Problem 23 - Calculus

Question

Find \(G^{\prime}(x).\) $$ G(x)=\int_{1}^{x^{2}} \sin t d t $$

Step-by-Step Solution

Answer

The derivative is \(G'(x) = 2x \sin(x^2)\).

Step 1: Identify the function format

The given function is in the form of a definite integral where the upper limit is a function of \(x\). We have \(G(x) = \int_{1}^{x^2} \sin t \, dt\). This integral represents an accumulation of area under the curve \(\sin t\) from \(1\) to \(x^2\).

Step 2: Apply the Fundamental Theorem of Calculus with Chain Rule

The Fundamental Theorem of Calculus part 1 states that if \(F(t)\) is an antiderivative of \(f(t)\), then \(\frac{d}{dx} \int_{a}^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)\). Here, \(g(x) = x^2\) and \(f(t) = \sin t\).

Step 3: Differentiate using the theorem

Differentiate the function \(G(x)\). According to the theorem, the derivative of \(G(x)\) is given by \(\sin(x^2) \cdot \frac{d}{dx}(x^2)\).

Step 4: Calculate the derivative of the inner function

Compute the derivative of \(x^2\) with respect to \(x\), which is \(2x\).

Step 5: Multiply to find the final derivative

Multiply the result from the previous step by \(\sin(x^2)\). Thus, \(G'(x) = \sin(x^2) \cdot 2x\).

Key Concepts

Definite Integrals

A definite integral calculates the total accumulation of a function over a specific interval. Imagine you want to know how much of a certain quantity accumulates between two points, 1 and a variable bound in this case. With our function, this accumulation is visualized as the area under the curve defined by \(\sin t\) from 1 to \(x^2\). Unlike indefinite integrals, which result in a general antiderivative with a constant, definite integrals provide a precise numerical value.
This exercise uses the limits of 1 and \(x^2\), making it a perfect candidate for applying the Fundamental Theorem of Calculus. This theorem bridges the concept of differentiating a function and integrating one, which is pivotal to solving problems involving these kinds of integrals. When the upper limit of the integral is a function of \(x\), like \(x^2\), we need to apply additional rules like the chain rule to account for the dependency on \(x\).

Chain Rule

The chain rule is a technique used to differentiate composite functions, namely, functions within functions. When the definite integral has a variable as an upper limit that is independent of the integrand, like our \(x^2\) in \(G(x) = \int_{1}^{x^2} \sin t \, dt\), this rule is vital. Given a function \(g(x)\) that acts as an upper limit, finding the derivative \(\frac{d}{dx}\) requires that we take into account the derivative of this inner function.

• Identify \(g(x) = x^2\).
• Calculate \(g'(x) = 2x\).
• Use the chain rule to adjust the integral's derivative, multiplying by \(g'(x)\).

By following these steps, the chain rule helps sum up the influences of the nested function inside, ensuring no part of \(g(x)\) is overlooked.

Derivative Calculations

Derivative calculations are at the heart of finding rates of change, such as how fast an area is accumulating in our exercise. Once the Fundamental Theorem of Calculus is applied, allowing us to relate back to the original function within the integral, we take its derivative.
This involves understanding both the derivative of the inner limit, \(g(x) = x^2\), which we determined to be \(2x\), and the direct application of the trigonometric function \(\sin(x^2)\). Combining these gives the final derivative: \(G'(x) = \sin(x^2) \cdot 2x\).

• First, differentiate the inner function: \(\frac{d}{dx}(x^2) = 2x\).
• Then, apply the chain rule to find the derivative of \(G(x)\).
• Finally, multiply \(\sin(x^2)\) by \(2x\) to complete the derivative.

This calculation encapsulates both the nature of the composite function and the integral involved, culminating in an accurate reflection of the changing area.