Problem 8
Question
Evaluate the integral by computing the limit of Riemann sums. $$\int_{0}^{3}\left(x^{2}+1\right) d x$$
Step-by-Step Solution
Verified Answer
The integral of the function \(x^{2} + 1\) from 0 to 3 is 12.
1Step 1: Definition of a Riemann sum
The definite integral of a function f from a to b is defined as the limit of the Riemann sum as the number of subintervals increases without bound. It can be calculated using the formula: \[\int_{a}^{b} f(x) dx = \lim_{{n \to \infty}} \sum_{i=1}^{n} f(x_{i}) \Delta x\] where \(x_{i} = a + i \Delta x\), \(\Delta x = \frac{b-a}{n}\), and \(n\) is the number of subintervals.
2Step 2: Apply the definition to our function
For our function \(f(x) = x^{2}+1\), from 0 to 3, we have: \[\int_{0}^{3} \left(x^{2}+1\right) dx = \lim_{{n \to \infty}} \sum_{i=1}^{n} f(x_{i}) \Delta x\] Where \(x_{i} = 0 + i \Delta x = i \Delta x\), \(\Delta x = \frac{3-0}{n} = \frac{3}{n}\), and \(n\) is the number of subintervals.
3Step 3: Simplify the Riemann sum
Transfer limit to each term separately: \[\lim_{{n \to \infty}} \sum_{i=1}^{n} f(x_{i}) \Delta x = \lim_{{n \to \infty}} \sum_{i=1}^{n} (\Delta x \cdot x_{i}^{2} + \Delta x)\] Now plug in values for \(x_{i}\) and \(\Delta x\), and simplify.
4Step 4: Evaluate the limit
Now calculate the sum, and then the limit. The sum of first \(n\) squares and the sum of first \(n\) numbers have known formulas which are \(\frac{n(n+1)(2n+1)}{6}\) and \(\frac{n(n+1)}{2}\) respectively. After putting these values and solving the limit, final answer is obtained.
Key Concepts
Definite IntegralLimit of a FunctionSum of SquaresCalculus Integration
Definite Integral
A definite integral represents the signed area under a curve plotted between two limits, from \( a \) to \( b \), on the x-axis. It is expressed as \( \int_{a}^{b} f(x) \, dx \).
To find a definite integral, you can use the concept of Riemann sums. Riemann sums involve breaking the area under the curve into smaller strips and calculating the sum of areas of these strips. As we increase the number of strips, their width becomes infinitesimally small, and the sum approaches the definite integral.
To find a definite integral, you can use the concept of Riemann sums. Riemann sums involve breaking the area under the curve into smaller strips and calculating the sum of areas of these strips. As we increase the number of strips, their width becomes infinitesimally small, and the sum approaches the definite integral.
- The definite integral gives you the net area, considering sections above the x-axis as positive and those below as negative.
- It helps find quantities like distance, area, and total accumulation over an interval.
Limit of a Function
In mathematics, the limit of a function describes the behavior of the function as its input approaches a particular point. It is not specifically bound to finite numbers; limits can be infinite, representing unbounded growth.
The idea is central in defining integrals and derivatives. When used with Riemann sums, as the number of subintervals increases indefinitely, the limit of the sum closely approximates the area under the curve, which is the integral.
To practically use limits:
The idea is central in defining integrals and derivatives. When used with Riemann sums, as the number of subintervals increases indefinitely, the limit of the sum closely approximates the area under the curve, which is the integral.
To practically use limits:
- Consider how a function behaves as its input gets closer and closer to a certain value.
- Limits help solve problems in calculus that involve instantaneous rates of change and accumulated quantities.
Sum of Squares
The sum of squares is a mathematical formula to add up squares of numbers in a sequence. It is used to simplify Riemann sums when calculating definite integrals. The formula for the sum of the squares of the first \( n \) integers is given by:
\[ \frac{n(n+1)(2n+1)}{6} \]
In the context of Riemann sums, the sum of squares formula helps to compute the integral of functions, particularly when the function involves terms like \( x^2 \).
\[ \frac{n(n+1)(2n+1)}{6} \]
In the context of Riemann sums, the sum of squares formula helps to compute the integral of functions, particularly when the function involves terms like \( x^2 \).
- This formula is especially useful in breaking down complex Riemann sums into easier components.
- Using it can significantly reduce calculation time and errors when evaluating integrals.
Calculus Integration
Calculus integration is a fundamental operation in calculus, opposite to differentiation. It involves finding the original function from its derivative, which is known as an antiderivative. Integration is used to calculate areas under curves, solve differential equations, and model various physical phenomena.
The two primary types of integrals are definite and indefinite integrals. Definite integrals compute net areas, while indefinite integrals represent a family of functions with an arbitrary constant.
The two primary types of integrals are definite and indefinite integrals. Definite integrals compute net areas, while indefinite integrals represent a family of functions with an arbitrary constant.
- Integration is essential in fields like physics, engineering, and economics to solve real-world problems.
- Techniques like substitution and integration by parts expand our ability to find complex integrals.
Other exercises in this chapter
Problem 8
Evaluate the indicated integral. $$\int \sin ^{3} x \cos x d x$$
View solution Problem 8
Use Part I of the Fundamental Theorem to compute each integral exactly. $$\int_{0}^{8}\left(\sqrt[3]{x}-x^{2 / 3}\right) d x$$
View solution Problem 8
Approximate the area under the curve on the given interval using \(n\) rectangles and the evaluation rules (a) left endpoint (b) midpoint (e) right endpoint. $$
View solution Problem 8
Write out all terms and compute the sums. $$\sum_{i=6}^{8}\left(i^{2}+2\right)$$
View solution