I don't know how to solve this with the absolute value.
Find the value of c: \[ \int_{0}^{c} |x(1-x)|dx=0 \]
Please, if possible, make a solution step-by-step. Thank you and sorry for the poor english.
Calculus Apostol, 1.26 exercise 21 b
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$\begingroup$
calculus
integration
absolute-value
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1If the absolute value is what's throwing you off, break in into to cases where $0 \le x \le 1$ and $|x(1-x)| = x(1-x)$ and where $x \ge 1$ where $|x(1-x)| = x(x-1)$. – 2017-02-16
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0One easy value is $c=0$ since $\int_{0}^{0} f(x)dx=0$ for any integrable function $f$ and your function being the absolute value of a continuous function si surely integrable. – 2017-02-16
1 Answers
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HINT:
If $c\le 1$, then $\frac{c^2}{2}-\frac{c^3}{3}=0$.
If $c>1$, then $\frac{1}{2}-\frac{1}{3}+\frac{c^3}{3}-\frac{c^2}{2}-\frac{1}{3}+\frac{1}{2}=0 \Rightarrow \frac{c^2}{2}-\frac{c^3}{3}=\frac{1}{3}$.
Now try it out.
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0I get it. Thank you so much, I'm a begginer in calculus. – 2017-02-16