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$\displaystyle\int \limits_{-1}^{1}\big(x\sin{\pi}x\big)dx $?

a) 2

b) -2

c)1

d)0

I've reduced it to

$=2\displaystyle\int \limits_{0}^{1}\big(x\sin{\pi}x\big)dx$ as it is an even function.

Do I have to apply LIATE rule from here?

Update 1: After applying integration by part..

$$ = 2\left[ {x\int\limits_0^1 {\sin } \pi xdx - \int\limits_0^1 {\left( {\frac{d}{{dx}}x\int\limits_0^1 {\sin } \pi xdx} \right)} dx} \right]$$

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2  
Do you know integration by parts? –  Paul Jan 20 '12 at 11:21
    
Ya.. LIATE rule comes in that only..I also applied that rule.. –  rohan-patel Jan 20 '12 at 11:24
    
@Paul: Do I have use substitution here? –  rohan-patel Jan 20 '12 at 11:41
    
Show us what happens when you use integration by parts. –  Gerry Myerson Jan 20 '12 at 11:43
1  
Notice that answers b and d are out, since you hae a positive integrand over an interval. –  ncmathsadist Jan 20 '12 at 11:47
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2 Answers

up vote 6 down vote accepted

Here's the integration by parts formula $$ \int f(x) g'(x) dx = f(x)g(x) - \int f'(x) g(x) dx. $$

I would suggest not relying on mnemonics such as the "LIATE rule", and instead think about what's going on in the formula. Note that on the left hand side of the above formula, you need to find an antiderivative of $f(x)g'(x)$ while on the right hand side, you need to find an antiderivative of $f'(x) g(x)$.

Let's look at your integral now. It might be best to find the indefinite integral first as it reduces chance of making errors, and then compute the definite integral.

When using integration by parts to evaluate $$ \int x\sin(\pi x)\,dx, $$

you essentially say

"well, finding an antiderivative of $x\sin(\pi x)$ is hard. But if I differentiate one part and antidifferentiate the other, could I deal with what results (that is, would the integration by parts formula prove useful)?"

Let's see. There are two choices:

$\ \ \underbrace{x\vphantom{(}}_f \underbrace{\sin(\pi x)}_{g'} \longrightarrow \underbrace{\vphantom{(}1}_{f'}\cdot\underbrace{{-\cos(\pi x)\over\pi}}_g$

and

$\ \ \underbrace{ x}_{g'} \underbrace{\sin(\pi x)}_f \longrightarrow \underbrace{{x^2\over2}}_g \cdot\underbrace{\pi \cos(\pi x)}_{f'}$

The former option is the best, we can integrate ${-\cos(\pi x)\over\pi} $.

So, using the integration by parts formula with $f(x)=x$ and $g'(x)=\sin(\pi x)$:

$$\eqalign{ \int \underbrace{x\vphantom{( }}_f \,\underbrace{\sin(\pi x)}_{g'} &= \underbrace{x\vphantom{(1\over2}}_{f} \underbrace{-\cos (\pi x)\over \pi}_{g} -\int \underbrace{1\vphantom{(1\over2}}_{f'} \cdot\underbrace{-\cos (\pi x)\over \pi}_{g}\, dx\cr &={-x\cos(\pi x)\over\pi}+{1\over\pi}\cdot{1\over\pi}\sin(\pi x)+C. } $$ So $$ \int_0^1 x\sin(\pi x)\, dx= \Bigl[{-x\cos(\pi x)\over\pi}+{1\over\pi}\cdot{1\over\pi}\sin(\pi x)\Bigr]_{x=0}^{x=1} ={1\over\pi}-{0}={1\over\pi}. $$


Note: either I made a mistake somewhere, or for your problem, you'd say "none of the above".

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I think options are wrong.. Two guys getting the same answer so most probably my options are wrong.. Thanks.. –  rohan-patel Jan 20 '12 at 13:08
    
But why integrate by parts while the whole point of the exercise was to teach you some elementary properties of the definite integral! –  user21436 Jan 20 '12 at 15:35
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@Kanna, how do you know what the point of the exercise was? Unless you were the lecturer who assigned it.... –  Gerry Myerson Jan 20 '12 at 23:34
    
This is a pretty good and easy solution which is suitable when asked in multiple choice question.. Thanks again. –  rohan-patel Jan 21 '12 at 16:12
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That you have reduced your integral to what you have given is a good job. I'll illustrate a technique, you often will have to use.

Set $I$ equals to the integral at hand. (i.e.)

$$I=\int_0^1{x \sin(\pi x)}$$

Now, I leave it to you (You can ask me how to prove this after some try!) to prove,

$$\int_a^b{f(x)dx}=\int_a^b{f(a+b-x)dx}$$

Now what would this property mean for this integral? This means, $$I=\int_0^1(1-x)\sin[\pi(1-x)]dx$$

Now, simplifying, you'll have that $$I=\int_0^1(1-x)\sin(\pi-\pi x)dx$$ That $\sin(\pi-y)=\sin y$, gives you, $$I=\int_0^1\sin(\pi x) dx-\int_0^1x\sin(\pi x) dx$$ $$2I=\dfrac{2}{\pi}$$ giving you $\boxed{I=\dfrac{1}{\pi}}$.

So, the answer to your question is $\dfrac{2}{\pi}$. which is your (e) None of these =)

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1  
I think your $\int1\,dx$ should be $\int\sin(\pi x)\,dx$. –  Gerry Myerson Jan 20 '12 at 12:16
    
@Sampath : Ok..Thanks..You showed me right direction.. I'll try to comprehend it.. –  rohan-patel Jan 20 '12 at 12:23
    
@Gerry Fixed it. But funny, isn't it? Or I went crazy. Now is everything fine? –  user21436 Jan 20 '12 at 12:32
    
@Kanna, looks good to me. –  Gerry Myerson Jan 20 '12 at 23:35
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