Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{3} g(x) d x, \text { where } g(x)=\left\\{\begin{array}{ll} 3 x^{2}+4 x+1 & \text { if } x \leq 2 \\ 2 x+5 & \text { if } x > 2 \end{array}\right.$$

Short answer

Answer: The value of the definite integral of the piecewise function g(x) on the interval [1, 3] is 22.

Step-by-step solution

Split Integral into Two Parts

Since the function g(x) is piecewise-defined, we will break the integral into two parts, according to the definition of g(x): $$\int_{1}^{3} g(x) dx = \int_{1}^{2} (3x^2 + 4x + 1) dx + \int_{2}^{3} (2x + 5)dx $$

First Integral: Calculate Antiderivative

First, let's calculate the antiderivative of the function in the first integral, which is \((3x^2+4x+1)\) with respect to x: $$F(x) = \int(3x^2 + 4x + 1)dx = x^3 + 2x^2 + x + C$$

First Integral: Apply the Fundamental Theorem of Calculus

Now, let's apply the Fundamental Theorem of Calculus to evaluate the first integral: $$\int_{1}^{2} (3x^2 + 4x + 1) dx = F(2) - F(1) = (8 + 8 + 2 - C) - (1 + 2 + 1 - C) = 12$$

Second Integral: Calculate Antiderivative

Now we'll do the same for the second integral. The antiderivative of \((2x + 5)\) with respect to x is: $$G(x) = \int (2x + 5) dx = x^2 + 5x + D$$

Second Integral: Apply the Fundamental Theorem of Calculus

Applying the Fundamental Theorem of Calculus to the second integral, we get: $$\int_{2}^{3} (2x + 5) dx = G(3) - G(2) = (9 + 15 + D) - (4 + 10 + D) = 10$$

Combine the Two Integrals

Now, we combine the two integrals to get the final result: $$\int_{1}^{3} g(x) dx = \int_{1}^{2} (3x^2 + 4x + 1) dx + \int_{2}^{3} (2x + 5)dx = 12 + 10 = 22$$

Key concepts

Piecewise Functions

Piecewise functions are like a series of instructions that only apply to certain parts of an equation. In math, these functions are defined by different expressions depending on the value of the input variable, usually denoted as \(x\). For instance, consider a function that behaves one way for \(x \leq 2\) and differently for \(x > 2\), just like the given function \(g(x)\) in the exercise.

• When \(x \leq 2\), the function \(g(x) = 3x^2 + 4x + 1\).
• When \(x > 2\), the function \(g(x) = 2x + 5\).

By breaking the function into these two parts, it helps us manage and solve different behaviors in a straightforward way. This approach is crucial when evaluating definite integrals of such piecewise functions, ensuring each segment gets the correct calculations.

Definite Integrals

Definite integrals are a way to find the "total accumulation" or "net area" under a curve over a specific interval. This process is carried out using limits of integration, which indicate the points under which calculations will take place. In the exercise, the definite integral from 1 to 3 for the function \(g(x)\) was calculated.
Here’s how it works:

• The limits of integration are the numbers at the bottom and top of the integral symbol, representing the interval over which the function is evaluated.
• For piecewise functions, the integral is split at the breaks in the function to account for different expressions over different intervals.

Using the Fundamental Theorem of Calculus, the definite integral of each part was calculated separately and then added together. This helped to find the exact area under the curve, or in this case, the total value of the function \(g(x)\) from 1 to 3, resulting in a final answer of 22.

Antiderivatives

Antiderivatives are functions that "undo" the process of differentiation. For any given function, its antiderivative returns the original function whose derivative produced the given function. In simpler terms, if you differentiate an antiderivative, you end up with the original function again.
In the exercise, we found antiderivatives for each piece of our piecewise function:

• For \(3x^2 + 4x + 1\), the antiderivative is \(x^3 + 2x^2 + x + C\).
• For \(2x + 5\), the antiderivative is \(x^2 + 5x + D\).

These antiderivatives allowed the application of the Fundamental Theorem of Calculus, which states that if you have an antiderivative, you can compute definite integrals by evaluating the antiderivatives at the bounds of integration and finding their difference. This calculation efficiently captures the total result of a piecewise function over a defined interval.