Problem 59
Question
Complete the following steps for the given integral and the given value of \(n\) a. Sketch the graph of the integrand on the interval of integration. b. Calculate \(\Delta x\) and the grid points \(x_{0}, x_{1}, \ldots, x_{n},\) assuming a regular partition. c. Calculate the left and right Riemann sums for the given value of \(n\). d. Determine which Riemann sum (left or right) underestimates the value of the definite integral and which overestimates the value of the definite integral. $$\int_{0}^{1}\left(x^{2}+1\right) d x$$
Step-by-Step Solution
Verified Answer
Answer: For the integral \(\int_{0}^{1}\left(x^{2}+1\right) d x\), the left Riemann sum (\(L_n\)) underestimates the value of the integral because it chooses the lower bound of the subinterval. In contrast, the right Riemann sum (\(R_n\)) overestimates the value of the integral since it chooses the upper bound of the subinterval. This behavior is due to the increasing nature of the function \(f(x)=x^2+1\) on the interval [0, 1].
1Step 1: Sketch the graph of the function on the interval [0, 1]
Plot the function \(f(x)= x^2 + 1\) on the interval [0, 1]. This is a simple parabola, with vertex at \((0, 1)\). The graph increases from left to right, with a minimum value at the vertex.
2Step 2: Calculate \(\Delta x\) and the grid points \(x_0, x_1, \ldots, x_n\)
We are given the interval [0, 1], and we need to partition it into \(n\) equal pieces for the Riemann sums computation. The size of each subinterval, or \(\Delta x\), can be calculated using the following formula:
$$\Delta x = \frac{b-a}{n}$$
For our case \(a=0\), \(b=1\), so our \(\Delta x\) will be:
$$\Delta x = \frac{1-0}{n} = \frac{1}{n}$$
The grid points are the values of \(x\) that are the endpoints of these subintervals, which are:
$$x_i = a + i *\Delta x$$
For our case, we find \(x_i\) by:
$$x_i = 0 + i * \frac{1}{n}$$
3Step 3: Calculate the left and right Riemann sums
For the left Riemann sum, we sum up the value of the function at the left endpoint of each subinterval, multiply it by \(\Delta x\), and sum all these products:
$$L_{n} = \Delta x\sum_{i=0}^{n-1}f(x_i)$$
For the right Riemann sum, we sum up the value of the function at the right endpoint of each subinterval:
$$R_{n} = \Delta x\sum_{i=1}^{n}f(x_i)$$
By substitution, we get:
$$L_{n} = \frac{1}{n}\sum_{i=0}^{n-1}\left[\left(0 + i * \frac{1}{n}\right)^2 + 1\right]$$
$$R_{n} = \frac{1}{n}\sum_{i=1}^{n}\left[\left(0 + i * \frac{1}{n}\right)^2 + 1\right]$$
4Step 4: Determine which Riemann sum underestimates and which overestimates the value of the definite integral
To determine which Riemann sum underestimates and which overestimates the definite integral, we need to look at the graph of the function \(f(x)=x^2+1\). We have established that it is an increasing parabola on the given interval [0, 1].
Since the function is increasing, the left Riemann sum will always choose the lower bound of the subinterval, and thus will underestimate the definite integral, while the right Riemann sum will choose the upper bound of the subinterval, and so it will overestimate the definite integral. Therefore, \(L_n\) underestimates the value of the integral and \(R_n\) overestimates the value of the integral.
Key Concepts
Definite IntegralPartition of IntervalRiemann Sum Underestimation and OverestimationCalculus
Definite Integral
The definite integral is a fundamental notion in calculus representing the signed area under a curve within a specified interval. It's symbolically represented as \( \int_a^b f(x)\,dx \) where \( f(x) \) is a function and \( [a,b] \) represents the interval over which the integration takes place.
Imagine a scenario where you track the distance covered by a cyclist over time. You can plot their speed against time to form a curve. The area under this curve between the start and end time gives the total distance traveled. In calculus, this concept is precisely what the definite integral calculates; how much 'stuff' accumulates between two points, whether it's distance, volume, or some other quantity. The beauty of the definite integral lies in its ability to account for functions that change rates or directions - like a cyclist speeding up or slowing down. In our exercise, the definite integral \( \int_0^1 (x^2+1)\,dx \) computes the area under the curve \( x^2+1 \) from \( x=0 \) to \( x=1 \) on a graph.
