Discovering Patterns in Integrals: Investigating Families of Functions Using CAS

[lunus]

Homework Statement

Hi, I am doing this discovery project called patterns in integrals i found in my calculus textbook. I have to use a CAS (I'm using Maple) to investigate indefinite integrals of families of functions. Then by observing the patterns that occur in the integrals, i have to first guess, and then prove, a general formula for the integral of any member of the family. there are four different familes and I am done with three of them, but stuck on the last one. I would appreciate any help. The question and what I have done so far is on the pdf attachment.

The Attempt at a Solution

My attempt is on the pdf file i attached.

 

[Gib Z]

Please post the actual family here, since we have to wait for a mentor to approve the file first. I'm sure we can help you though, so that's some reassurance for you =] Welcome to Physicsforums!

 

[lunus]

(a) use a CAS to evaluate the following integrals (I used maple)

$$\int{xe^{x}}dx = \left( x-1 \right) {e^{x}}$$
$$\int{x^{2}e^{x}}dx = \left( 2-2\,x+{x}^{2} \right) {e^{x}}$$
$$\int{x^{3}e^{x}}dx = \left( -6+6\,x-3\,{x}^{2}+{x}^{3} \right) {e^{x}}$$
$$\int{x^{4}e^{x}}dx = \left( 24-24\,x+12\,{x}^{2}-4\,{x}^{3}+{x}^{4} \right) {e^{x}}$$
$$\int{x^{5}e^{x}}dx = \left( -120+120\,x-60\,{x}^{2}+20\,{x}^{3}-5\,{x}^{4}+{x}^{5} \right) {e^{x}}$$

(b) based on the patterns of your responses in part (a), guess the value of $$\int{x^{6}e^{x}}dx$$ Then use your CAS to check your answer.

This was my guess: $$e^{x}(x^{6}-6x^{5}+30x^{4}-120x^{3}+360x^{2}-720x+720)$$ and maple returned the same answer.

(c) based on the pattern in parts (a) and (b), make a conjecture as to the value of the integral
$$\int{x^{n}e^{x}}dx$$
when n is a positive integer

This is what i came up with: $$\sum_{i=0}^{n}\frac{x!}{i!}n!e^{x}$$
Now this is where I am stuck because i know this is not correct.
I figured it has something to do with factorial or series.

(d) use mathematical induction to prove the conjecture you made in part (c)

 

[arildno]

Hmm..not quite sure you got that right, here's a better approach:
Define a sequence as follows:
$$F_{n}=\int{x}^{n}e^{x}=x^{n}e^{x}-nF_{n-1}$$
Thus, we have:
$$F_{n}+nF_{n-1}=x^{n}e^{x}$$

Assume a solution as follows:
$$F_{n}=(-1)^{n}n!e^{x}{\sum_{i=0}^{n}a_{i}x^{i}$$
Thus, inserting in our difference equation, we get:
$$e^{x}n!(-1)^{n}a_{n}x^{n}=x^{n}e^{x}\to{a}_{n}=\frac{(-1)^{n}}{n!}$$

Therefore, we get:
$$F_{n}=}=(-1)^{n}n!e^{x}{\sum_{i=0}^{n}\frac{(-x)^{i}}{i!}$$

 

[lunus]

I appreciate your help arildno. thank you!

 

[lunus]

Hmm..not quite sure you got that right, here's a better approach:
Define a sequence as follows:
$$F_{n}=\int{x}^{n}e^{x}=x^{n}e^{x}-nF_{n-1}$$
Thus, we have:
$$F_{n}+nF_{n-1}=x^{n}e^{x}$$

Assume a solution as follows:
$$F_{n}=(-1)^{n}n!e^{x}{\sum_{i=0}^{n}a_{i}x^{i}$$
Thus, inserting in our difference equation, we get:
$$e^{x}n!(-1)^{n}a_{n}x^{n}=x^{n}e^{x}\to{a}_{n}=\frac{(-1)^{n}}{n!}$$

Therefore, we get:
$$F_{n}=}=(-1)^{n}n!e^{x}{\sum_{i=0}^{n}\frac{(-x)^{i}}{i!}$$

I have one more question. I am a little loss on how to use mathematical induction to prove this, can u help me. thank you.

 

[arildno]

The mathematical induction step is taken care of by setting up the difference equation, valid for all n

 

[lunus]

thanks

 

[lunus]

ive tried but no luck, I am not good with mathematical induction.