Problem 13
Question
Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=3 x^{2}, \quad 0 \leq x \leq 2$$
Step-by-Step Solution
Verified Answer
The area under the curve is 8.
1Step 1: Understand the Problem
We need to find the area under the curve of the function \( f(x) = 3x^2 \) from \( x = 0 \) to \( x = 2 \). This means we will calculate the definite integral of the function over this interval.
2Step 2: Set Up the Integral
To find the area under \( f(x) = 3x^2 \) from \( x = 0 \) to \( x = 2 \), we set up the integral: \[ \int_{0}^{2} 3x^2 \, dx \]
3Step 3: Integrate the Function
We find the antiderivative of \( 3x^2 \). The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). So, the antiderivative of \( 3x^2 \) is \( x^3 \). This gives: \[ \int 3x^2 \, dx = x^3 + C \]
4Step 4: Evaluate the Definite Integral
Apply the limits of integration to evaluate the definite integral: \[ x^3 \bigg|_0^2 = (2)^3 - (0)^3 \] This simplifies to \( 8 - 0 = 8 \).
5Step 5: Conclusion
So, the area under the curve \( f(x) = 3x^2 \) from \( x = 0 \) to \( x = 2 \) is \( 8 \).
Key Concepts
Area Under a CurveAntiderivativeFundamental Theorem of Calculus
Area Under a Curve
Understanding the area under a curve is a fundamental concept in calculus. It helps us to determine the region enclosed by the graph of a function and the x-axis over a specific interval. This concept is often visualized as the "shade" between the curve of a function and the x-axis. In practical terms, this is used to calculate physical quantities, like distance or the amount of a substance over a given range.
To find the area under the curve of a function like \(f(x) = 3x^2\) from \(x = 0\) to \(x = 2\), you use the process of integration. Integration sums up an infinite number of tiny rectangles under the curve to find the total area. This is what makes the integral so powerful – it turns the area calculation into a mathematical procedure.
To find the area under the curve of a function like \(f(x) = 3x^2\) from \(x = 0\) to \(x = 2\), you use the process of integration. Integration sums up an infinite number of tiny rectangles under the curve to find the total area. This is what makes the integral so powerful – it turns the area calculation into a mathematical procedure.
- Integration involves setting up a definite integral, such as \(\int_{0}^{2} 3x^2 \, dx\), where the limits of integration (0 and 2) define the interval for calculating the area.
- After integrating, applying the limits helps determine the area between the curve and the x-axis across that interval.
Antiderivative
An antiderivative, often referred to as the indefinite integral, reverses the operation of differentiation. Thus, finding the antiderivative of a function is like asking, "what function would have given this derivative?"
In our example, with \(f(x) = 3x^2\), the antiderivative helps us build towards calculating the definite integral later. When a function is in the form \(x^n\), its antiderivative is \(\frac{x^{n+1}}{n+1}\). This means for \(3x^2\), the antiderivative is \(x^3\).
Moreover, antiderivatives include a constant \(C\) because adding any constant to a function doesn't change its derivative. In definite integrals, this constant cancels out as it's evaluated at the boundaries, making it practically unnecessary in this specific area context. However, understanding it plays a role in solving indefinite integrals.
In our example, with \(f(x) = 3x^2\), the antiderivative helps us build towards calculating the definite integral later. When a function is in the form \(x^n\), its antiderivative is \(\frac{x^{n+1}}{n+1}\). This means for \(3x^2\), the antiderivative is \(x^3\).
Moreover, antiderivatives include a constant \(C\) because adding any constant to a function doesn't change its derivative. In definite integrals, this constant cancels out as it's evaluated at the boundaries, making it practically unnecessary in this specific area context. However, understanding it plays a role in solving indefinite integrals.
- For \(3x^2\), the antiderivative is written as \(x^3 + C\), but when calculating definite integrals, \(C\) is omitted.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, making it a cornerstone of calculus. It states that if a function is continuous on an interval, the integral of a function over that interval can be calculated using its antiderivative.
This theorem essentially tells us that to find the definite integral, you need only find the antiderivative of the function and then evaluate it at the boundaries of the interval. Specifically, if \(F\) is an antiderivative of \(f\), then
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]This formula transforms the task of finding definite integrals into something manageable and formulaic.
This theorem essentially tells us that to find the definite integral, you need only find the antiderivative of the function and then evaluate it at the boundaries of the interval. Specifically, if \(F\) is an antiderivative of \(f\), then
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]This formula transforms the task of finding definite integrals into something manageable and formulaic.
- For \(f(x) = 3x^2\), with the antiderivative \(x^3\), we can use the theorem to find \( \int_{0}^{2} 3x^2 \, dx \) by evaluating \(x^3\) from 0 to 2.
- The result, \((2)^3 - (0)^3 = 8\), gives the area under \(f(x)\) from \(x = 0\) to \(x = 2\).
Other exercises in this chapter
Problem 12
Evaluate the limit, if it exists. $$\lim _{x \rightarrow 1} \frac{x^{3}-1}{x^{2}-1}$$
View solution Problem 12
Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=\sqrt{1+2 x} \text { at }(4,3)$$
View solution Problem 13
Find the limit. $$\lim _{x \rightarrow-\infty}\left(\frac{x-1}{x+1}+6\right)$$
View solution Problem 13
Evaluate the limit, if it exists. $$\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}$$
View solution