The maximum value of $\int_0^1(f(x))^3 dx$ is $\frac pq$ if $|f(x)|\leq 1$ and $\int_0^1 f(x)dx=0$. Find $p+q$, given that $p,q$ are relatively prime.  This question has been previously asked before  but all the answers are too complicated for me as I haven't been taught inequalities in integrals. So if anyone could solve it in a simpler way it would be really helpful. Edit: the answer is $p+q=5$. Edit 2: Since the question is/was closed, this is to provide &quot;additional context&quot;. The source of the problem is: Advanced Problems in Mathematics for JEE Main &amp; Advanced by Vikas Gupta and Pankaj Joshi, Subjective Exercise of Chapter 5, Indefinite and Definite Integration, Question Number 12. My work:  If $g(x)\leq f(x)$, for $a\leq x\leq b$, then $\int_a^b g(x)dx\leq \int_a^b f(x)dx$  My gut says that the problem will probably use this, because when $0\leq f(x)\leq 1, \left(f(x)\right)^3 \leq f(x)$ and the inequality reverses when $-1\leq f(x)\leq 0$. But I'm not able to put it to work. Relevance: Following is a picture showing the syllabus I have to study under Integral Calculus.  UNIT9:INTEGRAL CALCULUS Integral as an anti - derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of the type $\int \frac{d x}{x^{2} \pm a^{2}}, \int \frac{d x}{\sqrt{x^{2} \pm a^{2}}}, \int \frac{d x}{a^{2}-x^{2}},  \int \frac{d x}{\sqrt{a^{2}-x^{2}}}, \int \frac{d x}{a x^{2}+b x+c}$ $\int \frac{d x}{\sqrt{a x^{2}+b x+c}}, \int \frac{(p x+q) d x}{a x^{2}+b x+c}, \int \frac{(p x+q) d x}{\sqrt{a x^{2}+b x+c}}$ $\int \sqrt{a^{2} \pm x^{2}} d x \int \sqrt{x^{2}-a^{2}} d x$ Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals. Evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form.  (Transcribed from ) Since it does not involve higher level inequalities used by the solutions in the attached link, and the fact that this problem is posed to us under the above syllabus, there has to exist a simple solution (though lengthy, probably) to the question.
39182
Find the maximum value of $\int_0^1 f^3(x)dx$ given that $-1 \le f(x) \le 1$ and $\int_0^1 f(x)dx = 0$   I could not find a way to solve this problem. I tried to use the cauchy-schwarz inequality but could not proceed further  $$\int_0^1 f(x) \cdot f^2(x) dx \le \sqrt{\left(\int_0^1f^4(x)dx\right) \left( \int_0^1 f^2(x) dx\right)}$$  Any hints/solutions are appreciated.
1567
On , one can see that   $$\|fg\|_r\leq \|f\|_p\|g\|_q,$$ when $$\frac{1}{r} = \frac{1}{p} + \frac{1}{q}, \ p,q\in(0,\infty].$$  I have two questions,   Is the statement true? If so, is it stated in some references? I need it for my thesis.
46266
Let $f\in C[a,b]$. Assume that $\min_{[a,b]}f=m&gt;0$ and $M=\max_{[a,b]}f$. Which one is true?  a. $$\frac{1}{M}\int_a^bf(x)dx+m\int_a^b\frac{1}{f(x)}dx\geq 2\sqrt{\frac{m}{M}}(b-a)$$  b. $$\int_a^bf(x)dx\int\frac{1}{f(x)}dx\geq (b-a)^2$$  c. $$\int f(x)dx\int_a^b\frac{1}{f(x)}dx\leq (b-a)^2$$  For the first one  $$\frac{1}{M}\int _a^bf(x)+m\int _a^b\frac{1}{f(x)}dx\geq {\frac{m}{M}}(b-a)+{\frac{m}{M}}(b-a)$$  and the last two I am not getting. How to find these?
1466
Could someone take me through the steps of decomposing  $$\frac{2x^2+11x}{x^2+11x+30}$$ into partial fractions?   More generally, how does one use partial fractions to compute integrals $$\int\frac{P(x)}{Q(x)}\,dx$$ of rational functions ($P(x)$ and $Q(x)$ are polynomials) ?    This question is asked in an effort to cut down on duplicates. See .  Also see .
