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There was this 1 question on the last test that I am puzzeled about. The question states to compare 2 integrals without calculation.

$I_1 = \int_0^{\pi/2}\sin^{10}x \,dx$ and

$I_2 = \int_0^{\pi/2}\sin^{2}x\, dx$

I would say $I_1=I_2$ because they have the same lower and upper integral limit and because they are periodic functions, am I correct on this one ?

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    $\begingroup$ If $x$ ranges between $0$ and $\pi/2$, then $\sin^{10} x \leq \sin^2 x$. $\endgroup$
    – Gibbs
    Jun 8 '18 at 15:18
  • $\begingroup$ In fact, on any interval $[a,b]$ of positive length, $$\int_a^b\sin^{10}x \,dx <\int_a^b\sin^{2}x \,dx $$ $\endgroup$
    – zhw.
    Jun 8 '18 at 16:38
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No sorry friend it is $I_{10}<I_{2}$

The reason why is because on the interval $(0,\frac{\pi}{2})$ we have that

$0<\sin(x)<1$. Thus we know that when we raise a real number $r$ between $0$ and $1$ to a power that number gets smaller more generally $\lim_{n \to \infty} r^{n}=0$

Now Let's recall that an integral is the area boundd by a curve. If we instead consider the Riemann sum from the statements I've made above we can conclude that for every rectangle with width $dx$ the height in $I_{10}$ is smaller than $I_{2}$ and thus the area of the rectangle is smaller and thus we may conclude

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Hint: for any $x \in [0, \pi/2]$, we have $0 \leq \sin x \leq 1$, and so $0 \leq \sin^{10} x \leq \sin^2 x$.

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You are wrong. In fact, $I_2<I_1$ because $0\leqslant\sin^{10}x\leqslant\sin^2x$ and you only have $\sin^{10}x=\sin^2x$ when $x=0$ or $x=\frac\pi2$. Besides, both functions are continuous.

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  • $\begingroup$ @Gibbs Thanks to your non-stupid comment, I've edited my answer. Thank you. $\endgroup$ Jun 8 '18 at 15:23

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