Problem 7 - Single Variable Calculus

Question

Find the derivative of the function. $$ F(x)=\int_{x}^{\pi} \sin 2 t d t $$

Step-by-Step Solution

Answer

The derivative of the function \(F(x) = \int_{x}^{\pi} \sin{(2t)} \, dt\) is \(F'(x) = -\sin{(2x)}\).

Step 1: Computing the integral

First, calculate the integral of the given function: $$F(x) = \int_{x}^{\pi} \sin{(2t)} \, dt$$ Now, we'll find the antiderivative of $$\sin{(2t)}$$ with respect to $$t$$: $$-\frac{1}{2}\cos{(2t)}$$ Now, we'll evaluate this antiderivative at the endpoints $$\pi$$ and $$x$$: $$F(x) = -\frac{1}{2}\cos{(2\pi)} + \frac{1}{2}\cos{(2x)}$$ $$F(x) = 1 - \frac{1}{2}\cos{(2x)}$$

Step 2: Applying the Fundamental Theorem of Calculus

Now, apply the first part of the FTC to find the derivative of the function: $$F'(x) = -\frac{d}{dx}\left(\frac{1}{2}\right)\cos{(2x)}$$

Step 3: Finding the derivative

Finally, we'll find the derivative of $$\frac{-1}{2}\cos{(2x)}$$ with respect to $$x$$: $$F'(x) = \frac{1}{2}\cdot (-2)\sin{(2x)}$$ $$F'(x) = -\sin{(2x)}$$ So, the derivative of the given function is: $$F'(x) = -\sin{(2x)}$$

Key Concepts

Integration

Integration is essentially the process of finding the antiderivative or the area under the curve of a given function. When we integrate a function like \( \sin(2t) \), we are looking for another function whose derivative will give us the original function back.
Let's consider

• The antiderivative of \( \sin(2t) \) involves manipulating the sine function into a form that can be easily integrated.
• This is achieved by using knowledge of standard integral formulas, which in this case, leads to \(-\frac{1}{2}\cos(2t)\).

The final step of calculating this integral is substituting the limits of integration which are \(x\) and \(\pi\). The result gives us a function \(F(x)\) without any remaining terms in terms of \(t\). Hence, integration provides us with a way to express our original problem in a new, simplified form.

Differentiation

Differentiation is the process of finding the derivative of a function. It's the reverse operation of integration and is used to compute the rate of change or the slope of a function.
In the context of the given problem, once we have simplified the integral, the next task is to differentiate the new function, \(F(x) = 1 - \frac{1}{2}\cos{(2x)}\). Differentiation rules, such as the chain rule, allow us to find the derivative efficiently.
Make sure to closely follow these steps:

• Recognize that \( \frac{d}{dx}(\cos(2x)) = -2\sin(2x)\)
• Multiply this result by \(\frac{1}{2}\) to get \(\frac{-1}{2} \times -2\sin(2x)\), simplifying to \(-\sin(2x)\).

The significance of differentiation in this exercise lies in verifying our answer using the methods of calculus, giving us confidence that \(F'(x)\) represents the rate of change of \(F(x)\).

Trigonometric functions

Trigonometric functions are mathematical functions that relate angles to ratios of triangle sides. In calculus, they often appear in problems requiring the integration and differentiation of wave functions, such as sine and cosine.
In solving the problem, we encounter the sine and cosine functions:

• \(\sin(2t)\), which is the function we first integrate.
• \(\cos(2x)\), which emerges as part of the antiderivative of \(\sin(2t)\).

Understanding the behavior and properties of these trigonometric functions is crucial because they are periodic and have specific derivative and antiderivative forms, which are:\( \frac{d}{dx}\sin{(u)} = \cos{(u)}\) and \( \frac{d}{dx}\cos{(u)} = -\sin{(u)}\). These relationships are used to both integrate and differentiate trigonometric functions effectively.
Trigonometric identities and their derivatives simplify problems involving angular motion or waves in physics and engineering, underscoring their significance in mathematical applications.