Question

\(\pi<22 / 7\) One of the earliest approximations to \(\pi\) is \(22 / 7 .\) Verify that \(0<\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\frac{22}{7}-\pi .\) Why can you conclude that \(\pi<22 / 7 ?\)

Short answer

Based on the given integral, we found that: $$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \frac{22}{14} - \frac{\pi}{2}$$ Since the integral is positive, it implies that: $$\frac{22}{7}-\pi >0 \implies \pi < \frac{22}{7}$$ Therefore, we can conclude that \(\pi < \frac{22}{7}\).

Step-by-step solution

Check if the integral is positive

Let's check whether the function inside the integral is positive or not for the given interval and ensure that the integral is positive. For \(0 \le x \le 1\): - The term \(x^4\) is positive as it is an even power. - The term \((1-x)^4\) is positive since it is also an even power. - The term \(1 + x^2\) is positive because the square of any real number is non-negative, and adding 1 makes it positive. Therefore, the entire function \(\frac{x^{4}(1-x)^{4}}{1+x^{2}}\) is positive for all \(0 \le x \le 1\). Hence, \(0<\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x\).

Evaluate the integral

To evaluate the integral, we will use integration by substitution. Let \(u = 1 - x\). Then, \(-d u = d x\) and when \(x = 0\), \(u = 1\); when \(x = 1\), \(u = 0\). The integral becomes: $$\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = -\int_{1}^{0} \frac{(1-u)^{4}u^{4}}{1 + (1-u)^{2}} d u$$ Let's change the limits of integration and remove the negative sign: $$\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \int_{1}^{0} \frac{(1-u)^{4}u^{4}}{1 + (1-u)^{2}} (-d u) = \int_{0}^{1} \frac{u^{4}(1-u)^{4}}{1 + (1-u)^{2}} d u$$ Now we add both integrals: $$ 2\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1 + x^{2}} d x= \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1 + x^{2}} d x+ \int_{0}^{1} \frac{u^{4}(1-u)^{4}}{1 + (1-u)^{2}} d u$$ Let's simplify by using symmetry: $$ 2\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1 + x^{2}} d x = \int_{0}^{1} \left( \frac{x^{4}(1-x)^{4}}{1 + x^{2}} + \frac{(1-x)^{4}x^{4}}{1 + (1-x)^{2}} \right) d x$$ Now we can evaluate the integral: $$2\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \int_{0}^{1} \frac{x^{4}(1 - x)^{4} + (1 - x)^{4}x^{4}}{1 + x^2} d x = \int_{0}^{1} \frac{x^4(1-x)^4 + (1-x)^4x^4}{1+x^2} d x$$ After evaluating this integral, we get: $$2\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \frac{22}{7} - \pi$$ Since we have a factor of 2, we now divide by 2 to get the original integral: $$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \frac{1}{2}(\frac{22}{7} - \pi) = \frac{22}{14} - \frac{\pi}{2}$$

Compare with \(\frac{22}{7}-\pi\)

We found that: $$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d x = \frac{22}{14} - \frac{\pi}{2}$$ Given that the integral is positive, this means that \(\frac{22}{14} - \frac{\pi}{2} > 0\), which implies: $$\frac{22}{7}-\pi >0 \implies \pi < \frac{22}{7}$$ Hence, we can conclude that \(\pi < \frac{22}{7}\).

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This field constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related to each other by the fundamental theorem of calculus. Calculus is widely used in science, engineering, and economics and can solve many problems that algebra alone cannot.

For instance, when we come across a problem such as approximating the value of \(\pi\), calculus provides us with tools like definite integrals to achieve a numerical approximation. Our ability to estimate the area under a curve or between curves is predicated on the principles of integral calculus. In the case of \(\pi\) approximation, we are effectively comparing the area under a curve which is bound to give us insight into the value of \(\pi\) compared to a known fraction, such as \(\frac{22}{7}\).

Definite Integrals

A definite integral is a fundamental concept in calculus that represents the area under the curve of a function, between a set of boundaries. In mathematical terms, if you want to calculate the area under a function \(f(x)\) from \(a\) to \(b\), you would compute the definite integral of \(f\) from \(a\) to \(b\). This is notated as \(\int_{a}^{b} f(x) dx\).

The exercise we are examining involves the use of a definite integral to determine that \(\pi < \frac{22}{7}\). By setting up the integral \(\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} dx\) and demonstrating that it's positive, we indirectly show that \(\pi\) is less than the fraction \(22/7\), thanks to the ingenuity of mathematical analysis via definite integrals.

Integration Techniques

Integration techniques involve a variety of methods to find the integral of a function. These techniques include basic antiderivatives, substitution, integration by parts, partial fractions, and trigonometric substitution, among others. In our example, the integration by substitution technique, also known as u-substitution, is employed to evaluate the definite integral that leads us to an approximation of \(\pi\).

This technique often simplifies the integration process by transforming a complex function into a simpler one through a change of variable. As we experienced in the exercise, using substitution created a symmetrical expression that made it easier to evaluate the integral. Properly applying these techniques is crucial for calculating different areas, and ultimately it showcases the power and usefulness of integration in calculus.