Chapter 7: Problem 46
Use Part 2 of the Fundamental Theorem of Calculus to find the derivative. (a) \(\frac{d}{d x} \int_{0}^{x} \frac{d t}{1+\sqrt{t}}\) (b) \(\frac{d}{d x} \int_{1}^{x} \ln t d t\)
Short Answer
Expert verified
(a) \( \frac{1}{1 + \sqrt{x}} \), (b) \( \ln x \)
Step by step solution
01
Understanding the Fundamental Theorem of Calculus - Part 2
Let's recall that Part 2 of the Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \( \frac{d}{dx} \int_{a}^{x} f(t) \ dt = f(x) \). This means the derivative of the integral from a constant to \( x \) is simply the original function evaluated at \( x \).
02
Applying the Theorem to Part (a)
For part (a), we have the integral \( \int_{0}^{x} \frac{dt}{1 + \sqrt{t}} \). Here, \( f(t) = \frac{1}{1 + \sqrt{t}} \). Using the theorem, the derivative is \( f(x) = \frac{1}{1 + \sqrt{x}} \).
03
Applying the Theorem to Part (b)
For part (b), the integral given is \( \int_{1}^{x} \ln t \, dt \). In this case, \( f(t) = \ln t \). According to the theorem, the derivative is \( f(x) = \ln x \).
04
Conclusion
Using Part 2 of the Fundamental Theorem of Calculus, we have found the derivatives of the given integrals: for part (a), it is \( \frac{1}{1 + \sqrt{x}} \), and for part (b), it is \( \ln x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
derivatives
To begin with, let's discuss derivatives, a fundamental concept in calculus. Derivatives represent the rate at which a function is changing at any given point. Essentially, they tell us how a function's output value changes as its input value changes. In mathematical terms, if you have a function represented as \( f(x) \), the derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), provides you with a new function that describes this rate of change.
- Mathematically, the derivative of a function at a point captures the slope of the tangent line to the function's graph at that point.
- For physical applications, derivatives can represent different things; for example, in physics, they often denote velocity or acceleration.
- In the context of an integral, a derivative can help us understand how the accumulated area under a curve changes as the upper boundary of the integral moves.
antiderivatives
Antiderivatives, also known as indefinite integrals, are essentially the reverse operation of derivatives. When you take the antiderivative of a function, you're determining the function that, when differentiated, would give you the original function. If \( F(x) \) is an antiderivative of \( f(x) \), then \( F'(x) = f(x) \).
- The process of finding an antiderivative can be viewed as finding a function whose slope at any point corresponds to the value of the given function.
- Antiderivatives are not unique; given any antiderivative \( F(x) \), adding a constant \( C \) results in another valid antiderivative, \( F(x) + C \).
- These constants arise because the derivative of a constant is zero, meaning multiple functions can have the same derivative.
integrals
Integrals are a key concept in calculus, offering a way to calculate the area under a curve, among other things. There are two main types of integrals: definite and indefinite. Definite integrals compute the accumulated area under a curve between two bounds, while indefinite integrals provide an antiderivative of the function.
- Definite integrals are represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively. They provide a numeric value representing the net area.
- Indefinite integrals, shown as \( \int f(x) \, dx \), do not have bounds and instead yield a general function plus a constant \( C \).
- One key application of integrals, particularly definite integrals, is solving problems involving total accumulation, such as finding the total distance traveled given a velocity function.