Question

Evaluate the Riemann sums \(L_{4}\) and \(R_{4}\) for the following functions over the specified interval. Compare your answer with the exact answer, when possible, or use a calculator to determine the answer. \(y=3 x^{2}-2 x+1\) over \([-1,1]\)

Short answer

The exact integral is 2, with \(L_4 = 5.25\) and \(R_4 = 3.25\).

Step-by-step solution

Identify the interval and subintervals

We need to find the Riemann sums for the function \(y = 3x^2 - 2x + 1\) over the interval \([-1, 1]\). We will use 4 subintervals, so \(n=4\). The width of each subinterval, \( \Delta x \), is calculated by:\[ \Delta x = \frac{1-(-1)}{4} = \frac{2}{4} = 0.5 \]This means each subinterval is 0.5 units wide.

Calculate the Left Riemann Sum (L_4)

The Left Riemann Sum, \(L_4\), uses the left endpoints of the subintervals. The left endpoints for our interval \([-1, 1]\) with 4 subintervals are \(-1, -0.5, 0, 0.5\). The function values at these points are:\[ f(-1) = 3(-1)^2 - 2(-1) + 1 = 6 \]\[ f(-0.5) = 3(-0.5)^2 - 2(-0.5) + 1 = 2.75 \]\[ f(0) = 3(0)^2 - 2(0) + 1 = 1 \]\[ f(0.5) = 3(0.5)^2 - 2(0.5) + 1 = 0.75 \]The Left Riemann Sum is given by:\[ L_4 = \Delta x [f(-1) + f(-0.5) + f(0) + f(0.5)] \]Substitute the values: \[ L_4 = 0.5 [6 + 2.75 + 1 + 0.75] = 5.25 \]

Calculate the Right Riemann Sum (R_4)

The Right Riemann Sum, \(R_4\), uses the right endpoints of the subintervals. The right endpoints for our interval \([-1, 1]\) with 4 subintervals are \(-0.5, 0, 0.5, 1\). The function values at these points are:\[ f(-0.5) = 2.75 \]\[ f(0) = 1 \]\[ f(0.5) = 0.75 \]\[ f(1) = 3(1)^2 - 2(1) + 1 = 2 \]The Right Riemann Sum is given by:\[ R_4 = \Delta x [f(-0.5) + f(0) + f(0.5) + f(1)] \]Substitute the values: \[ R_4 = 0.5 [2.75 + 1 + 0.75 + 2] = 3.25 \]

Find the Exact Integral Value

To find the exact value, we need to compute the definite integral of the function from -1 to 1:\[ \int_{-1}^{1} (3x^2 - 2x + 1) \, dx \]First, find the antiderivative:\[ \int (3x^2 - 2x + 1) \, dx = x^3 - x^2 + x + C \]Now evaluate from -1 to 1:\[ \left[ x^3 - x^2 + x \right]_{-1}^{1} = (1 - 1 + 1) - ((-1)^3 - (-1)^2 + (-1)) = 1 - (-1) = 2 \]

Compare Riemann Sums with Exact Value

The Left Riemann Sum \(L_4\) is 5.25, the Right Riemann Sum \(R_4\) is 3.25, and the exact value from the integral is 2. Both the Riemann sums are approximations and differ from the exact integral. \(L_4\) overestimates the true integral value, while \(R_4\) underestimates it.

Key concepts

Definite Integral

The definite integral is a fundamental concept in calculus that finds the total accumulation of quantities, such as areas under curves, between two specific limits on the x-axis. In the context of a polynomial function such as the one given, it allows us to determine the area enclosed by the curve of the graph from -1 to 1.Definite integrals are denoted using the integral sign, with the lower limit at the bottom and the upper limit at the top. For example: \[\int_{-1}^{1} (3x^2 - 2x + 1) \, dx\]Here, the definite integral gives us a precise value by considering every single point within the given interval. This results in an accurate representation of the area beneath the polynomial curve. Understanding definite integrals is crucial, as they provide exact answers compared to the approximations from methods like Riemann sums.

Antiderivative

An antiderivative, also known as the indefinite integral, is essentially the reverse of differentiation. It involves finding a function whose derivative is the original function. For the given polynomial, the antiderivative helps in evaluating the definite integral.The connection between antiderivatives and definite integrals is critical because the Fundamental Theorem of Calculus establishes that if you know the antiderivative of a function, you can find the area under the curve exactly by subtracting the antiderivative values at the endpoints of the interval.In our example:\[\int (3x^2 - 2x + 1) \, dx = x^3 - x^2 + x + C\]Here, we've determined the antiderivative of the function. When we evaluate this from -1 to 1, we extract the exact area under the curve within the interval.

Polynomial Function

Polynomial functions are mathematical expressions consisting of variables raised to whole number powers with coefficients. They can represent simple linear equations or more complex curves.Our function:\[y = 3x^2 - 2x + 1\]is a quadratic polynomial. Quadratic polynomials form parabolic shapes on a graph, and their behavior is dictated by the coefficients and powers of the variables.Key characteristics of polynomial functions:

• Degree: Determines the shape of the graph.
• Zeros: Points where the function equals zero.
• Symmetry: Some polynomials display symmetry about the y-axis or origin.

Understanding polynomial functions helps us analyze their graphs and predict changes in behavior over different intervals or sections of the graph.

Interval

An interval represents a portion of the real line and is defined by two endpoints. In calculus, intervals are crucial in discussing functions over specific domains.In our exercise, we are focusing on the interval \([-1, 1]\), which defines the section of the x-axis we are interested in. Everything between these two numbers is included in the interval, providing us with limits for integration and approximation techniques like Riemann sums.When working with intervals:

• Endpoints determine the starting and stopping points for evaluation.
• Intervals can be open, closed, or half-open, impacting whether the endpoints are included.
• They can signify where a function is examined or what section holds relevance for calculations.

Choosing the correct interval is essential since it shapes the calculations and outcomes of mathematical operations and analysis.