Bader A.

asked • 02/11/18

What is ∫-7^-1 f(x)dx?

suppose that f(x) is an even function and let ∫0^1 f(x)dx = 5 and ∫0^7 f(x)dx = 1. What is ∫-7^-1 f(x)dx?
 
This is my work
 
I expand ∫-7^-1 f(x)dx to F(-1) - F(-7)
 
F(-1) - F(-7) is the same thing as F(1) - F(7) since it is an even function
 
plug in the values to get 5-1, which equals 4
 
I strongly believe that the answer is 4, however, on the test my teacher is saying it's -4. I talked to her about it and she did not listen. Could someone show me if I am right, or if the answer is indeed -4?
 
Also, I wasn't really sure how to write the definite integral on the keyboard, so if you didn't get what I mean by ∫-7^-1, I meant to say that the lower limit is -7 and the upper limit is -1

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Inactive Tutor answered • 02/11/18

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Disregard this answer!

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Inactive Tutor

The integral of an ODD function over a balanced interval is zero.  Example:
 
∫-11 xdx = 0
 
But the integral of an EVEN function over a balanced interval gives twice the integral from 0 to the upper limit.  Example:
 
∫-11 x2dx = (x3/3)|-11 = 1/3 - (-1/3) = 2/3 = 2∫01 x2dx
 
 
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02/11/18

Inactive Tutor

Therefore, since f(x) is even,
 
∫-77 f(x)dx = 2∫07 f(x)dx
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02/11/18

Inactive Tutor

Yes, right. I messed up.. Thanks
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02/11/18

Inactive Tutor

No problem.  Thank you for your many fine solutions.
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02/11/18

Inactive Tutor

Since f(x) is even ∫-77f(x)dx=2∫07f(x)dx=2.

2=∫-77f(x)dx=(∫-7-1+∫-10+∫07)f(x)dx= ∫-7-1f(x)dx+∫-10f(x)dx+1.
5=∫01f(x)dx=-∫-10f(-x)dx=-∫-10f(x)dx. ⇒∫-10f(x)dx=-5
∫-7-1f(x)dx=2+4=6.
Please disregard the previous answer. Thanks to Arturo O. I corrected my mistake. But there might be still some mistakes. Therefore, it would be goo to look through carefully.
 
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02/11/18

Inactive Tutor

Let us work this very carefully step by step.
 
f(x) is even, so
 
∫-70 f(x)dx = ∫07 f(x)dx = 1
 
and
 
∫-10 f(x)dx = ∫01 f(x)dx
 
But
 
∫-70 f(x)dx = ∫-7-1 f(x)dx + ∫-10 f(x)dx ⇒
 
∫07 f(x)dx = ∫-7-1 f(x)dx + ∫01 f(x)dx ⇒
 
1 = ∫-7-1 f(x)dx + 5 ⇒
 
∫-7-1 f(x)dx = 1 - 5 = -4
 
The teacher is correct.
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02/11/18

Inactive Tutor

You are absolutely right!  I messed it up for the second time.
Thank you Arturo!. You saved my life more twice.
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02/11/18

Inactive Tutor

To Bader from Scottsdale, AZ. Please disregard my answer and my comment above. They both were wrong!  
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02/11/18

Bader A.

Thank you guys, I now understand why the answer is -4
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02/11/18

Inactive Tutor

We have to thank Arturo O.!
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02/11/18

Inactive Tutor answered • 02/11/18

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As stated, the problem is literally asking for
 
∫(-1/7)f(x)dx = ?
 
∫(-1/7)f(x)dx = (-1/7)∫f(x)dx
 
Furthermore,
 
01 = 07 = 0, 
 
so the first two integrands in the first sentence of the problem statement are zero.  Is this what you really meant to say?
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Bader A.

I'm sorry for the confusion that I caused. What I meant is what is ∫a=-7 b=-1 f(x)dx, when f(x) is an even function, ∫a=0 b=1 f(x)dx=5 and ∫a=0 b=7 f(x)dx=1
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02/11/18

Inactive Tutor

If f(x) is even, then its integral F(x) is odd.  The indefinite integral of an even function is always an odd function, and the indefinite integral of an odd function is always an even function.  Your mistake was to assume that F(-1)-F(-7) is the same thing as F(1)- F(7).  It is true that
 
f(-x) = f(x),
 
since f(x) is even.  But its' indefinite integral F(x) is odd, so
 
F(-x) = -F(x).
 
Therefore 
 
F(-1) - F(-7) = -F(1) -[-F(7)] = F(7) - F(1).
 
I think you have what you need to figure it out now.
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02/11/18

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