49255
I read somewhere that   $$ \int_{0}^{\infty} \frac{x^3}{e^{x} - 1} = \frac{\pi^4}{15}$$  Does anyone see a way to prove this? My first idea was doing a contour integration and use the residue theorem, but that seems to be a lot of work so if anyone has a better idea I'd be glad to hear it.
31686
Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated.
111075
Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$
85118
If $x^4+ax^3+3x^2+bx+1 \ge 0$ for all real $x$ where $a,b \in R$. Find the maximum value of $(a^2+b^2)$.                                                                        I tried setting up inequalities in $a$ and $b$ but in the end had a hideous two variable expression whose maxima I had to calculate using Partial Derivatives. There must be a better way of doing it. Thanks!
5670
I'm stuck on this problem, and I dont know how to continue from here. It would be great if someone could just give me a hint of how to proceed from here.  The problem is,     "Let $X,Y,Z$ be independent and $N(1,1)$. Find $P(X+Y \geqq 3Z)$."   Now, the way I see it there are two obvious ways to proceed. First off, one can simply condition on the different variables and because of their independence obtain  $P(X+Y \geqq 3Z) = \displaystyle \int_{-\infty}^{\infty}P(X+Y \geqq 3Z | Z=z)f_Z(z)dz = \int_{-\infty}^{\infty}P(X+Y \geqq 3z)f_Z(z)dz $   $ (1)$  and by doing the exact same reasoning we get $P(X+Y \geqq 3z) = \int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dy$  and thus (with (1))  $P(X+Y \geqq 3Z) = \int_{-\infty}^{\infty}f_Z(z)\int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dydz$.  This is problematic however, since $X\sim N(1,1)$ means that we cannot express $F_X(3z-y)$ explicitly. If the variables would have had some other distribution with known cdf, the calculation could continue in a straight forward way from here. This is not the case right now, however. I did, however, try to use the fact that $F_X(3z-y) = \int_{0}^{3z-y}f_X(t)dt$. But then I couldnt make sense of the integration limits. So I doubt that this is the right way to go in solving this particular problem.   I was also thinking about using the fact that $X+Y =: W$ then $W\sim N(1+1,1+1)$ and would simplify the calculation above but yield the same problem. Maybe there is a nice trick one can use in calculating the integral above. Either way, I would really appreciate if someone could give me some hints as to how to continue from here.  Thanks!
30836
Calculate   $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$   In this problem, I'm unable to understand how to start.  I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use. Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.
41820
Consider a polynomial $(\alpha ,\beta &gt;0)$,$f(x)=x^3/\alpha+x^2/\beta+x/\gamma+1/\delta$.  If $|f(x)|\leq 1$ for $|x|\leq 1$ then $\max{(1/\alpha+1/\beta+|1/\gamma|+|1/\delta|)}$ is what?  Intuitively it seems that the value should be within the range $1$ to $10$....just plugged in random values.But I'm not able to solve it....
80232
Find all the primes $p$ so that $$x^3+x \equiv 1 \pmod p \tag{1}$$ has integer solutions.   We consider $x$ and $y$ are the same solutions iff $x\equiv y \pmod p.$  We can prove that $(1)$ cannot have exactly $2$ solutions unless $p=31$. When $p=31$, the solutions of $(1)$ are $x \equiv {17\text{ (double), } 28} \pmod {31}.$  My question is: when does $(1)$ have no solution, and when does $(1)$ have $1$ solution and when does $(1)$ have $3$ solutions? (As Ma Ming has pointed out, an equation of degree $3$ has no more than $3$ solutions.)  I have proved that if $p \not = 31$, and $a$ is a solution of $(1)$, then $(1)$ has $3$ solutions iff  $$\left(\frac{a-1}{p} \right)=\left(\frac{a+3}{p} \right),$$ here $\left( \frac{a}{p} \right)$ is the Jacobi symbol. Thanks in advance!
