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 \(f^2(x)=f(x)\cdot f(x)\).

## Comments

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TopNewestIn 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.) – Jon Haussmann · 3 years ago

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– Nicolae Sapoval · 3 years ago

Yes, you're definitely right. Thanks for pointing that out!Log in to reply

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 · 3 years ago

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