Question

Derivatives of integrals Simplify the following expressions. $$\frac{d}{d w} \int_{0}^{\sqrt{w}} \ln \left(x^{2}+1\right) d x$$

Short answer

Question: Calculate the derivative of the integral function \(F(w) = \int_{0}^{\sqrt{w}} \ln(x^2 + 1) dx\) with respect to variable w. Answer: The derivative of the given integral function with respect to w is \(F'(w) = \frac{1}{2 \sqrt{w}} \ln(w + 1)\).

Step-by-step solution

Apply Leibniz Rule

Let $$F(w) = \int_{0}^{\sqrt{w}} \ln(x^{2} + 1) dx$$ To find the derivative of F(w) with respect to w, we will apply Leibniz Rule, which states: $$\frac{d}{d w} \int_{a(w)}^{b(w)} f(x, w) d x = \frac{d b(w)}{d w} f(x, w) \Big \vert_{x=b(w)} - \frac{d a(w)}{d w} f(x, w) \Big \vert_{x=a(w)} + \int_{a(w)}^{b(w)} \frac{\partial f(x, w)}{\partial w}dx$$ Now, let's apply the Leibniz rule to our problem, we have: $$F'(w) = \frac{d}{d w} \int_{0}^{\sqrt{w}} \ln(x^{2} + 1) d x$$

Evaluate the first derivative

Using Leibniz Rule, we find the following: $$F'(w) = \frac{d (\sqrt{w})}{d w} \ln(\sqrt{w}^{2} + 1) - 0 + \int_{0}^{\sqrt{w}} \frac{\partial \ln(x^{2} + 1)}{\partial w} dx$$ Next, we need to find the derivative of \(\sqrt{w}\) with respect to w and the partial derivative of \(\ln(x^{2}+1)\) with respect to w.

Calculate derivative terms

Let's first find the derivative of \(\sqrt{w}\). $$\frac{d (\sqrt{w})}{d w} = \frac{1}{2 \sqrt{w}}$$ The second term in the Leibniz rule applied to our function is zero, since the lower bound of integration is a constant. Now, the integrand does not depend explicitly on w, so its partial derivative with respect to w is zero. Therefore, the integral term becomes zero, and we are left with: $$F'(w) = \frac{1}{2 \sqrt{w}} \ln(\sqrt{w}^{2} + 1)$$

Simplify expression

Now, we can simplify the expression: $$F'(w) = \frac{1}{2 \sqrt{w}} \ln(w + 1)$$ Thus, the derivative of the given integral function with respect to w is: $$\frac{d}{d w} \int_{0}^{\sqrt{w}} \ln \left(x^{2}+1\right) d x = \frac{1}{2 \sqrt{w}} \ln(w + 1)$$

Key concepts

Derivatives of Integrals

The concept of taking derivatives of integrals might seem a bit complex at first, but with the right tools, it becomes more approachable. One such tool is the Leibniz Rule, which helps in finding the derivative of an integral whose limits are functions of the variable in question. Consider the integral function \( F(w) = \int_{0}^{\sqrt{w}} \ln(x^{2} + 1) \, dx \). To find its derivative \( F'(w) \), use Leibniz Rule:

• Focus on the upper and lower limits of the integral, especially on how they depend on \( w \).
• Upper limit \(b(w)=\sqrt{w}\) is a function of \( w \), while the lower limit is a constant \( a(w)=0 \).
• Calculate the derivative of the upper limit with respect to \( w \) and evaluate the integrand at the upper limit.

By applying the Leibniz Rule, the derivative simplifies to terms that help clarify how changes in \( w \) affect the integral's value effectively.

Integral Calculus

Integral calculus provides powerful techniques for handling quantities that accumulate continuously. In our context, the integral \( F(w) = \int_{0}^{\sqrt{w}} \ln(x^{2} + 1) \, dx \) represents the accumulated effect of \( \ln(x^{2} + 1) \) over the interval from 0 to \( \sqrt{w} \).

• The main goal here is to express rates of change, captured through differentiation, while managing accumulation.
• The competence to evaluate integrals having variable limits of integration, and incorporating differentiation, is essential in understanding real-world dynamics, like changing boundaries in a physical process.
• Integral calculus's role simplifies complex added values into a single meaningful quantity, adaptable across varying bounds.

This notion of combining differentiation and integration is key to solving problems where changes happen continuously along an interval.

Simplification of Expressions

Mathematics often involves simplifying expressions to uncover clearer insights into their behavior or make them more usable in further calculations. In the derived expression for \( F'(w) \), simplification plays a crucial role:

• Original expression: \( F'(w) = \frac{1}{2 \sqrt{w}} \ln(\sqrt{w}^{2} + 1) \)
• Recognize \( \sqrt{w}^{2} = w \) and simplify \( \ln(w + 1) \)
• Final expression: \( F'(w) = \frac{1}{2 \sqrt{w}} \ln(w + 1) \)

This step-by-step simplification allows for a more intuitive understanding of results, transforming an initially cumbersome result into a version that is easier to interpret and apply. Simplifying methods aid in reducing errors, enhancing clarity, and improving computational efficiency in the analysis of problems.