Problem 171 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{-2}^{3}\left(x^{2}+3 x-5\right) d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \(-\frac{11}{6}\).

Step 1: Identify the Function to Integrate

The given integral is \( \int_{-2}^{3}(x^2 + 3x - 5) \, dx \). Identify the function that we need to integrate: \( f(x) = x^2 + 3x - 5 \).

Step 2: Find the Antiderivative

To apply the Fundamental Theorem of Calculus, first find the antiderivative \( F(x) \) of \( f(x) = x^2 + 3x - 5 \). The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \), the antiderivative of \( 3x \) is \( \frac{3x^2}{2} \), and the antiderivative of \( -5 \) is \( -5x \). Thus, the antiderivative \( F(x) \) is \( \frac{x^3}{3} + \frac{3x^2}{2} - 5x \).

Step 3: Evaluate the Antiderivative at the Upper Limit

Evaluate \( F(x) \) at the upper limit 3: \[ F(3) = \frac{3^3}{3} + \frac{3\cdot3^2}{2} - 5\cdot3 \] Simplifying, we get \[ F(3) = \frac{27}{3} + \frac{27}{2} - 15 = 9 + 13.5 - 15 = 7.5 \].

Step 4: Evaluate the Antiderivative at the Lower Limit

Evaluate \( F(x) \) at the lower limit -2: \[ F(-2) = \frac{(-2)^3}{3} + \frac{3\cdot(-2)^2}{2} - 5\cdot(-2) \] Simplifying, we get \[ F(-2) = \frac{-8}{3} + \frac{12}{2} + 10 = -\frac{8}{3} + 6 + 10 = \frac{28}{3} \].

Step 5: Calculate the Definite Integral

Apply the Fundamental Theorem of Calculus, Part 2, which states that \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]. Substitute the values obtained: \[ \int_{-2}^{3}(x^2 + 3x - 5) \, dx = F(3) - F(-2) \] \[ = 7.5 - \frac{28}{3} = \frac{45}{6} - \frac{56}{6} = -\frac{11}{6} \].

Key Concepts

Definite Integral

The definite integral of a function is a fundamental concept in calculus. It represents the area under the curve of a function between two specified points on the x-axis, known as the limits of integration. In essence, it calculates the net area, considering whether sections lie above or below the x-axis. Definite integrals are expressed in the form:

• \( \int_{a}^{b} f(x) \, dx \)

Where the values \(a\) and \(b\) are the lower and upper limits, respectively. This notation denotes that you integrate the function \(f(x)\) from \(x = a\) to \(x = b\). The utility of definite integrals extends beyond simple area calculations, as they are applied in physics, engineering, statistics, and more.
By using definite integrals, we can solve real-world problems involving total accumulation or consumption over an interval, such as finding the distance covered by a vehicle over time given its velocity.

Antiderivative

An antiderivative is essentially the "reverse" of taking a derivative. Finding an antiderivative means solving for a function whose derivative is the given function. In the exercise, the function \(f(x) = x^2 + 3x - 5\) requires determining an antiderivative to solve the integral.For each term in the function:

• The antiderivative of \(x^2\) is \(\frac{x^3}{3}\), since the derivative of \(\frac{x^3}{3}\) yields \(x^2\).
• The antiderivative of \(3x\) is \(\frac{3x^2}{2}\), reversing the derivative calculation.
• The antiderivative of \(-5\) is \(-5x\).

Thus, the complete antiderivative for this polynomial function is \(F(x) = \frac{x^3}{3} + \frac{3x^2}{2} - 5x\). The concept of antiderivatives is pivotal because it unlocks the ability to compute definite integrals using the Fundamental Theorem of Calculus.

Evaluate Integrals

Evaluating integrals, specifically definite integrals, involves applying the process of integration to find the exact area under a curve. By employing the Fundamental Theorem of Calculus, we can bypass the cumbersome task of summing an infinite number of small "slices". We do this by finding the antiderivative and computing its values at specified upper and lower limits.First, calculate the antiderivative \(F(x)\), then use it to evaluate \(F(b)\) and \(F(a)\), where \(b\) and \(a\) are the limits. The result is the difference:

• \(F(b) - F(a)\)

This effectively gives you the total "accumulation" or area within the boundaries set by \(a\) and \(b\). The evaluated integral from the example yields \(-\frac{11}{6}\), signifying the net signed area between \(-2\) and \(3\). This negative value indicates that there is more area below the x-axis than above, attributable to the function's range over the given limits.

Upper and Lower Limits

In integration, upper and lower limits are crucial as they define the segment of the x-axis over which the integration occurs. The lower limit, usually denoted as \(a\), is where the interval starts, while the upper limit \(b\) marks the end. These limits are the boundaries within which we calculate the area under a curve.In our example with the integral \(\int_{-2}^{3}(x^2 + 3x - 5) \, dx\), the lower limit is \(-2\) and the upper limit is \(3\). To solve the integral:

• Evaluate the antiderivative at the upper limit: \(F(3) = 7.5\).
• Evaluate the antiderivative at the lower limit: \(F(-2) = \frac{28}{3}\).

Finally, subtract \(F(a)\) from \(F(b)\) to find the net change. This illustrates not just calculating areas or sums but how these limits channel the integral's focus to a specific, finite section of the function.