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][0,1]f:[0,1]\rightarrow[0,1] is a continuous function, prove the following inequality (01f(x)dx)2201xf(x)dx\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)=(0tf(x)dx)220txf(x)dxF(t)=\left(\int^t_0f(x)\,dx\right)^2- 2\int^t_0xf(x)\,dx We know that F(0)=0F(0)=0, so to prove our inequality we simply need to prove taht F(t)F(t) is a decreasing fucntion. F(t)=2f(t)0tf(x)dx2tf(t)=2f(t)01(f(x)1)dxF^\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)f(x) is [0,1][0,1], so f(x)10f(x)-1\leq 0, subsequently by Theorem 1. 01(f(x)1)dx0\displaystyle\int^1_0(f(x)-1)\,dx\leq 0. So F(t)=2f(t)01(f(x)1)dx0,F^\prime(t)=2f(t)\int^1_0(f(x)-1)\,dx\leq 0, which means that F(t)F(t) is decreasing and F(1)F(0)=0F(1)\leq F(0)=0. Finaly we arrvie at the desired result (01f(x)dx)2201xf(x)dx.\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]Rf:[0,1]\rightarrow\mathbb{R}, there exists a(0,1)a\in(0,1) such that 0af(x)dx=0\displaystyle\int^a_0f(x)\,dx=0. Prove that

01f(x)dx1a2supx(0,1)f(x)\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]Rf:[0,1]\rightarrow\mathbb{R} be a continuous function satisfying following property 01f(x)dx=01xf(x)dx=1\int^1_0 f(x)\,dx=\int^1_0 xf(x)\,dx=1 Prove that 01f2(x)dx4\int^1_0 f^2(x)\,dx\geq 4

Hint. Try to search for a linear fucntion with the same proprties, also here f2(x)=f(x)f(x)f^2(x)=f(x)\cdot f(x).

Note by Nicolae Sapoval
5 years, 9 months ago

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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.

Dinesh Chavan - 5 years, 9 months ago

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In problem 4, I think the statement should be (01f(x)dx)2201xf(x)dx.\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)F'(t) makes sense. (Also, there exist counter-examples to the statement as is.)

Jon Haussmann - 5 years, 9 months ago

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

Nicolae Sapoval - 5 years, 9 months ago

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