Problem 15 - Calculus: Early Transcendental Functions

Question

Write the given (total) area as an integral or sum of integrals. The area between \(y=\sin x\) and the \(x\) -axis for \(0 \leq x \leq \pi\).

Step-by-Step Solution

Answer

Therefore, the total area under the curve \(y = \sin x\) from 0 to \(\pi\) is equal to 2.

Step 1: Define the Function

The function is defined as \(y = \sin x\).

Step 2: Set the Integration Limits

Integration is performed over the interval \(0 \leq x \leq \pi\). These values constitute the limits of integration. So, the lower limit is 0, and the upper limit is \(\pi\).

Step 3: Set the Integral up

The integral is set up to represent the area under the curve. This area can be found by evaluating the definite integral of \(y\) from 0 to \(\pi\), which can be mathematically represented as \(\int_0^\pi \sin x\, dx\).

Step 4: Evaluate the Integral

The next step is to evaluate the integral. The integral of \(\sin x\) with respect to \(x\) from 0 to \(\pi\) is \([- \cos x]_0^\pi\).

Step 5: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that, once an antiderivative is found, it can be evaluated at the upper limit and at the lower limit, and then subtract the two results (subtract the value at lower limit from the value at the upper limit). Hence, \([- \cos x]_0^\pi = - \cos \pi - (- \cos 0) = -(-1) - (-1) = 2\).

Key Concepts

Definite Integral

The definite integral is a core part of calculus, offering a way to calculate the total area under a curve within a specific interval. Think of it as summing up an infinite number of tiny rectangles under the curve of a function. In the context of the problem we're addressing, our function is given by \(y = \sin x\) and our interval—or, the set of limits—is from \(x = 0\) to \(x = \pi\).

To write this as a definite integral, you set up the integral with your limits of integration as bounds, shown here:

• The lower limit, 0, represents where the interval begins.
• The upper limit, \(\pi\), shows where the interval ends.

Put together, the definite integral is expressed as \(\int_0^\pi \sin x\, dx\). This notation represents evaluating the area under the curve \(y = \sin x\) from \(0\) to \(\pi\).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is pivotal in solving problems involving definite integrals. When you evaluate an integral like \(\int_0^\pi \sin x\, dx\), this theorem provides the link between antiderivatives and definite integrals.

Here's how it works: if you find an antiderivative \(F(x)\) of a function \(f(x)\), then the definite integral of \(f\) from \(a\) to \(b\) is equal to \(F(b) - F(a)\). This effectively allows you to "jump" from a derivative back to the original function to find the exact area.

• In our example, the antiderivative of \(\sin x\) is \(-\cos x\).
• To apply the theorem, evaluate \(-\cos x\) at both \(x = \pi\) and \(x = 0\).

Subtract these two results: \([-\cos \pi] - [-\cos 0] = 2\), giving the total area.

Antiderivative

An antiderivative, or the "opposite" of a derivative, is essential to finding a definite integral's value. Discovering the antiderivative helps in identifying the original function whose derivative gives the function we're interested in.

For the function \(y = \sin x\), the antiderivative is \(-\cos x\). Why \(-\cos x\)? Because when you take the derivative of \(-\cos x\), you return to \(\sin x\).

• To find this, you reverse the differentiation process.
• It's crucial to understand that finding an antiderivative is only a piece of solving a definite integral.

Once you have the antiderivative, you can apply it through the process outlined by the Fundamental Theorem of Calculus to solve the integral within given limits. This is how we arrive at the result of 2 from our integral \(\int_0^\pi \sin x\, dx\).