Solve Integral with TI-84 Plus

**kwikness** · February 6th 2011, 05:17 PM

It's been a long time since I've done any calculus, so I've forgotten how to use my calculator to solve integrals. I have to solve

$\int_1^x \! (1/t) \, \mathrm{d}t$

for a few different x values. How can I do this using a graphing calculator (TI-84 Plus)?

**dwsmith** · February 6th 2011, 05:20 PM

Originally Posted by kwikness

It's been a long time since I've done any calculus, so I've forgotten how to use my calculator to solve integrals. I have to solve

$\int_1^x \! (1/t) \, \mathrm{d}t$

for a few different x values. How can I do this using a graphing calculator (TI-84 Plus)?

You need a Ti - 89.

**kwikness** · February 6th 2011, 05:32 PM

Originally Posted by dwsmith

You need a Ti - 89.

A TI-84+ is recommended for the course. Here is the problem out of the book:

**dwsmith** · February 6th 2011, 05:35 PM

Originally Posted by kwikness

A TI-84+ is recommended for the course. Here is the problem out of the book:

Unless they have changed those calculators, they can't integrate with respect to a variable. If you had and integral bounded by values not at an asymptote, you could graph it, hit calculate, select integrate, and put in the bounds.

**pickslides** · February 6th 2011, 05:35 PM

Hi there, you are not required to evaluate that integral.

You need to make a table in the Ti-84 to solve the approximation of the intergal using simpson's rule.

Do you know what this rule is?

**kwikness** · February 6th 2011, 07:03 PM

Originally Posted by pickslides

Hi there, you are not required to evaluate that integral.

You need to make a table in the Ti-84 to solve the approximation of the intergal using simpson's rule.

Do you know what this rule is?

I looked up Simpson's rule, but I'm not sure how to apply it to doing this problem with a calculator. Does anyone have any advice?

**pickslides** · February 6th 2011, 07:20 PM

Originally Posted by kwikness

I looked up Simpson's rule,

Is this the rule?

$\displaystyle \int_a^b f(x)~dx \approx \frac{b-a}{6}\left[ f(a)+4f\left(\frac{b-a}{2}\right)+f(b)\right]$

If so you are given different upper bounds in the first row of the table.

For $x = 0.5$ then $\displaystyle \int_1^{0.5} f(x)~dx \approx \frac{0.5-1}{6}\left[ f(1)+4f\left(\frac{0.5-1}{2}\right)+f(1)\right]=\dots$

Using the calculator gives you the power to evaluate all these sums at the same time.

**TheEmptySet** · February 6th 2011, 07:36 PM

Originally Posted by pickslides

Is this the rule?

$\displaystyle \int_a^b f(x)~dx \approx \frac{b-a}{6}\left[ f(a)+4f\left(\frac{b-a}{2}\right)+f(b)\right]$

If so you are given different upper bounds in the first row of the table.

For $x = 0.5$ then $\displaystyle \int_1^{0.5} f(x)~dx \approx \frac{0.5-1}{6}\left[ f(1)+4f\left(\frac{0.5-1}{2}\right)+f(1)\right]=\dots$

Using the calculator gives you the power to evaluate all these sums at the same time.

That is Simpson's rule for $n=2$

The OP asked for Simpson's rule with $n=10$

In general we have

$n$ must be even and $\displaystyle \Delta x=\frac{b-a}{n}$

$x_i=a+i\Delta x, i=0,1,2,...,n$

$\displaystyle \int_{a}^{b}f(x)dx=\frac{\Delta x}{2}\left[ f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+2f(x_{n-2}+4f(x_{n-1})+f(x_n)\right]$

**kwikness** · February 13th 2011, 04:10 PM

Originally Posted by TheEmptySet

That is Simpson's rule for $n=2$

$\displaystyle \int_{a}^{b}f(x)dx=\frac{\Delta x}{2}\left[ f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+2f(x_{n-2}+4f(x_{n-1})+f(x_n)\right]$

How do I determine what to put in the denominator under delta x?
How do I get past leaving a 0 in the denominator when I evaluate X of 0? $f(x_0) = 1/0$
Also, I'm having trouble figuring out what the best way to evaluate this on a calculator would be. I have to evaluate it at .5, 1.5, ...

**skeeter** · February 13th 2011, 06:24 PM

Simpson's rule to approximate the value of a definite integral ...

for $x = 0.5$ ...

$\Delta x=\dfrac{0.5-1}{10} = -.05$

$\displaystyle \int_{1}^{0.5} \frac{1}{t} \, dt \approx \frac{-0.05}{3}\left[ \frac{1}{1}+4 \cdot \frac{1}{.95} + 2 \cdot \frac{1}{.90} + 4 \cdot \frac{1}{.85} + 2 \cdot \frac{1}{.80} + 4 \cdot \frac{1}{.75} + 2 \cdot \frac{1}{.70} + 4 \cdot \frac{1}{.65} + 2 \cdot \frac{1}{.60} + 4 \cdot \frac{1}{.55} + \frac{1}{.50}\right] = -0.6931502307$

now do the same for every x-value (upper limit of integration) in the table.

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