Playing with Integrals: Ineqaulitites and Integrals 2

Although my previous post on this topic didn't cause a discussion I'll try to continue. If no one will be interested in problems from this post too, I'll switch to another theme.

Today I would like to present several inequalities directly involving integrals.

Problem 4. If $$f:[0,1]\rightarrow[0,1]$$ is a continuous function, prove the following inequality $\left(\int^1_0f(x)\,dx\right)^2\leq 2\int^1_0xf(x)\,dx$

Solution. To start our proof we'll consider the following function $F(t)=\left(\int^t_0f(x)\,dx\right)^2- 2\int^t_0xf(x)\,dx$ We know that $$F(0)=0$$, so to prove our inequality we simply need to prove taht $$F(t)$$ is a decreasing fucntion. $F^\prime(t)=2f(t)\int^t_0f(x)\,dx-2tf(t)=2f(t)\int^1_0(f(x)-1)\,dx$ However, we know that range of $$f(x)$$ is $$[0,1]$$, so $$f(x)-1\leq 0$$, subsequently by Theorem 1. $$\displaystyle\int^1_0(f(x)-1)\,dx\leq 0$$. So $F^\prime(t)=2f(t)\int^1_0(f(x)-1)\,dx\leq 0,$ which means that $$F(t)$$ is decreasing and $$F(1)\leq F(0)=0$$. Finaly we arrvie at the desired result $\left(\int^1_0f(x)\,dx\right)^2\leq 2\int^1_0xf(x)\,dx.$

Now try your own techniques to solve the following problems.

Problem 5. For continuous, differentiable function $$f:[0,1]\rightarrow\mathbb{R}$$, there exists $$a\in(0,1)$$ such that $$\displaystyle\int^a_0f(x)\,dx=0$$. Prove that

$\left|\int^1_0f(x)\,dx\right|\leq\frac{1-a}{2}\sup_{x\in(0,1)}|f^\prime(x)|$

Hint. For this problem you might consider the Mean Value Theorem

Problem 6. Let $$f:[0,1]\rightarrow\mathbb{R}$$ be a continuous function satisfying following property $\int^1_0 f(x)\,dx=\int^1_0 xf(x)\,dx=1$ Prove that $\int^1_0 f^2(x)\,dx\geq 4$

Hint. Try to search for a linear fucntion with the same proprties, also here $$f^2(x)=f(x)\cdot f(x)$$.

Note by Nicolae Sapoval
4 years, 5 months ago

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In problem 4, I think the statement should be $\left( \int_0^1 f(x) \: dx \right)^2 \le 2 \int_0^1 xf(x) \: dx.$ This is the only way the expression for $$F'(t)$$ makes sense. (Also, there exist counter-examples to the statement as is.)

- 4 years, 5 months ago

Yes, you're definitely right. Thanks for pointing that out!

- 4 years, 5 months ago

Please do continue. I like your posts very much and I am always a fan of inequalities and a newcommer at Calculus.Please do such another ones.Thanks.

- 4 years, 5 months ago