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I have this math problem:

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First: I found the integral from $-1$ to $3$ of $|x^2+3|$ and got $64/3$

Second: I found the integral from $-1$ to $3$ of $|x|$ and got $5$

Finally: I subtracted $5$ from $64/3$ and got $49/3$

However $49/3$ is wrong, what did I do wrong?

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    $\begingroup$ Did you really mean the absolute values $|\cdots|$ ? $\endgroup$ – Han de Bruijn Sep 28 '14 at 19:12
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you have to use the form $$\int_a^b f(x)-g(x)dx$$ since the first integral contains a negative part of the function $y=x$ and this part gives negative area i.e. the value 5 is not true
The right one is $$\int_{-1}^3 x^2+3-xdx=52/3$$

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  • $\begingroup$ So, the answer is 52/3? $\endgroup$ – KFC Sep 28 '14 at 19:11

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