int_0^4 y(x) dx\approx (0\times \delta x) +(3 \times \delta x)$ $+ (4 \times \delta x) + (3 \times \delta x)=10 $ .

This may be compared with the result of doing the integral analytically, which
gives an area of $32/3 \approx 10.667$ .

RonL

**ThePerfectHacker** · March 23rd 2006, 06:04 AM

Originally Posted by roseh

-----------------------------------------------2. I have graph the function of y=1/x, where x goes from 0 to 5. Using the trapexoid rule with 4 equal intervals, I must approximate the area under the curve between x=1 and x=3.

You need to approximate,
$\int^3_1\frac{dx}{x}$
You are using 4 equal intervals thus, $n=4$ .
Next, you need equal widths which is $\Delta x=\frac{b-a}{n}=\frac{2}{4}=.5$ .
Now, by the trapezoidal rule,
$\frac{1}{2}[f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+$ $2f(a+3\Delta x)+f(a+4\Delta x)]$ $\Delta x$
But in your values,
$\frac{1}{2}[f(1)+2f(1.5)+2f(2)+2f(2.5)+f(3)].5$
Now, find the value $f(x)=\frac{1}{x}$ ,
$\frac{1}{2}[1+1.33+1+.8+.33].5\approx 1.11$
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Now, compare to the actual vale. This is what I am afraid of I do not know if you studied natrual logarithms. In that case, skip this part.
The actual value of,
$\int^3_1\frac{dx}{x}=\ln 3$
because this is the very definition of the natural logarithm. But $\ln 3\approx 1.09$

Notice that you error by using only 4 subintervals is only .02

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