Problem 47 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\int_{\sqrt{x}}^{0} \sin \left(t^{2}\right) d t$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{\sin(x)}{2\sqrt{x}} \).

Step 1: Identify the structure

The problem involves differentiating an integral with respect to a variable that is in the limits of integration. The integrand is \( \sin(t^2) \) and the integration is from \( \sqrt{x} \) to 0.

Step 2: Apply Leibniz Rule

Since we have a definite integral with a variable limit, use the Leibniz rule for differentiating under the integral sign. This 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, \( a(x) = \sqrt{x} \) and \( b(x) = 0 \).

Step 3: Compute the derivatives and evaluate at limit

Compute \( f(\sqrt{x}) = \sin((\sqrt{x})^2) = \sin(x) \) and note that \( f(0) = \sin(0^2) = 0 \). The derivative of the limits with respect to \( x \) are \( a'(x) = \frac{1}{2\sqrt{x}} \) and since \( b(x) = 0 \), \( b'(x) = 0 \).

Step 4: Substitute into the Leibniz Rule formula

Plug these values into the Leibniz Rule formula: \( \frac{d y}{d x} = -\sin(0) \cdot 0 + \sin(x) \cdot \left(-\frac{1}{2\sqrt{x}}\right) \). Simplify to get: \( \frac{d y}{d x} = -\frac{\sin(x)}{2\sqrt{x}} \).

Key Concepts

Differentiating Under the Integral Sign

Differentiating under the integral sign is a technique used when dealing with integrals whose limits or integrand depends on a variable. This often requires the use of Leibniz's rule, which provides a structured method to calculate derivatives in such cases.

• Leibniz's Rule: This is essential for differentiating these complex integrals. It states that for a function \( f(t) \) integrated with respect to \( t \) over limits \( a(x) \) to \( b(x) \), the derivative is given by: \[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]
• The formula can be thought of as involving two steps: Evaluate the integrand at the limits of integration, and then multiply by the derivative of the limits with respect to \( x \).

In our problem, we see this technique being used where the limits \( \sqrt{x} \) and \( 0 \) contribute to the derivative of \( y \) with respect to \( x \). Understanding this technique through the lens of the Leibniz rule helps in accurately differentiating such integrals.

Variable Limits of Integration

Variable limits of integration can complicate the process of differentiation, as they introduce dependencies on the variable we differentiate with respect to. In our exercise, the upper and lower limits of the integral expression involve the variable \( x \).

• Lower Limit: The lower limit is \( \sqrt{x} \), which changes as \( x \) changes, affecting the value of the integral.
• Upper Limit: In this example, the upper limit is \( 0 \), a constant, simplifying calculations as its derivative with respect to \( x \) is zero.

Variable limits require us to apply specific rules such as Leibniz's rule to differentiate properly. This involves computing the derivatives of the limits themselves, as they directly influence the outcome of the integral when \( x \) is a limit.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) bridges the world of calculus between differentiation and integration. This theorem is essential when dealing with integrals that depend explicitly on a variable not just in their limits, but embedded in the integrand as well.

• Parties: The theorem has two parts, where Part 1 links the process of finding the derivative of an integral with the integrand and its evaluation at boundary points.
• Integration to Differentiation: FTC allows us to find the derivative of integrals efficiently by focusing on the change in outputs with respect to shifts in input variable \( x \).

Though the FTC doesn’t directly apply to integrals with variable limits as our exercise shows, understanding it provides context for why differentiating under the integral sign, as we've done with Leibniz's rule, is necessary. Hence, forming a foundational basis to tackle such calculus problems effectively.