Imagine a scenario where you track the distance covered by a cyclist over time. You can plot their speed against time to form a curve. The area under this curve between the start and end time gives the total distance traveled. In calculus, this concept is precisely what the definite integral calculates; how much 'stuff' accumulates between two points, whether it's distance, volume, or some other quantity. The beauty of the definite integral lies in its ability to account for functions that change rates or directions - like a cyclist speeding up or slowing down. In our exercise, the definite integral \( \int_0^1 (x^2+1)\,dx \) computes the area under the curve \( x^2+1 \) from \( x=0 \) to \( x=1 \) on a graph.
Partition of Interval
To understand how we find the sum that approximates the area under a curve, we first need to discuss the concept of partitioning an interval. This involves dividing the interval into smaller 'pieces' to estimate the area under the curve within each piece. Consider partitioning as slicing a loaf of bread - you're segmenting it into smaller, more manageable pieces.
In our exercise, the interval \( [0,1] \) was chopped into \( n \) equal subintervals or partitions. Each partition has a width called \( \Delta x \) and the process of finding these partitions is crucial to calculating a Riemann sum. If we have a partitioned interval, we can more easily approximate the area underneath the curve of a function by adding up the areas of rectangles formed under each subinterval.
In our exercise, the interval \( [0,1] \) was chopped into \( n \) equal subintervals or partitions. Each partition has a width called \( \Delta x \) and the process of finding these partitions is crucial to calculating a Riemann sum. If we have a partitioned interval, we can more easily approximate the area underneath the curve of a function by adding up the areas of rectangles formed under each subinterval.
Riemann Sum Underestimation and Overestimation
Riemann sums are a bridge between algebra and calculus, helping us approximate areas under curves before we delve into the limitless precision of integrals. They give us a 'good enough' value, which becomes more accurate as we increase the number of partitions.
In our exercise, because the function \( f(x) = x^2 + 1 \) is always increasing in the interval \( [0,1] \) (becoming larger as \( x \) increases), important insights come into play: When you calculate a left Riemann sum, you use the left endpoint of each partition's subinterval; this will consistently capture a little less area than the actual curve - hence, an underestimation. Conversely, a right Riemann sum uses the right endpoint of each subinterval, grabbing a bit more area than exists under the curve, overestimating the true value. It's a balancing act - one side always takes a little less, the other a little more, like cutting slices of pie that are either too small or too large.
In our exercise, because the function \( f(x) = x^2 + 1 \) is always increasing in the interval \( [0,1] \) (becoming larger as \( x \) increases), important insights come into play: When you calculate a left Riemann sum, you use the left endpoint of each partition's subinterval; this will consistently capture a little less area than the actual curve - hence, an underestimation. Conversely, a right Riemann sum uses the right endpoint of each subinterval, grabbing a bit more area than exists under the curve, overestimating the true value. It's a balancing act - one side always takes a little less, the other a little more, like cutting slices of pie that are either too small or too large.
Calculus
Calculus, the mathematical study of change, is an indispensable tool in understanding the world around us. It's split into two branches: differential calculus, concerning rates of change and slopes of curves; integral calculus, involving accumulation of quantities and areas under or between curves.
It treats not just straight, predictable situations but also irregular ones - much like the uneven terrain of a mountain compared to a flat road. Through the use of derivatives and integrals, calculus allows us to solve complex problems in physics, engineering, economics, and many other fields. For example, in our Riemann sums exercise, we tapped into the power of integral calculus to understand how to approximate the value of a definite integral - a concept that can be applied to solve real-world problems such as finding the work done by a force or the fluid flow rate.
It treats not just straight, predictable situations but also irregular ones - much like the uneven terrain of a mountain compared to a flat road. Through the use of derivatives and integrals, calculus allows us to solve complex problems in physics, engineering, economics, and many other fields. For example, in our Riemann sums exercise, we tapped into the power of integral calculus to understand how to approximate the value of a definite integral - a concept that can be applied to solve real-world problems such as finding the work done by a force or the fluid flow rate.
Other exercises in this chapter
Problem 59
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