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Chapter 5: Problem 65

Simplify the following expressions. $$\frac{d}{d x} \int_{x}^{1} \sqrt{t^{4}+1} d t$$

Short Answer

Expert verified
Based on the step by step solution above, determine the derivative of the given integral expression with respect to x. Answer: The derivative of the given integral expression with respect to x is $$-\sqrt{x^4 + 1}$$

Step by step solution

01

Identify the functions involved in the integral

In our case, the following functions are involved: - The integrand function, $$f(t, x) = \sqrt{t^4 + 1}$$ - The lower limit of the integral, $$a(x) = x$$ - The upper limit of the integral, $$b(x) = 1$$
02

Calculate the partial derivative of the integrand function with respect to x

Since the integrand function does not contain x, its partial derivative with respect to x will be zero: $$\frac{\partial f(t, x)}{\partial x} = \frac{\partial}{\partial x} \sqrt{t^4 + 1} = 0$$
03

Calculate the derivatives of the limits with respect to x

The derivatives are: $$\frac{d a(x)}{d x} = \frac{d x}{d x} = 1$$ and $$\frac{d b(x)}{d x} = \frac{d (1)}{d x} = 0$$
04

Use Leibniz's rule to compute the derivative of the integral

Using the results from steps 2 and 3, we can now apply Leibniz's rule for the given integral: $$\frac{d}{d x} \int_{x}^{1} \sqrt{t^{4}+1} dt = \int_{x}^{1} 0 dt + \sqrt{1^4 + 1} \cdot 0 - \sqrt{x^4 + 1} \cdot 1$$ Simplifying the expression, we get: $$\frac{d}{d x} \int_{x}^{1} \sqrt{t^{4}+1} dt = -\sqrt{x^4 + 1}$$

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Key Concepts

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

Leibniz's Rule
Leibniz's Rule is a fundamental concept in calculus that helps us differentiate under the integral sign when the limits of the integral are functions of the differentiation variable. It is an extension of the Fundamental Theorem of Calculus. This rule is useful when you have an integral that depends on a variable, often seen when differentiating integrals with variable limits.Here are the steps for applying Leibniz's Rule:
  • Identify the integrand function, and determine if it depends on multiple variables.
  • Calculate the partial derivative of the integrand with respect to the variable of differentiation, if it depends on it.
  • Determine the derivatives of the limits with respect to the variable of differentiation.
  • Apply the Leibniz rule formula to find the derivative, which is:\[ \frac{d}{dx}\left( \int_{a(x)}^{b(x)} f(t, x) \, dt \right) = f(b(x), x) \cdot \frac{d b(x)}{d x} - f(a(x), x) \cdot \frac{d a(x)}{d x} + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt \]
These steps illustrate how Leibniz's Rule accounts for the variable nature of limits and potentially the integrand.
Derivative of an Integral
The derivative of an integral, particularly with variable limits, is a concept that finds its roots in the Fundamental Theorem of Calculus. This theorem connects differentiation and integration, showing that integration can be reversed by differentiation.When dealing with an integral of the form \( \int_{a(x)}^{b(x)} f(t) \, dt \), its derivative with respect to \( x \) involves:
  • Evaluating \( f(t) \) at the upper limit and multiplying by the derivative of the upper limit.
  • Evaluating \( f(t) \) at the lower limit and multiplying by the derivative of the lower limit, subtracting this product from the first.
This means that the derivative is reliant on both the function within the integral and the behavior of its variable limits. The process ensures we respect changes within the integral range, reflecting how they influence the derivative.
Integral Limits
Integral limits define the range over which the integration occurs. When these limits are functions of a variable, that variable plays a significant role in differentiation.In the problem given, the integral limits are \( a(x) = x \) (lower limit) and \( b(x) = 1 \) (upper limit). Because the integral's limits can change, how they change with respect to the variable of differentiation is crucial.Here's what you should consider:
  • If the upper or lower limit is a constant, its derivative with respect to any variable is zero, having no effect on differentiation.
  • If the limits are variables or functions of variables, find the derivative of each limit with respect to the variable of differentiation. This influences the derivative significantly, as seen in the use of Leibniz's Rule.
Understanding changes in integral limits and incorporating them in differentiation ensures you're accurately calculating how integrals behave when limits move or shift, as the variable changes.

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Most popular questions from this chapter

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{3} \frac{v^{2}+1}{\sqrt{v^{3}+3 v+4}} d v$$

Suppose \(f\) is continuous on \([a, b]\) with \(f^{\prime \prime}(x)>0\) on the interval. It can be shown that $$(b-a) f\left(\frac{a+b}{2}\right) \leq \int_{a}^{b} f(x) d x \leq(b-a) \frac{f(a)+f(b)}{2}$$. a. Assuming \(f\) is nonnegative on \([a, b],\) draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. Divide these inequalities by \((b-a)\) and interpret the resulting inequalities in terms of the average value of \(f\) on \([a, b]\).

Use a change of variables to find the following indefinite integrals. Check your work by differentiating. $$\int x^{3}\left(x^{4}+16\right)^{6} d x$$

Evaluate the following integrals. $$\int_{0}^{\pi / 6} \frac{\sin 2 y}{\sin ^{2} y+2} d y(\text {Hint}: \sin 2 y=2 \sin y \cos y .)$$

Use a change of variables to evaluate the following integrals. $$\int_{-1}^{1}(x-1)\left(x^{2}-2 x\right)^{7} d x$$
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