# Thread: Definite Integral, just plain evilness from my teacher?

1. ## Definite Integral, just plain evilness from my teacher?

Hi Forum
I started reviewing some of my calculus material and I found this unanswered question.

$\displaystyle \int_0^1 \! (x+2)(x+1)^5 \, \mathrm{d}x.$

Now, Wolfram tells me to actually calculate $\displaystyle (x+1)^5$
And I don't remember dealing with something like this before.
So, is there an easy way to solve this?
Maybe I just got a sadistic teacher hahahaha
Oh, I don't have any of that handy calculators, I usually do it all by hand.

I really hope something like this doesn't pop up in a test.

Thanks!

2. ## Re: Definite Integral, just plain evilness from my teacher?

use substitution, $\displaystyle t=x+1$

then the integral will become,

$\displaystyle \int_1^2 \!(t+1)t^5\, \mathrm{d}x$

now expand and solve.

Or simply manipulate the statement

$\displaystyle (x+2)(x+1)^5=[(x+1)+1](x+1)^5=(x+1)^6+(x+1)^5$

3. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hi Forum
I started reviewing some of my calculus material and I found this unanswered question.

$\displaystyle \int_0^1 \! (x+2)(x+1)^5 \, \mathrm{d}x.$

Now, Wolfram tells me to actually calculate $\displaystyle (x+1)^5$
And I don't remember dealing with something like this before.
So, is there an easy way to solve this?
Maybe I just got a sadistic teacher hahahaha
Oh, I don't have any of that handy calculators, I usually do it all by hand.

I really hope something like this doesn't pop up in a test.

Thanks!
let $\displaystyle u = x+1$

$\displaystyle du = dx$

$\displaystyle \int_0^1 [(x+1)+1](x+1)^5 \, dx = \int_1^2 (u+1)u^5 \, du$

finish it.

4. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hi Forum
I started reviewing some of my calculus material and I found this unanswered question.

$\displaystyle \int_0^1 \! (x+2)(x+1)^5 \, \mathrm{d}x.$

Now, Wolfram tells me to actually calculate $\displaystyle (x+1)^5$
And I don't remember dealing with something like this before.
So, is there an easy way to solve this?
Maybe I just got a sadistic teacher hahahaha
Oh, I don't have any of that handy calculators, I usually do it all by hand.

I really hope something like this doesn't pop up in a test.

Thanks!
No, your teacher is not sadistic or plain evil. S/he simply expects you to understand what has been taught and apply it to questions, some of which may differ slightly from routine simple ones.

5. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hi Forum
I started reviewing some of my calculus material and I found this unanswered question.

$\displaystyle \int_0^1 \! (x+2)(x+1)^5 \, \mathrm{d}x.$

Now, Wolfram tells me to actually calculate $\displaystyle (x+1)^5$
And I don't remember dealing with something like this before.
So, is there an easy way to solve this?
Maybe I just got a sadistic teacher hahahaha
Oh, I don't have any of that handy calculators, I usually do it all by hand.

I really hope something like this doesn't pop up in a test.

Thanks!
If the substitution is not obvious, you can use integration by parts.

$\displaystyle \displaystyle \int{u\,\frac{dv}{dx}\,dx} = u\,v - \int{v\,\frac{du}{dx}\,dx}$.

Here let $\displaystyle \displaystyle u = x + 2 \implies \frac{du}{dx} = 1$ and $\displaystyle \displaystyle \frac{dv}{dx} = (x + 1)^5 \implies v = \frac{(x + 1)^6}{6}$, which means

\displaystyle \displaystyle \begin{align*} \int{(x + 2)(x + 1)^5\,dx} &= \frac{(x + 2)(x + 1)^6}{6} - \int{\frac{1(x + 1)^6}{6}\,dx} \\ &= \frac{(x + 2)(x + 1)^6}{6} - \frac{(x + 1)^7}{42} + C \\ &= \frac{(x + 1)^6}{6}\left[x + 2 - \frac{x + 1}{7}\right] + C \\ &= \frac{(x + 1)^6}{6}\left[\frac{7x + 14 - (x + 1)}{7}\right] +C \\ &= \frac{(x + 1)^6}{6}\left(\frac{6x + 13}{7}\right) + C \\ &= \frac{(6x + 13)(x + 1)^6}{42} + C\end{align*}

6. ## Re: Definite Integral, just plain evilness from my teacher?

Hey Guys!
I was after easier ways to do this without a calculator.
That was what I meant to say.

Originally Posted by mr fantastic
No, your teacher is not sadistic or plain evil. S/he simply expects you to understand what has been taught and apply it to questions, some of which may differ slightly from routine simple ones.
mr fantastic, you got my intentions wrong.
I know what duty a teacher has. I just think that doing binomial expansion in the midst of a test is not a good teaching method.
Sorry if you got me wrong!

Thanks all for your replies, once more.
All the best.

7. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hey Guys!
I was after easier ways to do this without a calculator.
That was what I meant to say.

mr fantastic, you got my intentions wrong.
I know what duty a teacher has. I just think that doing binomial expansion in the midst of a test is not a good teaching method.
Sorry if you got me wrong!

Thanks all for your replies, once more.
All the best.
But expecting (or hoping) that you will be able to see that you don't need to is good teaching.

8. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by HallsofIvy
But expecting (or hoping) that you will be able to see that you don't need to is good teaching.
Am I sounding like a jerk here?
If so, I really don't mean it.
I am not to question my teacher's methods, I'm sure.
This isn't really complicated; I crossed the line in rage.
Calculus is a lot of information in little time.

9. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hi Forum
I started reviewing some of my calculus material and I found this unanswered question.

$\displaystyle \int_0^1 \! (x+2)(x+1)^5 \, \mathrm{d}x.$

Now, Wolfram tells me to actually calculate $\displaystyle (x+1)^5$
And I don't remember dealing with something like this before.
So, is there an easy way to solve this?
Maybe I just got a sadistic teacher hahahaha
Oh, I don't have any of that handy calculators, I usually do it all by hand.

I really hope something like this doesn't pop up in a test.

Thanks!
If you stuck go straight forward...

$\displaystyle (x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1$

$\displaystyle (x+2)(x+1)^5=(x+2)(x^5+5x^4+10x^3+10x^2+5x+1)=x^6+ 7x^5+20x^4+30x^3+25x^2+11x+2$

$\displaystyle \int_0^1 (x+2)(x+1)^5 dx=\int_0^1 (x^6+7x^5+20x^4+30x^3+25x^2+11x+2)dx$

10. ## Re: Definite Integral, just plain evilness from my teacher?

Originally Posted by Zellator
Hey Guys!
I was after easier ways to do this without a calculator.
That was what I meant to say.

mr fantastic, you got my intentions wrong.
I know what duty a teacher has. I just think that doing binomial expansion in the midst of a test is not a good teaching method.
Sorry if you got me wrong!

Thanks all for your replies, once more.
All the best.
I was alluding to the use of a substitution.