Problem 48 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=x \int_{2}^{x^{2}} \sin \left(t^{3}\right) d t$$

Step-by-Step Solution

Answer

\(\frac{d y}{d x} = \int_{2}^{x^2} \sin(t^3) \, dt + 2x^2 \sin(x^6)\).

Step 1: Identify the components of the function

The function given is a product of two functions: a monomial, \(x\), and an integral, \(\int_{2}^{x^2} \sin(t^3) \, dt\). Use the product rule for differentiation, which states \(\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'\). Let \(u = x\) and \(v = \int_{2}^{x^{2}} \sin(t^3) \, dt\).

Step 2: Differentiate the first function (monomial)

For \(u = x\), the derivative \(u' = \frac{d}{dx}(x) = 1\). This derivative will be used in the product rule.

Step 3: Differentiate the integral using Leibniz's Rule

Leibniz's Rule helps differentiate an integral with variable limits. For \(v = \int_{2}^{x^2} \sin(t^3) \, dt\), first set \(g(x) = \sin(t^3)\). Using the rule: \(\frac{d}{dx}\left(\int_{a(x)}^{b(x)} g(t) \, dt \right) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)\). Since \(a(x) = 2\) is constant, \(a'(x) = 0\). Thus, \(\frac{d}{dx}(v) = \sin((x^2)^3) \cdot \frac{d}{dx}(x^2)\) which simplifies to \(\sin(x^6) \cdot 2x\).

Step 4: Apply the product rule

Substitute the derivatives obtained into the product rule: \(\frac{d}{dx} (x \cdot \int_{2}^{x^2} \sin(t^3) \, dt) = 1 \cdot \int_{2}^{x^2} \sin(t^3) \, dt + x \cdot \sin(x^6) \cdot 2x\).

Step 5: Simplify the expression

Combine and simplify the terms: \(\frac{d}{dx}(y) = \int_{2}^{x^2} \sin(t^3) \, dt + 2x^2 \sin(x^6)\).

Key Concepts

Product Rule

The Product Rule is essential in calculus for differentiating a product of two functions. It provides a way to find derivatives when dealing with expressions where two different functions are multiplied together. The rule is expressed as:\[ \frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v' \]where:

• \(u\) and \(v\) are functions of \(x\).
• \(u'\) is the derivative of \(u\).
• \(v'\) is the derivative of \(v\).

Applying this rule involves identifying the two functions within a problem. In our exercise, we defined \(u = x\) and \(v = \int_{2}^{x^2} \sin(t^3) \, dt\). The next steps involve finding the derivatives of these individual functions and then applying the rule.
It's important to remember the product rule ensures all parts of the function come together properly. Thus, both \(u'\) and \(v'\) play critical roles in evaluating the overall derivative.

Leibniz's Rule

Leibniz's Rule helps differentiate an integral where the limits themselves are functions of \(x\). This rule is vital when dealing with integral calculus, especially in problems involving variable limits. It's used when the function you are differentiating is represented as an integral with variable limits. The rule is given by:\[\frac{d}{dx}\left(\int_{a(x)}^{b(x)} g(t) \, dt \right) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)\]
Consider a problem where \(a(x)\) is constant; here, \(a'(x) = 0\), which simplifies calculations. In the given activity, we treated \(v = \int_{2}^{x^2} \sin(t^3) \, dt\). We set \(b(x) = x^2\) and kept \(a(x) = 2\) as constant. Thus:

• \(g(b(x)) = \sin((x^2)^3)\)
• \(b'(x) = \frac{d}{dx}(x^2) = 2x\)

The derivative of \(v\) becomes \(\sin(x^6) \cdot 2x\). This part is crucial for the product rule's application.
Leibniz's rule facilitates handling integrals in differentiation, making problems with complex limits much more approachable.

Integral Calculus

Integral calculus involves the process of integration, which is essentially the reverse operation of differentiation. It focuses on finding the area under curves, among other applications. Integrals can be definite or indefinite:

• Definite integrals have limits and compute a numerical value representing the area under a curve between two points.
• Indefinite integrals represent a family of functions and include a constant of integration \(C\).

The exercise focuses on a definite integral from 2 to \(x^2\), represented as \(\int_{2}^{x^2} \sin(t^3) \, dt\). This type of integral evaluates based on its boundaries and integrates within those limits.
In calculus, especially in real-world applications, evaluating such integrals equips us to solve numerous physical problems, from calculating distance to computing probabilities.
Understanding integral calculus thus provides a foundation in understanding the flow and accumulation of quantities, central to mathematical analysis and many scientific disciplines.