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The question is

$\int_1^4 |a^2-x^2|dx=\cdots$ for $1<a<4$

Now that I am confuse since I don't have any idea on how to separate into several definite integrals. I mean, when $a=2$, then we can separate it into $\{1,2\}$ and $\{2,4\}$. However, when $a=3$, it will become $\{1,3\}$ and $\{3,4\}$, they're now different. How should I solve this? Thanks for any help!

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    $\begingroup$ From 1 to $a$ and from $a$ to 4. $\endgroup$
    – user376343
    Commented Jul 23, 2020 at 16:19
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    $\begingroup$ do you mean find an expression for $f(a)=\int^4_1|a^2-x^2|\,dx$ where $a$ ranges between $1$ and $4$? $\endgroup$
    – Mittens
    Commented Jul 23, 2020 at 19:32

1 Answer 1

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Separate in the two intervals $[1,a]$ and $[a, 4]$ as follows: $$ \int_1^4 |a^2-x^2| dx = \int_1^a (a^2-x^2)dx + \int_a^4 (x^2-a^2)dx = \left[ a^2x-\frac{x^3}{3} \right]_1^a + \left[ \frac{x^3}{3} - a^2x \right]_a^4 $$ $$ = \frac{2}{3}a^3 - a^2 + \frac{1}{3} + \frac{64}{3}-4a^2+\frac{2}{3}a^3 $$

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