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Magazine Chapter 6: Problem 71
Find \(\int_{-2}^{2} \frac{x^{3}}{3} d x\)
Short Answer
Expert verified
The integral evaluates to 0.
Step by step solution
01
Identify the integral type
The given integral is a definite integral of the function \( \frac{x^3}{3} \) over the interval \([-2, 2] \). We will first find an antiderivative and then use the Fundamental Theorem of Calculus.
02
Find the antiderivative
The function \( \frac{x^3}{3} \) can be expressed as \( \frac{1}{3} x^3 \). The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \). Thus, the antiderivative of \( \frac{1}{3} x^3 \) is \( \frac{1}{3} \cdot \frac{x^4}{4} = \frac{x^4}{12} \).
03
Apply the limits of integration
Apply the limits of integration \(-2\) and \(2\) to the antiderivative \( \frac{x^4}{12} \). This means computing \( \frac{(2)^4}{12} - \frac{(-2)^4}{12} \).
04
Calculate the definite integral
First, calculate \( \frac{(2)^4}{12} = \frac{16}{12} = \frac{4}{3} \). Then, calculate \( \frac{(-2)^4}{12} = \frac{16}{12} = \frac{4}{3} \). The difference between these two evaluations is \( \frac{4}{3} - \frac{4}{3} = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone of calculus that links the concept of differentiation with integration, two main operations in calculus. It comprises two main parts:
- The first part states that if a function is continuous over an interval \[a, b\], then the function has an antiderivative on that interval, and the definite integral of the function from \ a \ to \ b \ can be computed using that antiderivative.
- The second part asserts that the derivative of the integral of a function is the function itself, bridging a deep connection between the instantaneous rate of change and accumulation of quantities.
Antiderivative
The concept of an antiderivative is crucial when working with integrals. An antiderivative of a function is another function whose derivative is the original function. For example, in the exercise, we begin with the function \(\frac{x^3}{3}\). To integrate this function, we need to find its antiderivative.Here are the steps to find the antiderivative:
- Recognize the need to undo the derivative by increasing the power of each term and then dividing by the new power.
- The antiderivative of \(x^3\) is \(\frac{x^4}{4}\). For \(\frac{x^3}{3}\), you multiply the antiderivative of \(x^3\) by the constant \(\frac{1}{3}\), resulting in \(\frac{x^4}{12}\).
Limits of Integration
The limits of integration are the two numbers at the bounds of a definite integral. They define the interval over which you accumulate the area under the graph of the function. In our exercise, the limits are \(-2\) and \(2\).When you substitute these limits into the antiderivative, you essentially calculate the net area under the curve between these two points. Here's how it works:
- First, evaluate the antiderivative at the upper limit: \(\frac{(2)^4}{12}\).
- Then, evaluate it at the lower limit: \(\frac{(-2)^4}{12}\).
- Subtract the value of the function at the lower limit from the value at the upper limit to get the definite integral.
