Prove the inequalities without calculating the integrals

Question

$$ \int_{0}^{\frac{\pi}{2}} \sin^4x dx \le \int_{0}^{\frac{\pi}{2}} \sin^3xdx$$

I have tried to define 2 functions $ f, g:[0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ and say that $ f(x) = \sin^4x$ and $ g = \sin^3x$. After that, $ f(\frac{\pi}{2}) < g(0)$ since the greatest value of f should smaller than the smallest value that g can take in order for $f(x) < g(x)$. I thought that if the function $ f $ is less than or equal to the function $ g $, then this will be valid for the integrals of these 2 functions.

But $ f(\frac{\pi}{2}) = sin^4\frac{\pi}{2} = 1 $ and $g(0) = sin^30 = 0 $ and 1 is not less or equal to 0.

Where am I wrong?

Answer 1

You thought that a necessary condition is $f(x) \leq g(y), \, \forall x,y \in [0,\pi/2]$. It is only sufficient. The condition $f(x) < g(x)$ means, that $f(a) < g(a), \, \forall a \in D$. $D$ is the domain of definition of $f$ and $g$.

Answer 2

$$ \sin^3x-\sin^4x=\sin^3x(1-\sin x)>0,\quad 0<x<\frac{\pi}{2}. $$

Answer 3

You need compare the two functions on the same value $x$. For example for $x=0$ we have

$$f(0)\le g(0)$$ Notice that we say that $f\le g$ on a set $S$ if $$f(x)\le g(x)\quad\forall x\in S$$ and in this case we have

$$\int_Sf(x)dx\le \int_S g(x)dx$$

Answer 4

Since $0 \le \sin x \le 1$ on $[0, \pi/2]$, $\sin^4 x \le \sin^3 x$ on $[0,\pi/2]$. Therefore $\int_0^{\pi/2} \sin^4 x\, dx \le \int_0^{\pi/2} \sin^3 x\, dx$.