Question

Group Activity Use the Max-Min Inequality to find upper and lower bounds for the value of \(\int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { 4 } } d x\)

Short answer

The value of the integral lies between 1/2 and 1.

Step-by-step solution

Determining the Minimum and Maximum Values

The function is \(\frac { 1 } { 1 + x ^ { 4 } }\). For \(x\in[0,1]\), \(x^{4}\) will always be greater than or equal to 0 and less than or equal to 1. Therefore, \(1 + x ^ { 4 }\) lies in the interval \([1,2]\). The reciprocal of a number increases as the number decreases. Therefore, \(\frac { 1 } { 1 + x ^ { 4 } }\) attains its maximum value of 1 when \(x = 0\) and its minimum value of \(\frac { 1 } { 2 }\) when \(x = 1\).

Applying the Max-Min Inequality

According to the Max-Min Inequality, for any function \(f(x)\) continuous on [a, b], if \(m \leq f(x) \leq M\) for \(x\in[a, b]\), then the integral \(\int _ { a } ^ { b } f(x) d x\) lies between \(m(b-a)\) and \(M(b-a)\). Here, \(m = \frac { 1 } { 2 }\), \(M = 1\), \(a = 0\), and \(b = 1\). So, \(\frac { 1 } { 2 } (1 - 0) \leq \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { 4 } } d x \leq 1 (1 - 0)\).

Final Simplification

The integral can then be simplified as \(\frac { 1 } { 2 } \leq \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { 4 } } d x \leq 1\), which means the integral is bounded between 1/2 to 1.