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I have a question related to definite integrals and series from this site here.

It is written that the definite integral of $\max(x,1-x)dx$ from $0$ to $1$ is equal to $\frac34$:

$ \int_0^1 \max (x, 1-x) dx = \frac34$

but I have question, there are two different cases (I don't consider when $x$ is between $0$ and $1$, because in this case it is undefined which one is maximum), but in the second case, if $x<0$, then it is clear that $1-x$ is greater than $x$, so its integral is $x-x^2/2$, and after plugging values,we get $1/2$, and on the other hand, if $x>1$, then $x$ is maximum, its antiderivative is $x^2$ so we get $1/2$, so when is it equal to $3/4$? Please help me to understand this problem.

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    When you have "∫" with a number below and above it, you can say that you integrate from the lower number to the upper number with respect to the variable after the "d". So, you can read this as saying "the integral from 0 to 1 of the ma$x$imum of x and 1-x with respect to x." Thus, you x cannot equal anything less than 0 for this problem nor anything greater than 1, x can only take on values between, and including, 0 and 1.2012-08-16

3 Answers 3

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Since $1-x\geq x$ when $x\leq1/2$ and $x\geq 1-x$ when $x>1/2$, we have $\max (x, 1-x)=\left\{ \begin{array}{ll} 1-x, & \hbox{if }x\leq1/2; \\ x, & \hbox{if }x>1/2. \end{array} \right.$ Hence, $\int_0^1 \max (x, 1-x) dx =\int_0^{1/2} \max (x, 1-x) dx +\int_{1/2}^1 \max (x, 1-x) dx$ $=\int_0^{1/2} (1-x)dx +\int_{1/2}^1 x dx=(x-\frac{x^2}{2})\Big|_0^{1/2}+\frac{x^2}{2}\Big|_{1/2}^1=\frac{3}{4}.$

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    In a sense, this is the hard way. The integral is just the area of a square minus the area of a triangle. You don't need antiderivatives.2011-12-26
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This integral, viewed as the area under the graph of a function, is just the area of a square minus the area of a triangle---that's what the graph of the function looks like. So $ (\text{base}\times\text{height}) - \left(\frac12\times\text{base}\times\text{height}\right) = (1\times1)-\left(\frac12\times 1\times\frac12\right)=1-\frac14=\frac34. $

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This isn't that different from Paul's answer, but I wanted an excuse to use Iverson brackets. Recall that the Iverson bracket $[p]$ is equal to $1$ if condition $p$ is true, and $0$ if $p$ is false. We also have the relation $[\lnot p]=1-[p]$.

We then have

$\begin{align*} \max(x,1-x)&=x[x\geq 1-x]+(1-x)(1-[x\geq1-x])\\ &=1-x+(2x-1)\left[x\geq\frac12\right] \end{align*}$

Thus,

$\begin{align*} \int_0^1 \max(x, 1-x)\mathrm dx&=\int_0^1 \left(1-x+(2x-1)\left[x\geq\frac12\right]\right)\mathrm dx\\ &=\int_0^1 (1-x) \mathrm dx+\int_0^1 (2x-1)\left[x\geq\frac12\right]\mathrm dx\\ &=\int_0^1 (1-x) \mathrm dx+\int_{\frac12}^1 (2x-1)\mathrm dx\\ &=\left.\left(x-\frac{x^2}{2}\right)\right|_0^1+\left.\left(x^2-x\right)\right|_{\frac12}^1=\frac12+\frac14=\frac34 \end{align*}$