Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-26, find \(G^{\prime}(x)\). $$ G(x)=\int_{1}^{x^{2}+x} \sqrt{2 z+\sin z} d z $$

Step-by-Step Solution

Answer

\(G'(x) = (2x + 1) \sqrt{2(x^2 + x) + \sin(x^2 + x)}\).

Step 1: Understand the Problem

We are given an integral function \(G(x) = \int_{1}^{x^2 + x} \sqrt{2z + \sin z} \, dz\). We need to find its derivative, \(G'(x)\).

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \(F(x) = \int_{a}^{u(x)} f(t) \, dt\), then \(F'(x) = f(u(x)) \cdot u'(x)\). Here, \(f(z) = \sqrt{2z + \sin z}\) and \(u(x) = x^2 + x\).

Step 3: Differentiate the Upper Limit Function

Find the derivative of the upper limit function \(u(x) = x^2 + x\). The derivative is \(u'(x) = 2x + 1\).

Step 4: Apply the Chain Rule

According to the Fundamental Theorem, \(G'(x) = f(u(x)) \cdot u'(x)\). Substitute \(f(z)\) and \(u'(x)\): \(G'(x) = \sqrt{2(x^2 + x) + \sin(x^2 + x)} \cdot (2x + 1)\).

Step 5: Simplify the Expression

Substitute back to get the final expression: \(G'(x) = (2x + 1) \sqrt{2(x^2 + x) + \sin(x^2 + x)}\).

Key Concepts

Derivative of Integral Function

When dealing with an integral function like \( G(x) = \int_{1}^{x^2 + x} \sqrt{2z + \sin z} \, dz \), the goal is to find the derivative \( G'(x) \). This process usually involves the Fundamental Theorem of Calculus, which provides a straightforward method to differentiate integral functions with variable upper limits.

To find the derivative, remember: the theorem states that if you have an integral of the form \( F(x) = \int_{a}^{u(x)} f(t) \, dt \), its derivative \( F'(x) \) can be expressed as \( f(u(x)) \cdot u'(x) \). This approach converts the problem of differentiating the integral into a more manageable combination of working with the integrand and the upper limit of integration.

In our original problem, the integrand is \( f(z) = \sqrt{2z + \sin z} \), and we apply this structure to solve for the derivative of the integral function.

Chain Rule in Calculus

The chain rule is an indispensable part of calculus, especially when dealing with composite functions. In the context of our exercise, it plays a vital role in differentiating the integral with respect to the variable upper limit.

To apply it effectively, recognize that the derivative of the integral function depends on not just \( f(u(x)) \) but also on the derivative of the upper limit \( u(x) \). This is where the chain rule steps in:

• Identify \( u(x) \), in this case, \( x^2 + x \).
• Then find \( u'(x) \), the derivative. Here, \( u'(x) = 2x + 1 \).

This derivative is crucial because it scales the effect of the integrand evaluated at the upper limit. Think of it as adjusting for how quickly or slowly \( u(x) \) is changing as \( x \) varies. This refinement is why the chain rule is sometimes referred to as a method for understanding how functions "chain" or link together.

Upper Limit Differentiation

Differentiating with respect to the upper limit of an integral requires attention to how the variable limit affects the entire function. Our scenario involves an upper limit defined as \( u(x) = x^2 + x \), indicating that the integral's range is directly influenced by \( x \).

When we focus on this kind of differentiation, use the following steps:

1. **Identify changes due to the upper limit's behavior:** The derivative of the whole integral depends heavily on how \( u(x) \) behaves as a function of \( x \).
2. **Combine with the Fundamental Theorem:** By pairing the changes in \( u(x) \) with the original integrand examined at this new bound, the derivative \( G'(x) = (2x + 1) \sqrt{2(x^2 + x) + \sin(x^2 + x)} \) comes into form efficiently.

This approach showcases the power of combining calculus tools: the theorem simplifies integration differentiation, while the chain rule accounts for non-contextual changes due to upper limit shifts.