Question

Evaluate the following integrals. \(\int_{-1}^{1}\left(x^{3}-2 x^{2}+4 x\right) d x\)

Short answer

The integral evaluates to \(-\frac{7}{2}\).

Step-by-step solution

Identify the Antiderivative

To evaluate the integral \(\int_{-1}^{1}(x^3 - 2x^2 + 4x)\, dx\), start by finding the antiderivative of the integrand. The antiderivative of \(x^3\) is \(\frac{x^4}{4}\), the antiderivative of \(2x^2\) is \(-\frac{2x^3}{3}\), and the antiderivative of \(4x\) is \(2x^2\). Therefore, the antiderivative of the entire function is \(\frac{x^4}{4} - \frac{2x^3}{3} + 2x^2\).

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, evaluate the antiderivative from \(-1\) to \(1\). This means calculating \[\left[ \frac{x^4}{4} - \frac{2x^3}{3} + 2x^2 \right]_{-1}^{1}.\]

Evaluate at the Upper Limit

Substitute \(x = 1\) into the antiderivative: \(\frac{(1)^4}{4} - \frac{2(1)^3}{3} + 2(1)^2 = \frac{1}{4} - \frac{2}{3} + 2\). Simplify the expression: \(\frac{1}{4} - \frac{8}{12} + \frac{24}{12}\), which simplifies to \(\frac{1}{4} - \frac{2}{3} + 2 = \frac{15}{12} = \frac{5}{4}\).

Evaluate at the Lower Limit

Substitute \(x = -1\) into the antiderivative: \(\frac{(-1)^4}{4} - \frac{2(-1)^3}{3} + 2(-1)^2 = \frac{1}{4} + \frac{2}{3} + 2\). Simplify the expression: \(\frac{1}{4} + \frac{8}{12} + \frac{24}{12}\), which becomes \(\frac{57}{12} = \frac{19}{4}\).

Calculate the Definite Integral

Subtract the result from the lower limit from the result at the upper limit: \(\frac{5}{4} - \frac{19}{4} = -\frac{14}{4} = -\frac{7}{2}\). Therefore, \(\int_{-1}^{1}(x^3 - 2x^2 + 4x)\, dx = -\frac{7}{2}\).

Key concepts

Antiderivatives

Before diving into definite integrals like \(\int_{-1}^{1}(x^3 - 2x^2 + 4x)\, dx\), it's crucial to understand the concept of antiderivatives. This is because the first step in evaluating such integrals involves finding the antiderivative of the integrand.
Think of an antiderivative as essentially the reverse of differentiation. While differentiation involves breaking down a function to find its instantaneous rate of change, an antiderivative involves building a function that has a given derivative.
For example, if the derivative of \(x^4/4\) is \(x^3\), then \(x^4/4\) is an antiderivative of \(x^3\). Similarly, the antiderivative of \(2x^2\) becomes \(-\frac{2x^3}{3}\), and for \(4x\), it is \(2x^2\). Overall, finding an antiderivative is about reversing the derivative step.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a pivotal concept that bridges the gap between differentiation and integration. It states that if a function is continuous on an interval, then the definite integral over that interval can be found using its antiderivative.
In simpler terms, if you know the antiderivative of a function, you can use it to calculate the total accumulated value of the function over a specific interval. This saves significant effort as compared to summing infinitely many infinitesimal changes.
For instance, after finding the antiderivative \(\frac{x^4}{4} - \frac{2x^3}{3} + 2x^2\), you use it in the definite integral \(\int_{-1}^{1}(x^3 - 2x^2 + 4x)\, dx\) to evaluate the accumulated change over the interval \([-1, 1]\). The Fundamental Theorem simplifies this entire process and underscores how differentiation and integration are closely linked.

Evaluating Integrals

Once we have the antiderivative of a function, it's time to evaluate the integral using the limits of the definite integral. This involves substituting the upper and lower limits into the antiderivative and taking the difference.
For the integral \(\int_{-1}^{1}(x^3 - 2x^2 + 4x)\, dx\), you'll plug in the upper limit first, which in this case is \(1\), into \(\frac{x^4}{4} - \frac{2x^3}{3} + 2x^2\) to get \(\frac{5}{4}\).
Next, substitute the lower limit, \(-1\), which results in \(\frac{19}{4}\). The final value of the definite integral is found by subtracting these two results: \(\frac{5}{4} - \frac{19}{4} = -\frac{7}{2}\).
This step-by-step substitution and difference-taking process is essential for evaluating definite integrals. It provides a clear and systematic way to calculate the net change of a function over a given interval. The interplay of finding antiderivatives and applying the Fundamental Theorem demonstrates the beauty and utility of calculus.