- Evaluate \[ \int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx. \]
- Prove that \[ 0<\frac{22}{7}-\pi<\frac1{256}. \]
Solution
1. One has the identity\[ \frac{x^4(1-x)^4}{1+x^2} = x^6-4x^5+5x^4-4x^2+4 - \frac4{1+x^2}. \]
Multiplying the right-hand side by \(1+x^2\) gives \[ \begin{aligned} &(1+x^2) \left( x^6-4x^5+5x^4-4x^2+4 \right) -4\\ &\qquad= x^8-4x^7+6x^6-4x^5+x^4\\ &\qquad= x^4(1-x)^4. \end{aligned} \] Divide by \(1+x^2\).
2. The polynomial part integrates to\[ \int_0^1 \left( x^6-4x^5+5x^4-4x^2+4 \right)\,dx = \frac{22}{7}. \]
Termwise integration gives \[ \begin{aligned} &\frac17 - \frac46 + \frac55 - \frac43 + 4\\ &\qquad= \frac17-\frac23+1-\frac43+4\\ &\qquad= \frac17+3 = \frac{22}{7}. \end{aligned} \]
3. One has\[ 4\int_0^1\frac{dx}{1+x^2} = \pi. \]
Since \[ \frac{d}{dx}\arctan x = \frac1{1+x^2}, \] one obtains \[ 4\int_0^1\frac{dx}{1+x^2} = 4\left(\arctan1-\arctan0\right) = 4\cdot\frac\pi4 = \pi. \]
4. Therefore\[ \boxed{ \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx = \frac{22}{7}-\pi }. \]
3. Thisproves part (a).
5. For every \(x\in[0,1]\),\[ 0 \leq x(1-x) \leq \frac14. \]
Complete the square: \[ x(1-x) = \frac14 - \left(x-\frac12\right)^2. \] Nonnegativity follows from \(0\leq x\leq1\).
6. For every \(x\in[0,1]\),\[ 0 \leq \frac{x^4(1-x)^4}{1+x^2} < \frac1{256}, \]and the integrand is positive for \(0<x<1\).
By step 5, \[ x^4(1-x)^4 = \bigl(x(1-x)\bigr)^4 \leq \frac1{4^4} = \frac1{256}. \] Also \[ 1+x^2\geq1. \] The upper inequality is strict: equality in the numerator bound requires \(x=1/2\), where \(1+x^2>1\); at the points where the denominator equals \(1\), the numerator is \(0\). Positivity on \((0,1)\) is immediate.
7. Consequently,\[ \boxed{ 0 < \frac{22}{7}-\pi < \frac1{256} }. \]
By step 4, \[ \frac{22}{7}-\pi \] is the integral of the function in step 6 over an interval of length \(1\). Its strict positivity gives the left inequality, while its strict pointwise upper bound gives \[ \frac{22}{7}-\pi < \int_0^1\frac1{256}\,dx = \frac1{256}. \] This proves part (b).
8. Q.E.D.
Step 4 proves part (a), and step 7 proves part (b).