Question

Evaluate the following definite integrals. $$ \int_{-1}^{0}\left(1+x-x^{3}\right) d x $$

Short answer

Answer: The value of the definite integral is \(\frac{3}{4}\).

Step-by-step solution

Find the antiderivative of the function

To find the antiderivative of the function 1 + x - x^3, we will integrate each term individually with respect to x: $$ \int(1 + x - x^3) dx = \int 1 dx + \int x dx - \int x^3 dx $$ Now, find the antiderivative of each term: $$ \int 1 dx = x $$ $$ \int x dx = \frac{1}{2}x^2 $$ $$ \int x^3 dx = \frac{1}{4}x^4 $$ Then, combine the results to form the antiderivative, F(x): $$ F(x) = x + \frac{1}{2}x^2 - \frac{1}{4}x^4 $$

Apply the limits of integration

To find the definite integral, substitute the limits of integration into the antiderivative and subtract the results: $$ \int_{-1}^{0}\left(1+x-x^{3}\right) dx = F(0) - F(-1) $$ Now, evaluate the antiderivative at x = 0: $$ F(0) = 0 + \frac{1}{2}(0)^2 - \frac{1}{4}(0)^4 = 0 $$ And at x = -1: $$ F(-1) = -1 + \frac{1}{2}(-1)^2 - \frac{1}{4}(-1)^4 = -1 + \frac{1}{2} - \frac{1}{4} = \frac{-3}{4} $$

Calculate the final value of the integral

Now, find the definite integral by subtracting F(-1) from F(0): $$ \int_{-1}^{0}\left(1+x-x^{3}\right) dx = F(0) - F(-1) = 0 - \left(-\frac{3}{4}\right) = \frac{3}{4} $$ The value of the definite integral is \(\frac{3}{4}\).

Key concepts

Antiderivative

Understanding the concept of an antiderivative is crucial when dealing with definite integrals. An antiderivative of a function is essentially the reverse process of taking a derivative. It is a function whose derivative gives back the original function.

When you integrate each term of a function, you are essentially finding the antiderivative of that function. In the case of the function given in the exercise, we broke it down into simpler terms: the constant term 1, the linear term \(x\), and the cubic term \(x^3\).

• For \(1\), the antiderivative is \(x\), which essentially means the derivative of \(x\) with respect to \(x\) is \(1\).
• For \(x\), the antiderivative is \(\frac{1}{2}x^2\). This follows the power rule where you increase the power by 1 and then divide by the new exponent.
• For \(x^3\), the antiderivative is \(\frac{1}{4}x^4\). Again, applying the power rule helps in integrating this term.

The combination results in a single antiderivative \(F(x) = x + \frac{1}{2}x^2 - \frac{1}{4}x^4\) for the whole function. This function is used to evaluate the definite integral over a specific interval.

Limits of Integration

Limits of integration are the bounds between which you evaluate the definite integral. They are indicated as subscript and superscript next to the integral sign.

In this exercise, the limits of integration are \(-1\) and \(0\). These represent the starting and ending points on the x-axis over which we want to compute the total area under the curve defined by our function.

• First, we calculate the value of the antiderivative at the upper limit, \(x = 0\).
• Next, we calculate the value at the lower limit, \(x = -1\).
• Finally, we subtract the latter from the former to find the net area.

This illustrates that definite integrals provide a way to calculate accumulated values over a range of inputs. Understanding how to apply these limits accurately is vital in deriving the correct integral value.

Calculus Problems

Calculus can seem complex, but it is a powerful tool for solving a wide range of problems. At its core, calculus is about change and accumulation. Two primary operations in calculus — differentiation and integration — help in understanding these concepts.

In practical calculus problems involving definite integrals, like the one tackled here, the aim is often to find the total accumulated value over a specific range. When solving such problems, consider the following steps:

• Identify the function for integration, breaking it into simpler terms if necessary.
• Find the antiderivative for each term individually, before combining them.
• Apply the limits of integration to find specific values at endpoints.
• Subtract the values found to find the definite integral.

These steps form a foundation that can be applied to various problems in calculus. As you get more familiar with the process, tackling calculus problems becomes progressively less intimidating.