Prove the inequality using the basic properties of integrals: $\int_{a}^{b} \sin^2(x) dx< \int_{a}^{b} |\sin x|dx$ for $a<b$
At first I integrated $\sin^2(x)$, and then I integrated |sin x|. For $\sin x >0$ the integral is $\cos x$ but for $\sin x < 0$ the integral is $-\cos x$. If $a$ or $b$ were given it'd be easier since I can integrate $|\sin x|$ based on the limits but I'm lost here.
Another question is to prove that: $\int_{0}^{1}e^{-x^2}< 1$
Here it didn't take me long to realise that the integral can't be expressed using elementary functions, so I don't know what to do?
Edit: I think I can analyse this and come to a conclusion using one of property of integral. For x in [0,1] $e^-x^2$ is always smaller than the constant a in this interval, therefore the integral will be too. I need to know if I'm right and also if there is another way of doing it.
For any $0<y<1$, we have that $y<\sqrt y$. For $y=\sin^2 x$, we will get $$\sin^2 x<\sqrt{\sin^2 x}=|\sin x|.$$ Now integrate over $(0,1)$.
For the second part, by the Mean Value Theorem for integrals, there exists some $c\in(a,b)$ such that $$\int_a^bf(x)dx=f(c)(b-a),$$ which here becomes $$\int_0^1 e^{-x^2}dx=e^{-c^2}(1-0)$$ for some $c\in(0,1).$ But for any $c\in(0,1), e^{-c^2}<1$, so $\displaystyle\int_0^1 e^{-x^2}dx<1$.