A second-year college student's fantasy

The easiest way to do this question is to bind it between two functions.

Look at the rough graph of $x^{x}$ . All the points have been labelled and their respective coordinates given.

Now, observe that:

$Area(OBCE)+Area(\triangle DPC) <\int _{ 0 }^{ 1 }{ { x }^{ x }dx } <Ar(OBCE)+Ar(\triangle DPC)+Ar(\triangle ABP)$

$\Rightarrow \quad { x }^{ x }+\frac { 1 }{ 2 } (1-x)(1-{ x }^{ x })<\int _{ 0 }^{ 1 }{ { x }^{ x }dx } \quad <\frac { 1+{ x }^{ x } }{ 2 }$

Now, putting $x$ equal to our old friend $0.5$ means that ${ x }^{ x }\approx 0.7$ .

$\Rightarrow \quad 0.780<\int _{ 0 }^{ 1 }{ { x }^{ x }dx } \quad <0.85$

$\Rightarrow \quad 6.24<\quad 8\int _{ 0 }^{ 1 }{ { x }^{ x }dx } \quad <6.8$

Hence $\left\lfloor 8\int _{ 0 }^{ 1 }{ { x }^{ x }dx } \right\rfloor = \boxed{6}$

I know that this is a very crude method(and slightly incorrect), but it will work in this case. Why doesn't this happen to me in the exam! :/

This integral was studied by Johann Bernoulli and he called it the 'series mirabilis'; and with good reason:

$I=\int _{ 0 }^{ 1 }{ { x }^{ x }dx }=1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}...$

I am unaware of the existence of a closed form. But, for the purpose of this question, manually adding just the first 3 or 4 terms gives the required result. (Or you could just Wolfram-alpha it -_-)

$Note:$

For an outline of a proof of this result, see this .

A related series is $\int _{ 0 }^{ 1 }{ { x }^{ -x }}dx=1+\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}...$

Since exact answer is not required, I have used Numerical Integration . Here, $f(x)=x^x$ and $x$ varies from 0 to 1. Split this interval into two parts ,i.e., break at $0.5$ (you can split the interval into more parts in order to get more exact answer)

Interval size $h=0.5$ Now, we have three values of $x$ ,say :

$x_0=0$ which implies $\displaystyle y_0=f(x_0)=\lim_{x \to 0} x^x =1$

$x_1=0.5$ which implies $\displaystyle y_1=f(x_1)=0.5^{0.5}=0.7071$

$x_2=1$ which implies $\displaystyle y_2=f(x_2)=1^1=1$

Using the Trapezoidal Rule,

$\displaystyle I=\frac{h}{2}(y_0 + y_1 + 2y_1)$ $I=0.85355$ $\displaystyle 8I=6.828$

Thus, $\displaystyle \lfloor 8I\rfloor = 6$