Question

Suppose that \(u, v,\) and \(f\) are defined on \((-\infty, \infty)\). Let \(u\) and \(v\) be differentiable and \(f\) be continuous for all \(x\). Show that $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x) . $$

Short answer

Question: Prove that $\frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = f(v(x)) v^{\prime}(x) - f(u(x)) u^{\prime}(x)$ for differentiable functions \(u\) and \(v\), and a continuous function \(f\). Answer: To prove this, we can follow these steps: 1. Write down the expression for the integral using the Fundamental Theorem of Calculus: $\int_{u(x)}^{v(x)} f(t) d t = F(v(x)) - F(u(x))$, where \(F(t)\) is an antiderivative of \(f(t)\). 2. Differentiate both sides of this equation with respect to \(x\): $\frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = \frac{d}{d x} [F(v(x)) - F(u(x))]$. 3. Apply the Chain Rule to differentiate \(F(v(x))\) and \(F(u(x))\) with respect to \(x\): $\frac{d}{d x} F(v(x)) = f(v(x)) \cdot v'(x)$ and $\frac{d}{d x} F(u(x)) = f(u(x)) \cdot u'(x)$. 4. Substitute these expressions into the equation for the derivative of the integral, and rearrange the terms: $\frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = f(v(x)) v^{\prime}(x) - f(u(x)) u^{\prime}(x)$. Through these steps, we have proven the given expression for the derivative of the integral with respect to its integration limits.

Step-by-step solution

Expression to Prove

$$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x) . $$

Differentiate the Integral

Let's say \(F(t)\) is an antiderivative of \(f(t)\), i.e., \(F'(t) = f(t)\). Using the Fundamental Theorem of Calculus, we have: $$ \int_{u(x)}^{v(x)} f(t) d t = F(v(x)) - F(u(x)) $$ Now, differentiate both sides of this equation with respect to \(x\): $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = \frac{d}{d x} [F(v(x)) - F(u(x))] $$

Apply the Chain Rule

We have to differentiate \(F(v(x))\) and \(F(u(x))\) with respect to \(x\). For that, we will use the Chain Rule: $$ \frac{d}{d x} F(v(x)) = F'(v(x)) \cdot v'(x) $$ $$ \frac{d}{d x} F(u(x)) = F'(u(x)) \cdot u'(x) $$ Now, recall that \(F'(t) = f(t)\). We can rewrite the above derivatives as: $$ \frac{d}{d x} F(v(x)) = f(v(x)) \cdot v'(x) $$ $$ \frac{d}{d x} F(u(x)) = f(u(x)) \cdot u'(x) $$

Compute the Derivative

Now, we can substitute these expressions into the equation for the derivative of the integral, and get: $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = \left[ f(v(x)) \cdot v'(x) \right] - \left[ f(u(x))\cdot u'(x) \right] $$

Rearrange the Terms

Rearrange the terms, and we will obtain the expression we want to prove: $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t = f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x) . $$ This completes the proof, and we have demonstrated that the derivative of the integral with respect to its integration limits is equal to the given expression.

Key concepts

Real Analysis

Real analysis is a branch of mathematical analysis dealing with the set of real numbers and the functions defined on it. This area involves examining the behavior and properties of real-valued sequences, series, and functions. One of the fundamental concepts in real analysis is the notion of integration, which calculates the accumulation of quantities, such as area under a curve.

When we look at problems like the derivative of an integral with varying limits, real analysis provides the tools and theorems necessary to tackle such problems. Techniques from real analysis, such as limits, continuity, integration, and differentiation, are essential to solve and understand these kinds of problems. The Leibniz integral rule, which is the focus of this exercise, lies at the intersection of derivatives and integrals and is a prime example of real analysis in action.

Derivative of Integral

The derivative of an integral, especially when the integral has variable limits, becomes a question of understanding how the area under a curve changes as these limits themselves change. This concept may seem daunting at first, but it is crucial in both pure and applied mathematics.

To tackle this problem, we use a result known as the Leibniz integral rule, which tells us how to take the derivative of an integral whose limits are functions of the variable of differentiation. The rule states that under appropriate conditions, the derivative of the integral of a function with respect to its variable boundaries can be represented as the difference of the function evaluated at the upper and lower limits, each multiplied by the derivative of its respective limit. This relationship ties together the concepts of differentiation and integration in a beautifully symmetrical way.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to differentiate composite functions. When you have a function nested inside another function, like the integral with function limits in our exercise, the chain rule tells us how to find the derivative of this composite. It essentially unpacks the composite function into simpler parts that can be differentiated individually.

In our example, to differentiate an integral from a function of \(x\) to another function of \(x\), we must differentiate each limit with respect to \(x\) and then multiply by the derivative of the inner function with respect to that limit. The chain rule enables us to see the relationship between the rates of change of these nested functions and is used to derive the Leibniz rule for the derivative of an integral with variable limits.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of integration and differentiation, two cornerstone operations in calculus. It is divided into two parts: the first part provides a connection between the anti-derivative and the integral, while the second part allows us to evaluate the integral of a function via its antiderivative.

In the context of our exercise, we use the theorem to represent the definite integral as the difference of the anti-derivative evaluated at the upper and lower limits. Then, recognizing that the antiderivative's derivative is the original function, we can apply the theorem to find the derivative of integral expressions. Whether a student is grappling with definite integrals or exploring the derivative of an integral, the Fundamental Theorem of Calculus serves as an essential tool for understanding and connecting these concepts.