Question

For the following exercises, evaluate by any method. $$ \frac{d}{d x} \int_{x}^{1} \frac{d t}{t} $$

Short answer

The derivative is \(-\frac{1}{x}\).

Step-by-step solution

Understanding the Problem

The problem involves taking the derivative with respect to \(x\) of an integral with bounds \(x\) and \(1\). The integral is \(\int_{x}^{1} \frac{dt}{t}\).

Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus says that if \(F(t)\) is an antiderivative of \(f(t)\), then \(\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) b'(x) - f(a(x)) a'(x)\). Here, the lower bound \(x\) depends on \(x\), and the upper bound \(1\) is constant. Thus, we apply the theorem for \(\int_{x}^{1} \frac{dt}{t}\).

Substitute in the Fundamental Theorem of Calculus

The integrand \( \frac{1}{t} \) has the antiderivative \( \ln |t| \). The bounds are \(a(x) = x\) and \(b(x) = 1\). Substitute these into the formula to get \( \frac{1}{1} \cdot 0 - \frac{1}{x} \cdot 1 = 0 - \frac{1}{x} = -\frac{1}{x} \).

Compute the Derivative

Apply the result of the substitution to obtain the derivative: \(-\frac{1}{x}\). This represents the rate of change of the integral with respect to the lower limit \(x\).

Key concepts

Derivatives

The concept of derivatives is crucial in understanding how a function changes. When we talk about the derivative \( \frac{d}{dx} \), we are asking how a function changes as its input changes. In the exercise, the derivative is applied to an integral where the upper limit is constant and the lower limit is a variable, specifically \( x \).

• The derivative of an integral with respect to a variable bound essentially measures the rate of change of the integral's value as that bound changes.
• Using the Fundamental Theorem of Calculus, it's important to note the relationship between derivatives and integrals, which allows us to compute these rates of change effectively.

In this case, we're examining how a switch in the lower bound of the integral, from say \(a(x)\) to a point \(x\), affects the value of the integral itself. By understanding derivatives this way, we gain insights into dynamic systems and changes over time.

Integration by Parts

Integration by Parts is a technique used to solve certain integrals, closely related to the product rule for differentiation. However, in the original exercise, this technique isn't directly applied. But it's essential to understand why it's useful. Integration by parts essentially states:\[ \int u \, dv = uv - \int v \, du \]Here's how it can be applied in general:

• Pick the parts from the integral for \(u\) and \(dv\) strategically; this choice makes the integral \(\int v \, du\) easier to solve than the original.
• Use derivatives (for \(du\)) and antiderivatives (for \(v\)) during the process.
• Commonly used when the integrand is a product of two functions, like polynomial and exponential.

Even though the exercise at hand doesn't require integration by parts, being familiar with it enriches one's toolkit for evaluating more complex integrals.

Antiderivatives

Antiderivatives are the reverse process of derivatives. When a derivative shows how a function changes, an antiderivative takes a rate of change back to its original quantity form. Finding an antiderivative means finding a function whose derivative is the given function.In the given exercise, the antiderivative of \(\frac{1}{t}\) is \(\ln |t|\). This is crucially important in applying the Fundamental Theorem of Calculus:

• Identifying the correct antiderivative is the first step before substituting in the theorem.
• Once we have the antiderivative, we apply the integration limits to find the integral value, then differentiate it with respect to the lower boundary to solve the exercise.

Antiderivatives help bridge the concept between integration and differentiation. In this context, they're a method to express accumulated change, which is unwrapped back to rates of change through differentiation.