Question

Evaluate the definite integrals using the definition, as in Examples 3 and \(4 .\) $$ \int_{-2}^{1}(2 x+\pi) d x $$

Short answer

\(-3 + 3\pi\)

Step-by-step solution

Identify the Function and Limits

The given definite integral is \( \int_{-2}^{1}(2x+\pi) \, dx \). Here, the function under the integral is \( f(x) = 2x + \pi \), and the limits of integration are \( -2 \) to \( 1 \).

Find the Antiderivative

To evaluate the definite integral, we first need to find an antiderivative of the function \( f(x) = 2x + \pi \). The antiderivative of \( 2x \) is \( x^2 \) and the antiderivative of \( \pi \) is \( \pi x \). Thus, an antiderivative of \( f(x) \) is \( F(x) = x^2 + \pi x \).

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \), where \( F(x) \) is any antiderivative of \( f(x) \). Thus, we calculate \( F(1) \) and \( F(-2) \):\[ F(1) = 1^2 + \pi \times 1 = 1 + \pi \]\[ F(-2) = (-2)^2 + \pi \times (-2) = 4 - 2\pi \]

Calculate the Definite Integral

Now, substitute the values obtained in previous steps into the expression \( F(1) - F(-2) \):\[ F(1) - F(-2) = (1 + \pi) - (4 - 2\pi) \]Simplify the expression:\[ = 1 + \pi - 4 + 2\pi \]\[ = -3 + 3\pi \]

Conclude

Thus, the value of the definite integral \( \int_{-2}^{1}(2x+\pi) \, dx \) is \( -3 + 3\pi \).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It provides a method to evaluate definite integrals by relating them to the antiderivatives of functions. This theorem states that if you have a continuous function on a closed interval \([a, b]\), and if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral of \(f(x)\) from \([a, b]\) is calculated as \(F(b) - F(a)\).
For example, in the problem \( \int_{-2}^{1}(2x+\pi) \, dx \), the function is \(f(x) = 2x + \pi\). To solve this using the Fundamental Theorem, we first need to find an antiderivative \(F(x)\) for \(f(x)\). Once we have it, the theorem allows us to calculate the exact area under the curve from the lower limit \(-2\) to the upper limit \(1\).
This direct application of the antiderivative saves extensive manual summation, illustrating why the theorem is such a powerful tool in calculus.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. Finding an antiderivative for a function means determining another function whose derivative gives you the original function. This is central to computing definite integrals using the Fundamental Theorem of Calculus.
In the example \(f(x) = 2x + \pi\), we find an antiderivative by integrating each term separately. The antiderivative of \(2x\) is \(x^2\), visualizing the power rule in integration. The term \(\pi\) behaves as a constant during integration, meaning its antiderivative is simply \(\pi x\).
The combined antiderivative is \(F(x) = x^2 + \pi x\). With this antiderivative, we apply it in the context of the limits to solve for the definite integral. This approach highlights how antiderivatives simplify more complex integral calculations by leveraging the reverse process of differentiation.

Evaluating Integrals

Evaluating integrals involves determining the exact value of an integral. For definite integrals, this often includes three key steps: identifying the original function \(f(x)\), finding its antiderivative \(F(x)\), and applying the Fundamental Theorem of Calculus.
In the given solution, the process starts with recognizing \(f(x) = 2x + \pi \) with limits from \(-2\) to \(1\). After determining the antiderivative \(F(x) = x^2 + \pi x\), we substitute the limits into the expression \((F(b)-F(a))\) to find the value of the definite integral.
This process involves careful substitution and simplification. The resulting calculation for this particular integral gives us \(-3 + 3\pi\), representing the net area between the function and the x-axis over the specified interval. Such calculations are crucial for accurately capturing the behavior and accumulation of functions over specific ranges.

Limits of Integration

The limits of integration play a pivotal role in defining the range over which a definite integral is evaluated. These limits provide the boundaries a and b on the interval \([a, b]\) within which the integration occurs.
In the problem \( \int_{-2}^{1}(2x+\pi) \, dx \), the limits of integration are \(-2\) and \(1\). These tell us that we are interested in the area under \(f(x) = 2x + \pi\) from \(x = -2\) to \(x = 1\).
The use of these limits in the Fundamental Theorem of Calculus directs how we substitute the antiderivative values. The result \(F(1) - F(-2)\) precisely represents the cumulative value of the function across this set interval, making limits one of the core elements to evaluate the definite integral accurately.