I know I got this wrong it was on my final

[flyingpig]

Homework Statement

I will try to word it the best I can...

True or False? If True explain why. If False, provide a counterexample to explain why it is wrong, DO NOT CORRECT THE STATEMENT

Suppose f(x) is an odd function and then $$\int_{-a}^{a} f(x) dx = 0$$

The Attempt at a Solution

This is true, I wrote down, but my proof was wrong...I think. I checked the textbook and they did it differently. My textbook splited the integral from -a to 0 and 0 to a and did a substitution.

Here is how I did it

Let f(x) be odd

$$f(-x) = -f(x)$$

$$\int_{-a}^{a} f(-x) dx = \int_{-a}^{a} -f(x) dx$$

By the Fundamental Theorem of Calculus

$$F(-a) - F(+a) = -\left [ F(a)- F(-a)) \right ]$$

$$F(-a) - F(+a) = -F(a) + F(-a))$$

$$F(-a) - F(+a) = F(-a) - F(a)$$

$$0 = 0$$

Did I "prove" anything...?

 

[Mark44]

Assuming that the statement you were to explain (it didn't say to prove) or find a counterexample for was
If f(x) is odd, then
$$\int_{-a}^a f(x) dx = 0$$

Consider f(x) = 1/x and the interval [-1, 1].

f is an odd function, but is $$\int_{-1}^1 \frac{dx}{x} = 0 \text{?}$$

 

[flyingpig]

The integral diverges...

Perhaps the word "continuous" was thrown in there and I forgot to add it to this question.

Mark, I don't think even the TA would've realized it lol. The TAs at my university are all pretty dumb, except for one guy, but he's a grad student

 

[romsofia]

The integral diverges...

Perhaps the word "continuous" was thrown in there and I forgot to add it to this question.

Mark, I don't think even the TA would've realized it lol. The TAs at my university are all pretty dumb, except for one guy, but he's a grad student

What do mean by diverges? And based on what Mark proposed, then the statement would be false, right?

 

[flyingpig]

What do mean by diverges? And based on what Mark proposed, then the statement would be false, right?

Only if it is discontinuous. I think the function f(x) was said to be continuous, which is something I forgot to put down...

That or it never had the word "continuous". Either way, I am wrong lol. There goes my 100%...

 

[Mark44]

A lot hinges on whether the function was given to be continuous. If there was nothing stated, you can't assume that it is continuous.

Also, with regard your proof, concluding that 0 = 0 (or any other statement that is always true - a tautology) is not what you want to do.

 

[LCKurtz]

Homework Statement

I will try to word it the best I can...

True or False? If True explain why. If False, provide a counterexample to explain why it is wrong, DO NOT CORRECT THE STATEMENT

Suppose f(x) is an odd function and then $$\int_{-a}^{a} f(x) dx = 0$$

The Attempt at a Solution

This is true, I wrote down, but my proof was wrong...I think. I checked the textbook and they did it differently. My textbook splited the integral from -a to 0 and 0 to a and did a substitution.

Here is how I did it

Let f(x) be odd

$$f(-x) = -f(x)$$

$$\int_{-a}^{a} f(-x) dx = \int_{-a}^{a} -f(x) dx$$

By the Fundamental Theorem of Calculus

$$F(-a) - F(+a) = -\left [ F(a)- F(-a)) \right ]$$

$$F(-a) - F(+a) = -F(a) + F(-a))$$

$$F(-a) - F(+a) = F(-a) - F(a)$$

$$0 = 0$$

Did I "prove" anything...?

Yes. You proved that if f is odd, then 0 = 0.

 

[Mark44]

Yes. You proved that if f is odd, then 0 = 0.

flyingpig could just as well have proved that if 1 = 2, then 0 = 0, which is another true statement.

 

[flyingpig]

Our professor promised a few students (I, included) that if we score extremely well (I am guessing he hinted 100%, because the few students who were "nominated" were getting like 99.x% lol), he will just give the free 1%.

There goes my chance...

At least it's over 90%...