Problem 41
Question
Compute the indefinite integrals. $$ \int\left(\frac{1}{3} x^{2}-\frac{1}{2} x\right) d x $$
Step-by-Step Solution
Verified Answer
\( \int \left( \frac{1}{3} x^2 - \frac{1}{2} x \right)\, dx = \frac{1}{9}x^3 - \frac{1}{4}x^2 + C \)
1Step 1: Identify the Terms to Integrate
The integral is \( \int \left( \frac{1}{3} x^{2} - \frac{1}{2} x \right)\, dx \). There are two terms to integrate separately: \( \frac{1}{3} x^2 \) and \( -\frac{1}{2} x \).
2Step 2: Apply the Power Rule to the First Term
For the first term \( \frac{1}{3} x^2 \), apply the power rule of integration, \( \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \). So, \( \int \frac{1}{3} x^2\,dx = \frac{1}{3} \cdot \frac{x^{3}}{3} = \frac{1}{9}x^3 \).
3Step 3: Integrate the Second Term
For the second term \( -\frac{1}{2} x \), apply the power rule: \( \int x^1\,dx = \frac{x^{2}}{2} \). Thus, \( \int -\frac{1}{2} x\,dx = -\frac{1}{2} \cdot \frac{x^{2}}{2} = -\frac{1}{4}x^2 \).
4Step 4: Combine the Integrated Terms
Combine the results from steps 2 and 3 to form the indefinite integral, adding an arbitrary constant \( C \) at the end: \( \int \left( \frac{1}{3} x^2 - \frac{1}{2} x \right)\, dx = \frac{1}{9}x^3 - \frac{1}{4}x^2 + C \).
Key Concepts
Power Rule of IntegrationUnderstanding IntegrationThe Role of the Arbitrary Constant
Power Rule of Integration
The power rule is an essential technique in integration, allowing us to integrate polynomial functions with ease. It states that if you need to integrate a function in the form of \( x^n \), where \( n \) is not equal to -1, you can use the formula:
In our specific example, for the term \( \frac{1}{3}x^2 \), we enhanced the exponent from 2 to 3, resulting in the integrated term \( \frac{1}{9}x^3 \). Similarly, \( -\frac{1}{2}x \) is simplified by raising the power from 1 to 2, which gives us \( -\frac{1}{4}x^2 \). Both solutions neatly fit into the power rule framework.
- \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
In our specific example, for the term \( \frac{1}{3}x^2 \), we enhanced the exponent from 2 to 3, resulting in the integrated term \( \frac{1}{9}x^3 \). Similarly, \( -\frac{1}{2}x \) is simplified by raising the power from 1 to 2, which gives us \( -\frac{1}{4}x^2 \). Both solutions neatly fit into the power rule framework.
Understanding Integration
Integration can be seen as the reverse operation of differentiation. While differentiation divides a composite function to find its rate of change, integration collects all parts to find the original function or area under a curve.
- There are two types of integration: indefinite and definite.
- Indefinite integration yields a family of functions and includes an arbitrary constant \( C \).
- Definite integration returns a specific numerical value and doesn't include \( C \).
The Role of the Arbitrary Constant
In the world of indefinite integrals, the arbitrary constant \( C \) plays a crucial role. It emerges from the fact that differentiation of a constant gives zero, meaning any number added to a function will disappear when we differentiate back.
Thus, after performing indefinite integration, every solution includes a \( C \) because it represents all possible vertical shifts of the original function that could have existed. This constant accounts for the unknown initial conditions.
For instance, in our final answer of \( \int \left( \frac{1}{3} x^2 - \frac{1}{2} x \right)\, dx = \frac{1}{9}x^3 - \frac{1}{4}x^2 + C \), \( C \) could be any constant value. It ensures our fully integrated expression can shift up or down on the graph to match any initial configuration of the function.
Thus, after performing indefinite integration, every solution includes a \( C \) because it represents all possible vertical shifts of the original function that could have existed. This constant accounts for the unknown initial conditions.
For instance, in our final answer of \( \int \left( \frac{1}{3} x^2 - \frac{1}{2} x \right)\, dx = \frac{1}{9}x^3 - \frac{1}{4}x^2 + C \), \( C \) could be any constant value. It ensures our fully integrated expression can shift up or down on the graph to match any initial configuration of the function.
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