Question

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

Short answer

Answer: The derivative of the given integral with respect to \(x\) is $-\sqrt{x^4 + 1}$.

Step-by-step solution

Recognize the use of the Fundamental Theorem of Calculus Part 1

The integral in question involves a variable limit of integration. Specifically, it has the form: $$\int_{x}^{1} \sqrt{t^{4}+1} d t$$ In order to find the derivative of this integral with respect to \(x\), we need to apply the Fundamental Theorem of Calculus (FTC) Part 1, which says that: If a function \(F(x)\) is an antiderivative of \(f(x)\) on an interval \([a, b]\), then $$\int_a^b f(x) dx = F(b) - F(a)$$ In our case, we want to find the derivative of the integral, which will involve using FTC Part 2.

Apply the Fundamental Theorem of Calculus Part 2

The Fundamental Theorem of Calculus (FTC) Part 2 states that: If a function \(f(x)\) is continuous on an interval \([a, b]\) and \(F(x)\) is an antiderivative of \(f(x)\), then $$\frac{d}{dx}\int_a^x f(t) dt = f(x)$$ In our case, our integral has a variable lower limit of integration, so we actually need to apply the negative of FTC Part 2. That is, if \(h(x) = \int_x^a f(t) dt\), then $$\frac{dh}{dx} = -f(x)$$ Here, \(f(x) = \sqrt{t^4 + 1}\). So by applying the negative of FTC Part 2, we will find the derivative of the integral.

Compute the derivative and simplify the expression

Now, applying the negative of FTC Part 2 to the integral to find its derivative, we have: $$\frac{d}{dx}\left(\int_x^1 \sqrt{t^4 + 1} dt\right) = -\sqrt{x^4 + 1}$$ Thus, the simplified expression for the derivative of the integral is: $$-\sqrt{x^4 + 1}$$

Key concepts

Derivatives

Derivatives are a foundational concept in calculus that measure how a function changes as its input changes. Essentially, derivatives provide us with the rate of change or the slope of the function's graph at a specific point. In the context of the Fundamental Theorem of Calculus, derivatives link the concepts of differentiation and integration. Differentiation involves finding the derivative of a function, which can be thought of as a form of reverse integration.

• Derivatives help us understand the behavior of functions, showing us how they increase or decrease.
• They are crucial in various applications, from physics to economics, wherever a change is observed and analyzed.

Since derivatives provide a snapshot of a function's behavior at any given point, they are essential for solving instantaneous rate problems. In our exercise, we specifically look at the derivative of an integral, which gives insights into specific values this integral approaches as the variable changes.

Integrals

Integrals are central to calculus and can be best understood as a means of accumulating quantities, such as areas under curves. This concept serves as the opposite of differentiation, reversing the process to find the original function given its derivative. Integrals can be either definite, which provide a numerical value representing the accumulated quantity, or indefinite, providing a general formula for potential outcomes.

• An integral \(\int f(t) \, dt\) can be thought of as the accumulation of the function's values across a range.
• This accumulation helps in identifying total quantities and areas under plots.

In our exercise, the integral \(\int_{x}^{1} \sqrt{t^{4}+1} \, dt\) involves a limit that is variable, affecting how the integral behaves as we differentiate it.

Variable Limits of Integration

Variable limits of integration involve integration bounds that are functions of another variable rather than specific constants. These variable limits play a crucial role in how we apply the Fundamental Theorem of Calculus, particularly when differentiating integrals. With variable limits, any change in the integral’s bounds directly affects the value of the integral itself, necessitating the use of different rules when finding derivatives.

• Variable limits mean the bounds of integration depend on a variable, requiring adjustment to standard calculus theorems.
• In particular, they are key to using the Fundamental Theorem of Calculus effectively on integrals where limits are not fixed values.

In our original exercise, the variable limit of integration for the integral is "x", making it essential to apply the Fundamental Theorem of Calculus correctly. We ultimately use it to determine the derivative of the integral with respect to \(x\), considering the changing limits and their impacts.