79935
The question is as in the title: what's the minimum of $\int_0^1 f(x)^2 \: dx$, subject to $\int_0^1 f(x) \: dx = 0, \int_0^1 x f(x) \: dx = 1$? (Assume suitable smoothness conditions.)  A problem in the textbook for the course I am TEACHING (not taking) reduces to minimizing $w_1^2 + w_2^2 + w_3^2$ subject to $w_1 + w_2 + w_3 = 0, w_1 + 2w_2 + 3w_3 = 1$. Of course there's nothing special about the number $3$ here, and so one can ask for the minimum of $\sum_{i=1}^n w_i^2$ subject to $\sum_{i=1}^n w_i = 0, \sum_{i=1}^n iw_i = 1$. At least when $n = 3, 4, 5$, we get $w_i = c_n(i-(n+1)/2)$, for some constant $c_n$ which depends on $n$. So for fixed $n$, $w_i$ is a linear function of $i$. (This is a bit of an annoying computation, so I won't reproduce it here.)  So it seems like there should be a continuous analogue of this. If we have  $$ \int_0^1 f(x) \: dx = 0, \int_0^1 x f(x) \: dx = 1 $$ and $f(x)$ is linear, then we get $f(x) = 12(x-1/2)$, and $\int_0^1 f(x)^2 \: dx = 12$.  Is this the function satisfying these integral conditions with smallest $\int_0^1 f(x)^2 \: dx$? That is, is it the case that $$ \int_0^1 f(x)^2 \: dx \ge 12 $$ for every $f(x)$ satisfying the two conditions above and whatever smoothness conditions are necessary?  I've tagged this calculus-of-variations because that's what it looks like to me. But I don't know the calculus of variations, which is why I can't just solve the problem myself.
21345
Find the maximum value of $$(1-x)(2-y)(3-z)(x+\frac y2+\frac z3)$$ where $$x&lt;1,y&lt;2,z&lt;3,x+\frac y2+\frac z3&gt;0$$  I tried several cases to find a certain product. I tried 7 such cases. Three of them yield negative values while 4 gave postive fractional values. I chose the highest value which is 5/3. I already gave up and looked at the answer at the back of the questionnaire. The answer is $$\frac {243}{128}$$  Can anyone help me in answering this question?
62130
I am trying to prove Hardy's Inequality for integrals:     Let $f:(0,\infty) \rightarrow \mathbb R$ be in $L^p$. Let $F(x) = \frac{1}{x} \int_0 ^x f(t)dt$. Then $F \in L^p$ and $\|F\|_p \leq \frac{p}{p-1} \|f\|_p$.    I have seen proofs of this which use Haar measure and Fourier analysis, eg. here . However, I am wondering if there is a more elementary proof.   I have tried to deduce this using Jensen's inequality and Fubini's Theorem as follows:  Jensen's inequality implies that $\left|\frac{1}{x} \int_0 ^x f(t)dt)\right|^p \leq \frac{1}{x} \int_0 ^x |f(t)|^p dt$. Therefore   $$\int_0 ^\infty |F(x)|^p dx \leq \int_0 ^\infty \frac{1}{x} \int_0 ^x |f(t)|^p dtdx = \int_0 ^\infty \int_t ^\infty \frac{1}{x} |f(t)|^p dx dt.$$  But, this last integral is infinite, and I'm not sure how to get a sharper bound on $|F(x)|^p$. Any suggestions?
35672
Let $f$ be a continuously differentiable real valued function on $[0,1]$.  It is given that $\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}f(x) dx=0$     Find the minimum value of $\dfrac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_{0}^{1} f(x) dx \right)^2}$   I tried to use Cauchy-Schwarz to show that  $$\frac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_0^1 f(x) dx \right)^2} \ge \frac{\left( \int_0^1 \bigl| f(x)f'(x) \bigr| dx \right)^2}{ \left( \int_0^1 f(x) dx \right)^2} \ge \frac{ f(1)^2 - f(0)^2}{2 \left( \int_0^1 f^2 (x) dx \right)^2}$$ But I can't proceed from here.  Also, I don't know how to use the condition  $\int_{1/3}^{2/3}f(x) dx=0$
30543
Evaluate $$\int_0^1 f(x) dx$$ where  $$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$  I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to  $$\sin^2y\cdot \ln (\sin y) dy$$  Then I used the property of definite integrals that  $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$  Then too it wasn't getting simplified. I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?