Question

Find \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}} \int_{0}^{x}\left(\int_{1}^{\operatorname{sint}} \sqrt{1+\mathrm{u}^{4}} \mathrm{du}\right) \mathrm{dt}\)

Short answer

Question: Find the second derivative with respect to x of the given double integral: \(F(x) = \int_{0}^{x} \left(\int_{1}^{\operatorname{sint}} \sqrt{1+u^{4}} du\right) dt\) Answer: \(\frac{d^{2}}{dx^{2}} \int_{0}^{x}\left(\int_{1}^{\operatorname{sint}} \sqrt{1+\mathrm{u}^{4}} \mathrm{du}\right) \mathrm{dt} = \cos t\)

Step-by-step solution

Simplify the given expression

First, let's simplify the given expression. The double integral can be expressed as: \(F(x) = \int_{0}^{x} \left(\int_{1}^{\operatorname{sint}} \sqrt{1+u^{4}} du\right) dt\)

Perform the first differentiation

Now, perform the first differentiation of F(x) with respect to x: \(\frac{dF}{dx} = \frac{d}{dx} \int_{0}^{x} \left(\int_{1}^{\operatorname{sint}} \sqrt{1+u^{4}} du\right) dt\) Using Leibniz's rule for differentiation under the integral sign, we have: \(\frac{dF}{dx} = \int_{1}^{\operatorname{sint}} \sqrt{1 + u^{4}} du\)

Perform the second differentiation

Next, perform the second differentiation of \(\frac{dF}{dx}\) with respect to x: \(\frac{d^{2}F}{dx^{2}} = \frac{d}{dx} \int_{1}^{\operatorname{sint}} \sqrt{1 + u^{4}} du\) Again, using Leibniz's rule for differentiation under the integral sign, we have: \(\frac{d^{2}F}{dx^{2}} = \frac{d}{dx} \operatorname{sint} = \cos t\)

Final answer

Therefore, the second derivative with respect to x of the given double integral is: \(\frac{d^{2}}{dx^{2}} \int_{0}^{x}\left(\int_{1}^{\operatorname{sint}} \sqrt{1+\mathrm{u}^{4}} \mathrm{du}\right) \mathrm{dt} = \cos t\)

Key concepts

Leibniz's Rule

When we talk about Leibniz's rule, we're delving into a powerful method in calculus that guides us in differentiating integrals that have variable limits. Imagine we have an integral with limits that aren't just numbers, but expressions involving the variable we're differentiating with respect to. Leibniz's rule provides us with a systematic way to find the derivative of such integrals.

According to Leibniz's rule, if you have an integral like \( \int_{a(x)}^{b(x)} f(t, x) dt \), the derivative of this integral with respect to \( x \) is not only the integral of the partial derivative of \( f \) with respect to \( x \) from \( a(x) \) to \( b(x) \), but also includes evaluating \( f \) at the upper and lower limits and multiplying by the derivative of these limits with respect to \( x \).

This rule is indispensable in solving complex problems in calculus, such as finding the second derivative of a double integral, as seen in the IIT JEE advanced mathematics curriculum. Leibniz's rule simplifies the process by focusing only on the outer integral's variable limit and the inner function, bypassing the inner integral altogether in subsequent differentiations.

Differentiation Under the Integral Sign

Differentiation under the integral sign is a technique where we take the derivative of an integral without having to evaluate the integral first. This strategy is particularly useful when we're dealing with integrals that are hard to solve or when the integral involves a parameter. The technique leverages the understanding that differentiation and integration are inverses of each other, up to a constant.

According to this concept, if we have a function \( F(x) \) defined as an integral with \( x \) in its limits or integrand, we can differentiate \( F(x) \) with respect to \( x \) by treating the integral sign as a 'container' that remains intact. The process applied in our exercise is a direct application of this concept where the first and second derivatives are taken with respect to \( x \) while keeping the integral sign in place.

This approach is crucial when working with nested functions, such as double integrals, where we differentiate the outer function while treating the inner integral as a whole. It reduces the complexity of simultaneous integrations, making the path to the solution more accessible.

Integral Calculus

Integral calculus is the branch of mathematics that focuses on adding up small pieces to find the whole. It is often seen as the inverse of differential calculus, which is about breaking down a whole into small parts. Integral calculus helps us solve problems involving areas, volumes, displacements, and accumulated quantities.

When studying integral calculus, students learn about definite and indefinite integrals. Definite integrals calculate the accumulation of quantities and have limits—a lower and an upper bound. Indefinite integrals, on the other hand, represent antiderivatives of functions and don't have specified limits. The double integral in our given problem is an example of definite integrals extended into two dimensions, which is often used to calculate volumes under surfaces.

For IIT JEE aspirants and mathematics enthusiasts, mastering integral calculus is essential because it lays the foundation for more advanced topics in engineering, physics, and several other fields. Understanding the principles behind operations such as differentiating under the integral sign and deploying tools like Leibniz's rule is critical for solving higher-level problems.

IIT JEE Advanced Mathematics

IIT JEE Advanced Mathematics is a challenging and comprehensive curriculum that is part of the entrance examination for the prestigious Indian Institutes of Technology (IITs). This course encompasses a wide range of mathematical concepts including calculus, algebra, geometry, and trigonometry, all of which demand a high level of understanding and analytical skills.

In tackling problems from this curriculum, like the one we've discussed involving the second derivative of a double integral, students are expected to apply various mathematical techniques and rules adeptly. Leibniz's rule and differentiation under the integral sign are just two of the valuable tools that students need to be proficient with. These concepts not only enable students to solve complex calculus problems but also help develop a deep comprehension of the underlying mathematical principles.

For students aspiring to excel in the IIT JEE Advanced examinations, a strong grasp on integral calculus and the ability to manipulate integrals and derivatives are indispensable. It's not merely about finding the right solution but also about appreciating the elegance and efficiency of mathematical methods, such as the ones illustrated in our example problem.