Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\pi} f(x) d x, \text { where } f(x)=\left\\{\begin{array}{ll} \sin x+1 & \text { if } x \leq \pi / 2 \\ 2 \cos x+2 & \text { if } x > \pi / 2 \end{array}\right.$$

Short answer

$$f(x) = \begin{cases} \sin x + 1,\quad &x \le \pi / 2 \\ 2 \cos x + 2,\quad &x > \pi / 2 \end{cases} $$ Using the Fundamental Theorem of Calculus, we find that the definite integral of the piecewise function from 0 to 𝜋 is: $$\int_{0}^{\pi} f(x) dx = \frac{3\pi}{2} + 1$$

Step-by-step solution

Identify the integrand for both pieces of the function

Since we are given a piecewise function, we should identify the integrand for both pieces: $$f(x) = \begin{cases} \sin x + 1,\quad &x \le \pi / 2 \\ 2 \cos x + 2,\quad &x > \pi / 2 \end{cases} $$

Split the integral based on the given conditions

We will now split the integral into two parts, according to the conditions of the piecewise function: $$\int_{0}^{\pi} f(x) dx = \int_{0}^{\pi/2} (\sin x + 1) dx + \int_{\pi/2}^{\pi} (2\cos x +2) dx$$

Apply the Fundamental Theorem of Calculus for each integral

Now, we will find the antiderivative for each function and apply the Fundamental Theorem of Calculus. For the first integral: $$F_1(x) = \int (\sin x + 1) dx = -\cos x + x + C_1$$ For the second integral: $$F_2(x) = \int (2\cos x + 2) dx = 2\sin x + 2x + C_2$$ Now, applying the theorem: $$\int_{0}^{\pi/2} (\sin x + 1) dx = F_1\parens{\frac{\pi}{2}} - F_1(0) = -\cos\parens{\frac{\pi}{2}} + \frac{\pi}{2} - \parens{-\cos(0) + 0}$$ $$\int_{\pi/2}^{\pi} (2\cos x +2) dx = F_2(\pi) - F_2\parens{\frac{\pi}{2}} = 2\sin{\pi} + 2\pi - \parens{2\sin\parens{\frac{\pi}{2}} + 2\cdot\frac{\pi}{2}}$$

Calculate and sum the integrals

Now, it's time to calculate the individual integrals and sum them for the overall result: $$\int_{0}^{\pi/2} (\sin x + 1) dx = -\cos\parens{\frac{\pi}{2}} + \frac{\pi}{2} - \parens{-\cos(0) + 0} = -0 + \frac{\pi}{2} - \parens{-1 + 0} =\frac{\pi}{2} + 1$$ $$\int_{\pi/2}^{\pi} (2\cos x +2) dx = 2\sin{\pi} + 2\pi - \parens{2\sin\parens{\frac{\pi}{2}} + 2\cdot\frac{\pi}{2}} = 0 + 2\pi - \parens{2 + \pi} =\pi$$ Now, sum the results: $$\int_{0}^{\pi} f(x) dx = \parens{\frac{\pi}{2} + 1} + \pi = \frac{3\pi}{2} + 1$$ The definite integral of the piecewise function is: $$\int_{0}^{\pi} f(x) dx = \frac{3\pi}{2} + 1$$

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) plays a crucial role in connecting differentiation and integration, two cornerstone concepts in calculus.

It consists of two parts: the first part gives us the derivative of the integral function, and the second part, often used in computing definite integrals, states that if we know an antiderivative of a function, we can evaluate the definite integral of the function over an interval [a, b] by taking the difference of the values of the antiderivative at the endpoints.

Mathematically, if F(x) is the antiderivative of f(x), then:
\[\int_{a}^{b} f(x) dx = F(b) - F(a)\]
This powerful tool means that instead of finding the area under the curve by summing infinitesimally small rectangles (which can be computationally complex), we can simply evaluate the antiderivative at the upper and lower bounds of our interval to find the difference.

In the context of our example with the piecewise function f(x), the FTC simplifies the process: after finding the antiderivatives for each piece of the function, the definite integral from 0 to \pi can be found by evaluating the antiderivatives at \pi/2 and \pi, and subtracting the value of the antiderivative evaluated at 0.

Piecewise functions

Piecewise functions are mathematical creatures that may behave differently depending on the input value. These functions can be broken down into pieces, each defined by a particular sub-domain of the entire domain.

In essence, a piecewise function is like a chameleon, changing its expression based on where you are in the domain. In the case of our integral exercise:
\[f(x) = \begin{cases}\sin x + 1, & x \le \pi / 2 \2 \cos x + 2, & x > \pi / 2\end{cases}\]
f(x) changes its formula at \pi/2. To deal with piecewise functions in integration, you split the integral into segments where each piece of the function applies. This means evaluating separate integrals for each segment and then combining their results for the final answer.

The ability to handle such functions is necessary because not all relationships in mathematics (and the natural world!) are uniform across their entire domain. As a result, piecewise functions are a more realistic model for representing complex phenomena.

Antiderivative

The antiderivative, also known as the indefinite integral, represents a function F(x) whose derivative is the function f(x). When we take the derivative of F(x), we get back to our original function f(x).

Finding the antiderivative is crucial for evaluating definite integrals – you can think of it as discovering the original quantity when given the rate of change.
\[F(x) = \int f(x) dx\]
When working with the Fundamental Theorem of Calculus, we leverage antiderivatives to evaluate definite integrals efficiently. It's essential to remember that antiderivatives are not unique; they include an arbitrary constant, C, because derivative of a constant is zero.

For example, the antiderivative of f(x) = \sin(x) is F(x) = -\cos(x) + C. In the context of definite integrals, this constant cancels out because we take the difference between the upper and lower bounds, rendering C irrelevant and simplifying the evaluation process.

Understanding antiderivatives can uncover a deeper appreciation for how the accumulation of quantities, like distance traveled over time given a speed function, translates into the practical calculation of total change.