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I'm given the following information..

$$ \int_0^2 f(x)dx = 3 $$

$$ \int_0^4 f(x)dx = 10 $$

$$ \int_0^2 g(x)dx = -5 $$

$$ \int_0^4 g(x)dx = 8 $$

And I've been asked to solve 5 following questions using the above information. The first 3 are easy, but I would like the last two checked if possible.

  1. $ \int_2^0 f(x) + 2g(x) dx $


$ - (3) + 2(-5) = -13$

  1. $ \int_2^2 \sqrt{f(x) + g(x)} dx $


Unsure, but is this just $0$ since it goes from $2$ to $2$?

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up vote 5 down vote accepted

$\int_2^0 (f(x)+2g(x))\ dx=-\int_0^2 f(x)\ dx-2\int_0^2 g(x)\ dx=-3-2(-5)=7$. The other integral is $0$.

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I'm known to mix my signs...I really need to pay attention when dealing with signs. Thanks for the quick response! – StrugglingWithMath Dec 10 '12 at 17:34

For the first one you have listed, notice $$\begin{align}\int_2^0(f(x)+2g(x))dx&=-\int_0^2(f(x)+2g(x))dx\\&=-\left(\int_0^2f(x)dx+2\int_0^2g(x)dx\right)\\&=-(3+2(-5))\\&=7.\end{align}$$ Your reasoning on the second equation is correct, so the value of the integral is $0$.

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Thank you for the quick answer, it is much appreciated – StrugglingWithMath Dec 10 '12 at 17:34

I only see two questions, not 3. But anyway, the first is wrong as you have an integral from $2$ to $0$ (it's $-3-2(-5)=7$)... while the second is correct.

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Thank you for the response, there were a total of 5 but I know the first 3 are correct and just wanted to check the last two. Thanks again! – StrugglingWithMath Dec 10 '12 at 17:35

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