Problem 22 - The Calculus with Analytic Geometry

Question

\(D_{x} \int_{1}^{x^{2}} \sqrt{1+t^{4}} d t\)

Step-by-Step Solution

Answer

\(2x \sqrt{1+x^8}\).

Step 1 - Understand the Problem

We need to find the derivative with respect to x of the integral from 1 to \(x^2\) of \(\sqrt{1+t^4}\) dt.

Step 2 - Use the Leibniz Rule

Apply the Leibniz rule for differentiation under the integral sign. The rule states \[\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).\] Here, \(a(x) = 1\) and \(b(x) = x^2\).

Step 3 - Compute Derivatives of Limits

Since \(a(x) = 1\) is a constant, \(a'(x) = 0\), and for \(b(x) = x^2\), we have \(b'(x) = 2x\).

Step 4 - Evaluate the Integral at the Limits

The integral simplifies to \(\sqrt{1+(x^2)^4} - \sqrt{1+1^4} \). Since \(a(x) = 1\) is constant, it contributes nothing.

Step 5 - Multiply by Derivative of Upper Limit

Using the Leibniz rule, we get \( \sqrt{1+(x^2)^4} \cdot 2x = \sqrt{1+x^8} \cdot 2x \).

Key Concepts

Leibniz Rule

In calculus, the Leibniz Rule helps us differentiate an integral where the limits of integration are functions of the variable we’re interested in differentiating. It is crucial when the limits are not fixed and depend on the variable. This rule elegantly connects differentiation and integration. The rule states:
\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b\'(x) - f(a(x)) \cdot a\'(x).
Here's a breakdown of the components:

• \(a(x)\) and \(b(x)\) are the lower and upper limits as functions of \(x\).
• \(f(t)\) is the integrand, a function in terms of \(t\).
• \(b\'(x)\) and \(a\'(x)\) are the derivatives of the limits with respect to \(x\).

This rule allows us to tackle integrals that aren’t over fixed intervals, making it versatile and powerful.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing they are essentially inverse operations. It consists of two parts:

1. First part: If \(F\) is an antiderivative of \(f\), then \(\int_{a}^{b} f(t) dt = F(b) - F(a)\).
2. Second part: If \(F(x) = \int_{a}^{x} f(t) dt\), then \(\frac{d}{dx} F(x) = f(x)\).

The first part shows how to compute a definite integral if you know an antiderivative of the integrand. The second part states that the process of integrating a function and then differentiating it will yield the original function.
This dual nature is fundamental in solving many calculus problems involving area under curves and rates of change.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative measures how the function's output value changes as its input value changes. If \(y = f(x)\), the derivative is often written as \(\frac{dy}{dx}\) or \(f\'(x)\).
Key notions include:

• The derivative represents the slope of the function's graph at any point.
• It is a foundational tool for understanding rates of change in various applications, such as physics, biology, and economics.

Let's revisit the step-by-step solution through this concept. We found:

• \(a(x) = 1\) and \(b(x) = x^2\)
• \(a\'(x) = 0\) and \(b\'(x) = 2x\)

Using these, we applied the Leibniz Rule to differentiate \( \int_{1}^{x^2} \sqrt{1+t^{4}} dt \):

1. Evaluate the function at the upper limit: \( f(b(x)) = \sqrt{1+x^8}\)
2. Multiply by the derivative of the upper limit: \( \sqrt{1+x^8} \cdot 2x = 2x\sqrt{1+x^8}\)