- #1

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## Homework Statement

The letter W is defined as

## Homework Equations

## The Attempt at a Solution

Now, I had this question on a test and got it wrong. I get 0.93, which I think is correct. Apparently the answer is 0.24. I call BS on my teacher.

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- Thread starter b_roberts
- Start date

- #1

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The letter W is defined as

Now, I had this question on a test and got it wrong. I get 0.93, which I think is correct. Apparently the answer is 0.24. I call BS on my teacher.

- #2

gabbagabbahey

Homework Helper

Gold Member

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Now, I had this question on a test and got it wrong. I get 0.93, which I think is correct. Apparently the answer is 0.24. I call BS on my teacher.

Well, you're incorrect, as any calculus capable calculator will confirm. So, instead of directing profanities towards your teacher, perhaps you should post your calculations so we can point out where your error(s) are.

- #3

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- #4

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Who said anything about the area under a curve? It says compute a definite integral, not "Find the area under the curve."

- #5

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The definite integral is the area under the curve...

- #6

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That's a common misconception. The definite integral can get you the area under the curve, but that doesn't mean it is.

- #7

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- #8

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Remember the fundamental theorem of Calculus:

For a sufficiently nice function,

[tex]

\int_a^b f(x) dx = F(b) - F(a)

[/tex]

Where a and b are the bounds and F is the anti-derivative.

It doesn't care about negative or positive. All it cares about is the value of F at b and a.

- #9

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Alright. That makes sense. Thank you.

- #10

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That's why

[tex] \int_0^{2\pi} \cos x \ dx[/tex]

evaluates to zero. There is the same amount of area above the x-axis as there is below it.

- #11

gabbagabbahey

Homework Helper

Gold Member

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The definite integral is the area under the curve...

That's

If you were asked to find the area between the curve [itex]f(x)=4\sin x\cos x[/itex] and the x-axis on the interval [itex]x\in [1,2][/itex], then your method would be correct. However, that wasn't what you were asked.

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