/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 Prove that \(\frac{d}{d x}\left[... [FREE SOLUTION] | 91Ó°ÊÓ

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Prove that \(\frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x)\).

Short Answer

Expert verified
Therefore, \( \frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=f(v(x))v'(x) - f(u(x))u'(x) \), successfully demonstrating the valid application of both the chain rule and the Fundamental Theorem of Calculus.

Step by step solution

01

Setup

Start by setting up the proof and explaining the concepts that will be used. The Fundamental Theorem of Calculus (Part 1) states that if \( f \) is continuous on the interval \([a, b]\) and \( F \) is an antiderivative of \( f \) on \([a, b]\), then \( \int_{a}^{b} f(t) dt = F(b) - F(a) \). The other useful concept is the chain rule, \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \).
02

Apply Fundamental Theorem of Calculus Part 1

Apply the Fundamental Theorem of Calculus Part 1: \( \frac{d}{dx} \int_{u(x)}^{v(x)} f(t) dt = \frac{d}{dx} [F(v(x)) - F(u(x))] \), where \( F(t) \) is the integral of \( f(t) \).
03

Apply Chain Rule

Differentiate \( F(v(x)) - F(u(x)) \) using the chain rule: \( \frac{d}{dx} [F(v(x)) - F(u(x))] = F'(v(x))v'(x) - F'(u(x))u'(x) \). We know that \( F'(t) = f(t) \), since \( F(t) \) is the integral of \( f(t) \). Therefore, the derivative should be the original function: \( f(v(x))v'(x) - f(u(x))u'(x) \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus that facilitates the differentiation of composite functions. Imagine you're trying to find the rate at which one quantity changes with respect to another, but there's a third variable mediating their relationship. This scenario is representative of a composite function – one function nestled within another.

Mathematically, if you have two functions, say, g(x) and f(u), where u = g(x), the chain rule helps to find the derivative of the composite function f(g(x)) with respect to x. It can be expressed as:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \]
or in a more intuitive sense, the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function. When applying the chain rule to the exercise at hand, it allows us to differentiate functions that have bounds as other functions of x, which is a concrete example of this rule's pivotal role in calculus.
Antiderivative
An antiderivative, also known as an indefinite integral, is essentially the reverse process of differentiation. If you have a function f(x), an antiderivative F(x) is a function whose derivative gives you back f(x). In notation, if \( F'(x) = f(x) \), then F(x) is an antiderivative of f(x).

This concept is critical when dealing with integration, as it represents the accumulation of the values of f(x). Knowing the antiderivative allows us to compute definite integrals over an interval, which provides the area under the curve of f(x) from one point to another. In the context of our exercise, recognizing that an antiderivative of the integrand is needed to apply the Fundamental Theorem of Calculus is a key step.
Integration
Integration is the act of finding an integral, which encompasses finding the antiderivative, the area under a curve, or the accumulation of quantities. It is often described as the opposite operation to differentiation, and in many ways, it stitches together infinitesimal pieces to find a whole.

There are two main types of integrals: indefinite and definite. Indefinite integrals give us a family of functions (antiderivatives), while definite integrals result in a numerical value representing area or accumulation.
\[ \int f(x) dx = F(x) + C \]
This equation represents the indefinite integral, where C is the constant of integration. For definite integrals, we have:
\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \]
In the given exercise, we deal with definite integration as we find the net change of the function f(t) between the variable limits u(x) and v(x).
Differentiation
Differentiation is the process of finding the derivative of a function, which measures the rate at which a function's value changes as its input changes. In simpler terms, it tells us about the slope of the function at any point.

The derivative of a function f(x) at a point x is represented as f'(x) or \( \frac{df}{dx} \). Calculating derivatives is central to calculus and many real-world applications like motion, optimization, and rates of change. In our particular exercise, we differentiate an integral, which showcases how differentiation and integration are interconnected operations − the foundational concept behind the Fundamental Theorem of Calculus.

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Most popular questions from this chapter

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