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\(7-30\) Find the first partial derivatives of the function. $$F(\alpha, \beta)=\int_{\alpha}^{\beta} \sqrt{t^{3}+1} d t$$

Short Answer

Expert verified
\( \frac{\partial F}{\partial \beta} = \sqrt{\beta^3 + 1} \) and \( \frac{\partial F}{\partial \alpha} = -\sqrt{\alpha^3 + 1} \).

Step by step solution

01

Understanding the Function

The function given is a definite integral \( F(\alpha, \beta) = \int_{\alpha}^{\beta} \sqrt{t^3 + 1} \ dt \). Our task is to find the first partial derivatives of this function with respect to \( \alpha \) and \( \beta \).
02

Applying Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that if \( F(\alpha, \beta) = \int_{\alpha}^{\beta} f(t) \ dt \), then the partial derivative of \( F \) with respect to \( \beta \) is \( \frac{\partial F}{\partial \beta} = f(\beta) \) and with respect to \( \alpha \) is \( \frac{\partial F}{\partial \alpha} = -f(\alpha) \).
03

Finding Partial Derivative with Respect to \( \beta \)

Using the Fundamental Theorem of Calculus for \( \beta \), we have: \( \frac{\partial F}{\partial \beta} = \sqrt{\beta^3 + 1} \).
04

Finding Partial Derivative with Respect to \( \alpha \)

For \( \alpha \), using the theorem, we have: \( \frac{\partial F}{\partial \alpha} = -\sqrt{\alpha^3 + 1} \). This negative sign arises because differentiation is with respect to the lower limit of integration.

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is an essential principle bridging the concepts of differentiation and integration. It essentially states that if you have an integral of a function from one point to another, the derivative of this integral with respect to the upper limit returns the integrand evaluated at that limit. This is incredibly useful because it simplifies the process of finding derivatives of integral-based functions.
The theorem comes in two parts. The first part states that if you take the derivative of an integral with variable limits, the derivative is equal to the integrand function evaluated at that variable's limit. More formally, if you have an integral \( \int_{a}^{b} f(t) \ dt \), and you differentiate with respect to \( b \), you get \( f(b) \).
The second part focuses on the antiderivative. It reveals that any function that can be expressed as an integral of another function has an antiderivative linked to it. This dual nature of the theorem is why it's considered a cornerstone in calculus.
Definite Integrals
Definite integrals are integral expressions evaluated over specific limits. Unlike indefinite integrals, which include an arbitrary constant representing a family of functions, definite integrals output a specific numerical value. This value represents the area under the curve of the function from one point to another.
In practical terms, the definite integral \( \int_{a}^{b} f(t) \ dt \) gives the net area between the
  • curve \( y = f(t) \)
  • the x-axis
  • the vertical lines \( x=a \) and \( x=b \)
This application is crucial in various fields such as physics and engineering, where calculating total quantities over a specific interval is essential.
In the context of the Fundamental Theorem of Calculus, definite integrals enable a straightforward process of evaluating the accumulation of a quantity, solidifying its use in more advanced mathematical problems.
Multivariable Calculus
Multivariable calculus is an extension of traditional calculus that deals with functions of multiple variables. It broadens the scope of calculus to accommodate more complex systems that depend on several input variables.
In multivariable calculus, concepts like partial derivatives and multiple integrals come into play. Partial derivatives allow us to see how a function that depends on several variables changes concerning one of those variables while keeping others constant. This is very useful in optimization problems, physics, and engineering, where situations frequently involve more than one influencing factor.
When working with functions that involve integration, such as our original exercise where the function is defined with definite integrals, it becomes necessary to understand how each variable influences the function. The partial derivatives provide this insight by showing how changes in one variable affect the integral's value. This is an essential skill for interpreting and manipulating multivariable functions.

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Most popular questions from this chapter

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