Problem 24 - Calculus

Question

Find \(G^{\prime}(x).\) $$ G(x)=\int_{1}^{x^{2}+x} \sqrt{2 z+\sin z} d z $$

Step-by-Step Solution

Answer

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

Step 1 - Identify the Function

We begin with identifying the given function: \[ G(x) = \int_{1}^{x^2 + x} \sqrt{2z + \sin z} \, dz \]This is an integral function where the limits depend on the variable \(x\).

Step 2 - Apply the Fundamental Theorem

To find the derivative \(G'(x)\), we apply the Fundamental Theorem of Calculus Part 2, which states that if \(F(t)\) is an antiderivative of \(f(t)\), then:\[\frac{d}{dx} \left( \int_{a}^{u(x)} f(t) \, dt \right) = f(u(x)) \cdot \frac{du}{dx}.\] For our problem, \(f(z) = \sqrt{2z + \sin z}\) and \(u(x) = x^2 + x\).

Step 3 - Differentiate the Upper Limit

Calculate the derivative of the upper limit \(u(x) = x^2 + x\):\[ \frac{du}{dx} = \frac{d}{dx}(x^2 + x) = 2x + 1. \]

Step 4 - Use the Fundamental Theorem Result

Substitute \(u(x)\) and \(\frac{du}{dx}\) back into the theorem:\[ G'(x) = \sqrt{2(x^2 + x) + \sin(x^2 + x)} \cdot (2x + 1). \]

Step 5 - Simplify the Expression

Finally, express the derivative:\[ G'(x) = (2x + 1) \cdot \sqrt{2(x^2 + x) + \sin(x^2 + x)}. \]

Key Concepts

Understanding Integral Functions

Integral functions are an essential component in calculus and play an important role in the Fundamental Theorem of Calculus. These functions are defined as an integral with variable limits of integration, particularly where one or both limits are expressed in terms of another variable.
In the exercise, the function is defined as \[ G(x) = \int_{1}^{x^2 + x} \sqrt{2z + \sin z} \, dz \]This expression tells us that we are dealing with an integral function since the upper limit is a function of the variable \(x\).
Important points to remember about integral functions include:

• The integral function represents the accumulation of the area under the curve of the integrand as the limits change.
• The variable of integration (here \(z\)) is a different variable from the given variable \(x\), making it a definite integral with a specific focus on the interval \([1, x^2+x]\).

Understanding how the variable upper limit produces different outputs based on \(x\) helps us see the connection between integration and differentiation.

Derivatives and Their Function

A derivative represents the rate at which a function is changing at any point. It is central to calculus, and understanding derivatives is key to applying many calculus concepts.
For a function \(f(x)\), the derivative, noted as \(f'(x)\), measures how \(f\) changes as \(x\) shifts. In the context of our exercise, we need to find the derivative of the integral function \(G(x)\).
Some basic yet critical points about derivatives:

• The process of finding a derivative is called differentiation.
• For complex functions, differentiation rules like the chain rule become crucial.
• Converting the integral form of a function to its derivative helps to better interpret behavior and trends of functions.

Being able to differentiate an integral function allows us to predict and quantify changes that depend deeply on another variable, as demonstrated in this exercise.

Differentiating the Upper Limit

Differentiating with an upper limit is a critical application of the Fundamental Theorem of Calculus. Here, we deal with integrals where the upper bounds are not constants but functions of another variable, often \(x\).
When faced with a function like \[ G(x) = \int_{1}^{x^2+x} \sqrt{2z + \sin z} \, dz \]we use the second part of the Fundamental Theorem of Calculus. This theorem provides that if you have an integral with a variable upper limit \(u(x)\), the derivative is calculated by multiplying the derivative of the integrand evaluated at \(u(x)\) by \(u'(x)\): \[ \frac{d}{dx} \left( \int_{a}^{u(x)} f(t) \, dt \right) = f(u(x)) \cdot \frac{du}{dx} \]In our exercise, \(u(x) = x^2 + x\) and \( \frac{du}{dx} = 2x + 1 \).
This method allows us to handle functions where the upper limit varies, giving precise control over calculations in complex scenarios.