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Chapter 6: Problem 1

In Problems \(1-14\), find \(\frac{d y}{d x}\) \(y=\int_{0}^{x} 2 t^{2} d t\)

Short Answer

Expert verified
The derivative is \(\frac{dy}{dx} = 2x^2\).

Step by step solution

01

Recognize the Function Type

The given function \(y\) is defined as a definite integral with a variable upper limit, \(x\). This form suggests the use of the Fundamental Theorem of Calculus.
02

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \(y = \int_{a}^{x} f(t)\, dt,\) then \(\frac{dy}{dx} = f(x).\) In this problem, \(f(t) = 2t^2\). Therefore, the derivative of \(y\) with respect to \(x\) is \(f(x) = 2x^2\).
03

Write Down the Derivative

Based on Step 2, we can immediately write the derivative: \(\frac{dy}{dx} = 2x^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral, in simple terms, evaluates the area under a curve from one point to another. It requires an integrand, which is a function describing how the value changes. You also need limits of integration, which set the start and end points.

In this case, the function is described by the integrand \(2t^2\), and the limits are from \(0\) to \(x\). By integrating from \(0\) to \(x\), we can determine how the area under the curve of \(2t^2\) changes as \(x\) moves. This makes definite integrals essential for understanding how cumulative quantities change over a range.
  • Integral Limits: These are the boundaries over which you calculate the integral, such as \(0\) to \(x\).
  • Integrand: This is the function, \(2t^2\), whose area under the curve you want to find.
Derivatives
Derivatives provide a way to measure how a quantity changes instantaneously as another quantity varies. When you have a function like \(y = \int_{0}^{x} 2t^2 \, dt\), taking the derivative \(\frac{dy}{dx}\) describes how \(y\) changes as \(x\) changes.

In our exercise, the derivative simplifies the work of the definite integral, revealing the rate of change directly. Using the derivative, we get the exact rate without needing to calculate the entire area change. Here, the process involves taking the derivative of the integrand evaluated at the upper limit; this is elegantly shown by the Fundamental Theorem of Calculus.
  • Instantaneous Rate: The specific change at each point \(x\), calculated using derivatives.
  • Fundamental Theorem: It connects the process of taking integrals with derivatives, simplifying calculations.
Variable Limits of Integration
The concept of variable limits means using functions that have an upper limit that can change or vary. In our instance, the upper limit is \(x\). This implies that as \(x\) changes, so does the region we are evaluating.

When dealing with such problems, the Fundamental Theorem of Calculus becomes very valuable. It states that if you differentiate a definite integral, where the upper limit is a variable, you get back the integrand evaluated at that upper limit. So here, differentiating \(\int_{0}^{x} 2t^2 \, dt\) yields the integrand \(2x^2\), showing that our integral's effect is tracked efficiently.
  • Dynamic Limits: Variable limits change as the input variable changes, reflecting a broader or narrower scope of evaluation.
  • Practical Use: This helps model and predict circumstances where outcomes change with respect to another variable.

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