Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(G^{\prime}(x)\). $$ G(x)=\int_{0}^{x}\left(2 t^{2}+\sqrt{t}\right) d t $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the derivative of the function \( G(x) = \int_{0}^{x}(2t^2 + \sqrt{t}) \, dt \). This is a definite integral with variable upper limit \( x \).

Step 2: Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(t) \) is an antiderivative of \( f(t) \), then the derivative of the integral from 0 to \( x \) is \( f(x) \). Mathematically, if \( G(x) = \int_{a}^{x} f(t) \, dt \), then \( G'(x) = f(x) \).

Step 3: Apply the Fundamental Theorem

In our case, \( f(t) = 2t^2 + \sqrt{t} \). Therefore, using the theorem, the derivative is simply the integrand evaluated at \( x \), which is \( G'(x) = 2x^2 + \sqrt{x} \).

Key Concepts

Definite Integrals

Definite integrals are a foundational concept in calculus that relate to finding the net area under a curve between two specific points. When calculating a definite integral, you determine the total change or accumulated quantity over that interval.
In our exercise, the definite integral is represented by \[\int_{0}^{x}(2t^2 + \sqrt{t}) \, dt \]. The limit from \(0\) to \(x\) signifies that we are interested in the area under the curve of the function \(2t^2 + \sqrt{t}\), starting from \(t = 0\) to \(t = x\). This is a specific requirement because the upper limit is not a constant but a variable \(x\).
This type of integral with a variable limit is a powerful tool because it allows us to analyze how the accumulated area or total quantity changes as the variable \(x\) changes. It connects to the idea of accumulation with respect to a moving boundary, which is essential for applications in physics and other sciences.

Derivatives

The concept of a derivative is core to calculus. Derivatives represent the rate of change of a function. In simpler terms, it tells us how a function changes as its input changes.
In the context of the problem, we seek the derivative of the function \(G(x)\), which is defined by the integral of \(2t^2 + \sqrt{t}\) from \(0\) to \(x\). According to the Fundamental Theorem of Calculus, the derivative \(G'(x)\) is directly linked to the function within the integral.
When finding \(G'(x)\), you evaluate the integrand, \(2t^2 + \sqrt{t}\), at \(t = x\). Therefore, the derivative \(G'(x) = 2x^2 + \sqrt{x}\). This process shows how derivatives provide insight into the instantaneous rate of change, enabling us to understand the behavior and trends of the function \(G(x)\) as \(x\) varies.

Calculus

Calculus is the mathematical study of change. It is broadly divided into differential calculus and integral calculus. Each plays a critical role in understanding and describing real-world phenomena.
Differential calculus focuses on the concept of the derivative, looking at rates of change and slopes of curves. Integral calculus, on the other hand, deals with taking integrals to find areas under curves, which represent accumulated quantities or total change over an interval.
In our exercise, both aspects of calculus are utilized. We deal with a definite integral, allowing us to calculate the accumulated area from \(0\) to \(x\) under the curve of \(2t^2 + \sqrt{t}\). Furthermore, by employing the Fundamental Theorem of Calculus, we leverage differential calculus to find how this accumulation changes instantaneously as \(x\) varies, by evaluating the derivative \(G'(x)\).
Overall, calculus provides us with the tools to understand and model dynamic systems, predict behavior and outcomes, and solve complex problems in diverse fields including science, engineering, and economics.