Problem 72 - Calculus Early Transcendentals

Question

Without evaluating integrals, prove that $$ \int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x=\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x $$

Step-by-Step Solution

Answer

Short Answer: To prove that the given integrals are equal without evaluating them, we found the derivatives of the functions within the integrals and used the property of integration stating that the derivative of an integral is equal to the function itself. We found that the derivatives of the functions (12sin(πx^²) and x¹⁰(2-x)³) inside the integrals are equal, so the integrals must be equal up to a constant.

Step 1: Find the derivatives of the functions

Let's find the derivative of the function inside the first integral. Function: \(12\sin(\pi x^2)\) We can use the chain rule, so we have: \( \frac{d}{dx}(12 \sin(\pi x^2)) = 12 \frac{d}{dx}\sin(\pi x^2) \) Since the derivative of \(\sin(x)\) is \(\cos(x)\), we have: \( 12 \frac{d}{dx}\sin(\pi x^2) = 12 \cos(\pi x^2) \frac{d}{dx}(\pi x^2) \) Now, taking the derivative of \(\pi x^2\): \( \frac{d}{dx}(\pi x^2) = 2\pi x \) Now multiply them together: \( 12 \cos(\pi x^2) (2\pi x) = 24\pi x \cos(\pi x^2) \) Now, let's find the derivative of the function inside the second integral. Function: \(x^{10}(2-x)^3\) We can use the product rule, so we have: \( \frac{d}{dx}( x^{10}(2-x)^3) = x^{10}\frac{d}{dx} (2-x)^3 + (2-x)^3\frac{d}{dx} x^{10}\) Now, taking the derivative of \((2-x)^3\) and \(x^{10}\): \( \frac{d}{dx} (2-x)^3 = -3(2-x)^2 \) \( \frac{d}{dx} x^{10} = 10x^9 \) Now multiply and add them together: \(x^{10}(-3(2-x)^2) + (2-x)^3(10x^9) = -3x^{10}(2-x)^2 + 10x^9(2-x)^3 \)

Step 2: Analyze the integrals of the derivatives

We have two integrals: 1. \(\int_{0}^{2} \frac{d}{d x}(12 \sin \pi x^{2}) d x = \int_{0}^{2} 24\pi x \cos(\pi x^2) d x\) 2. \(\int_{0}^{2} \frac{d}{d x}(x^{10}(2-x)^{3}) d x = \int_{0}^{2} (-3x^{10}(2-x)^2 + 10x^9(2-x)^3) d x\) To prove that the two integrals are equal without evaluating them, we can use the property of integrals: If two functions, when integrated over the same interval, have the same derivative, then their integrals are equal up to a constant.

Step 3: Use the property of integrals to prove the equality

We know that the derivative of the first integral is equal to the function inside the second integral, and vice versa. Based on the property of integrals stated in Step 2, since the derivatives of both functions are equal, the given integrals are equal up to a constant. Therefore, we can conclude that: \( \int_{0}^{2} \frac{d}{d x}(12 \sin \pi x^{2}) d x = \int_{0}^{2} \frac{d}{d x}(x^{10}(2-x)^{3}) d x \)

Key Concepts

Chain Rule

When you come across a function that is composed of another function, the chain rule will be your best friend to find the derivative. In the problem, we needed to find the derivative of the function \(12 \sin(\pi x^2)\). This function is a composition: \(\sin(g(x))\), with \(g(x) = \pi x^2\). The chain rule tells us that to find \(\frac{d}{dx} f(g(x))\), you multiply the derivative of the outer function \(f\) evaluated at the inner function \(g(x)\), by the derivative of the inner function \(g(x)\).

For our function, this meant taking the derivative of \(\sin(\pi x^2)\), which is \(\cos(\pi x^2)\), and then multiplying it by the derivative of \(\pi x^2\), which is \(2\pi x\). So using the chain rule:

• The outer derivative is \(12 \cdot \cos(\pi x^2)\)
• The inner derivative is \(2 \pi x\)

Putting them together gives: \(24\pi x\cos(\pi x^2)\), which is the derivative of our original function using the chain rule.

Product Rule

The product rule is crucial when you are dealing with the derivative of a product of two functions. In the exercise, we needed to differentiate \(x^{10}(2-x)^3\). Here, we have two separate functions multiplying each other: \(u(x) = x^{10}\) and \(v(x) = (2-x)^3\). The product rule states that the derivative of a product \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\).

We applied this rule as follows:

• The derivative of \(x^{10}\), denoted \(u'(x)\), is \(10x^9\).
• The derivative of \((2-x)^3\), denoted \(v'(x)\), is \(-3(2-x)^2\).

Putting these in the product rule results in: \[ x^{10}(-3(2-x)^2) + (2-x)^3(10x^9) = -3x^{10}(2-x)^2 + 10x^9(2-x)^3 \], which is the derivative of the product of these two functions.

Derivative

Derivatives are a fundamental tool in calculus, representing the rate at which a function is changing at any given point. They tell us about the slope of the tangent line to the function's graph at any point. In our exercise, we focused on finding the derivatives of two complex functions: \(12 \sin(\pi x^2)\) and \(x^{10}(2-x)^3\).

To perform these derivative calculations efficiently, we utilized two vital rules in calculus:

• Chain Rule: Useful when dealing with composite functions like \(12 \sin(\pi x^2)\).
• Product Rule: Necessary when differentiating products of functions, as seen with \(x^{10}(2-x)^3\).

Understanding how derivatives work and applying these rules allows us to analyze the behavior of functions for various mathematical applications, such as improving accuracy in calculations and predictions in real-world scenarios.

Integral Equality

Integral equality in calculus states that if two functions have the same derivative over an interval, their integrals over that interval are equal, up to a constant. This concept was key in our problem, where we needed to prove the equality between two integrals without evaluating them.

The functions under the integrals \( \int_{0}^{2} \frac{d}{d x}(12 \sin \pi x^{2}) d x \) and \( \int_{0}^{2} \frac{d}{d x}(x^{10}(2-x)^{3}) d x \) were derived to give us \(24\pi x \cos(\pi x^2)\) and \(-3x^{10}(2-x)^2 + 10x^9(2-x)^3\) respectively. Despite appearing different, the process of taking derivatives transforms these functions into expressions that are equivalent over any continuous interval from 0 to 2.

Therefore, integral equality lets us assert that these integrals are equal, showing the deep connection between derivatives and integrals, often referred to as the "Fundamental Theorem of Calculus." This theorem demonstrates the reversibility of integration and differentiation and is foundational to calculus.