Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\pi}(1-\sin x) d x$$

Short answer

Answer: The value of the definite integral is \(\pi-2\).

Step-by-step solution

Find the antiderivative of the given function

To find the antiderivative of \((1-\sin x)\), we need to find the antiderivative of each term separately. The function is split into two parts: 1 and \(-\sin x\) 1. For 1, the antiderivative is x, as the derivative of x is 1. 2. For \(-\sin x\), the antiderivative is \(\cos x\), as the derivative of \(\cos x\) is \(-\sin x\). Now, we combine these two antiderivatives to get the antiderivative of \((1-\sin x)\): $$F(x) = x + \cos x$$

Evaluate F(b) and F(a)

Now, we need to evaluate F(x) at the limits a and b, which are 0 and \(\pi\). So we need to find F(\(\pi\)) and F(0). 1. F(\(\pi\)) = x + \(\cos x\) = \(\pi+\cos(\pi)\) = \(\pi -1\) 2. F(0) = x + \(\cos x\) = \(0+\cos(0)\) = 1

Apply the Fundamental Theorem of Calculus

Now we can apply the Fundamental Theorem of Calculus to find the definite integral: $$\int_{0}^{\pi}(1-\sin x) d x = F(b) - F(a)$$ Substituting the values we found in step 2: $$\int_{0}^{\pi}(1-\sin x) d x = (\pi-1) - 1 = \pi-2$$ The definite integral of the function from 0 to \(\pi\) is \(\pi-2\).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a beautiful bridge between two central areas of calculus: differentiation and integration. It consists of two main parts. The part that is most relevant to solving definite integrals states that if you have a continuous function and you want to find the definite integral over an interval \[a, b\], you can use its antiderivative.
Here's how it works in simple terms:

• Find an antiderivative of the function, which is essentially a function whose derivative yields your original function.
• Evaluate this antiderivative at the upper limit of the integration interval (b) and at the lower limit (a).
• Subtract the value at the lower limit from the value at the upper limit.

This process allows you to find the accumulated area under the curve of the function from a to b. It's like "undoing" the process of differentiation, and that's why it involves antiderivatives.

Antiderivative

Finding the antiderivative is a crucial step when using the Fundamental Theorem of Calculus. An antiderivative of a function is another function whose derivative is the original function.
For example, consider the function \((1 - \sin x)\). We find the antiderivative of each part separately:

• The antiderivative of 1 is simply \(x\). When you differentiate \(x\), you get 1.
• The antiderivative of \(-\sin x\) is \(\cos x\), because the derivative of \(\cos x\) is \(-\sin x\).

Thus, the antiderivative of \((1 - \sin x)\) is \(F(x) = x + \cos x\). This is a key step since it allows you to compute the definite integral over the desired interval.

Evaluation of Limits

After finding the antiderivative, you evaluate it at specific points, known as the limits of integration. These limits are the boundaries of the area under the curve you are calculating.
In our problem, the limits are 0 and \(\pi\). Here's what you do next:

• Evaluate the antiderivative \(F(x) = x + \cos x\) at the upper limit \(b = \pi\): \(F(\pi) = \pi + \cos(\pi) = \pi - 1\).
• Evaluate at the lower limit \(a = 0\): \(F(0) = 0 + \cos(0) = 1\).

Subtract the value at the lower limit from the value at the upper limit: \(F(\pi) - F(0) = (\pi - 1) - 1 = \pi - 2\). This gives the result of the definite integral, which represents the total accumulation of the function from 0 to \(\pi\).