Let $f'(x)={{192x^3}\over {2+\sin^4(\pi x)}}$ for all real $x$ and $f(1/2)=0$, $m\leq \int_{1/2}^1f(x)\leq M$. Find m and M 
Let $$f'(x)={{192x^3}\over {2+\sin^4(\pi x)}}$$ for all real $x$ and $f(1/2)=0$
  
  If $$m\leq \int_{1/2}^1f(x)\leq M$$ then find $m$ and $M$

I am going to post a solution now. Please give a different nice solution. Please do comment on the solution..
 A: Try the following for a better bound:
$$\frac{192x^3}3 \leqslant \frac{192x^3}{2+\sin^4(\pi x)} \leqslant \frac{192x^3}2$$
$$\implies \int_{\frac12}^x\frac{192x^3}3 dx \leqslant \int_{\frac12}^x\frac{192x^3}{2+\sin^4(\pi x)} dx \leqslant \int_{\frac12}^x\frac{192x^3}2 dx$$
$$\implies 16x^4-1 \leqslant f(x) \leqslant 24x^4-\frac32$$
$$\implies \int_{\frac12}^1 (16x^4-1)dx \leqslant  \int_{\frac12}^1 f(x) dx \leqslant  \int_{\frac12}^1 (24x^4-\frac32)dx$$
$$\implies \frac{13}5 \leqslant  \int_{\frac12}^1 f(x) dx \leqslant  \frac{39}{10}$$
A: I saw a solution. We are given only $f'(x)$ and the limits $1/2$ and $1$. We have to bound something. See how $f'(x)$ is varying. It is continuously increasing because the numerator is increasing and denominator is decreasing($\sin$ is decreasing).
$$f'(1/2)\leq f'(x) \leq f'(1)$$
$$8\leq f'(x)\leq 96$$
Using Fundamental Theorem of Calculus
$$\int_{1/2}^x8\leq \int_{1/2}^xf'(x)\leq \int_{1/2}^x96$$
$$8x-4\leq f(x)\leq 96x-48$$
Again using the theorem,
$$\int_{1/2}^18x-4\leq \int_{1/2}^1f(x)\leq \int_{1/2}^196x-48$$
$$1\leq \int_{1/2}^1f(x)\leq 12$$
A: The following approach provides a GREAT improvement in bounds:
$$
f'(x) = \dfrac{192 x^3}{2+\sin^4(\pi x)} \quad\wedge\quad f(0.5) = 0 \quad\quad\Rightarrow\quad\quad f(x) = \int\limits_{0.5}^{x} \dfrac{192 x^3}{2+\sin^4(\pi x)} dx
$$
With some math magic:
$$
f(x) = \int\limits_{0.5}^{x} \dfrac{192 x^3}{2+(\pi x)^4\mbox{sinc}^4(\pi x)} dx
$$
You will see that the function $\mbox{sinc}(\pi x)$ remains almost linear in the range $[0.5,1]$, and it's a convex function in that interval. So, you can define a function $s(x)$ passing through the endpoints of the interval and ensure that the following equation is true:
$$ 
\begin{align}
s(x) &= \dfrac{\mbox{sinc}(\pi 0.5) - \mbox{sinc}(\pi 1)}{0.5 - 1} \cdot (x - 1) + \mbox{sinc}(\pi 1) \\
s(x) &= 1.27324 - 1.27324 x \\
\end{align}
$$
$$ \therefore \forall x\in[0.5,1]: \quad \mbox{sinc}(\pi x) < s(x) $$
By the mean value theorem, you can get the lower bound because the function is continuously decreasing. You can estimate their displacement $x_0$ with:
$$
\left.\dfrac{d}{dx} \mbox{sinc}(\pi x) \right|_{x_0} = \left. \dfrac{d}{dx} s(x) \right|_{x_0}
\quad\Rightarrow\quad \dfrac{\pi x_0 \cos(\pi x_0) - \sin(\pi x_0)}{(\pi x_0^2)} = - 1.27324
$$
$$\therefore\quad x_0  \approx 0.8296 \quad\Rightarrow\quad s(x) +\mbox{sinc}(\pi x_0) - s(x_0) < \mbox{sinc}(\pi x) $$
Taking a margin of error to ensure boundary condition:
$$
\mbox{sinc}(\pi x_0) - s(x_0) = -0.02123... \approx -0.022 \quad\Rightarrow\quad S(x) = s(x) -0.022
$$
$$S(x) = 1.25124 - 1.27324 x$$
$$ \therefore \forall x\in[0.5,1]: \quad S(x) < \mbox{sinc}(\pi x) $$
You can now replace the function $\mbox{sinc}(\pi x)$ by $s(x)$ and $S(x)$, without violating the bounds:
$$
\forall x\in[0.5,1]:\quad
\int\limits_{0.5}^{x} \dfrac{192 x^3}{2+(\pi x)^4 s(x)^4} dx < f(x) < \int\limits_{0.5}^{x} \dfrac{192 x^3}{2+(\pi x)^4 S(x)^4} dx
$$
And from here on, Copperfield!:
$$
\quad \int\limits_{0.5}^{1} \int\limits_{0.5}^{x} \dfrac{192 x^3}{2+(\pi x)^4 s(x)^4} dx dy < \int\limits_{0.5}^{1} f(x) dx < \int\limits_{0.5}^{1} \int\limits_{0.5}^{x} \dfrac{192 x^3}{2+(\pi x)^4 S(x)^4} dx dy
$$
Each integral can be solved by partial fractions, obtaining:
$$3.26715... <  \int\limits_{0.5}^{x} f(x) dx < 3.36639...$$
