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This question was on a calculus 1 test:

Given $\int_1^5 f(x) dx = 5$ and $\int_1^5 g(x) dx = \frac{2}{3}$,

Evaluate $\int_5^1 f(x) g(x) dx$

The following was my (naive) approach (using properties more suited to limits):

$\int_5^1 f(x) g(x) dx = -\int_1^5 f(x) g(x) dx = -5\bigl(\frac{2}{3}\bigr) = -\frac{10}{3}$

But I'm unsure whether integrals distribute over products in that way, and think the question was likely misprinted. However, I was awarded full marks for that answer. So is there a way that integral of a product can be the product of their integrals for unknown functions?

rigel
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    In general, $\int f \cdot g \neq \int f \cdot \int g$. So you should not have been given credit for your answer. – Anurag A May 12 '20 at 10:14
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    I'm thinking the problem contains some kind of typo. I would not expect any particular way of getting $\int fg$ just from knowing $\int f$ and $\int g$. Otherwise, things like H"older's inequality or worse would be absolutely trivial (and we would have an equality instead). – I was suspended for talking May 12 '20 at 10:17
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    The given information is not sufficient to work out the product integral. I'm guessing your teacher realized that and gave you full marks because it was their error. – Paul May 12 '20 at 10:18
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    It’s only true in rare cases. https://math.stackexchange.com/questions/2250993/when-the-integral-of-products-is-the-product-of-integrals – Vishu May 12 '20 at 10:18
  • I'll bet a chocolate fish that the last integrand was supposed to be $f(x)+g(x).$ – B. Goddard May 12 '20 at 11:03

1 Answers1

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As the comments point out, the integral cannot be evaluated with only the given information and the property used only holds true in rare cases.

So, $\int_5^1 f(x) g(x) dx = -\int_1^5 f(x) g(x) dx \ne -\int_1^5f(x)dx \int_1^5g(x)dx = -5\bigl(\frac{2}{3}\bigr) = -\frac{10}{3}$

rigel
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