Question

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=x^{2}-x^{3} $$ $$ [0,1] $$

Short answer

The solution involves applying the Midpoint Rule on the given function to compute the approximate area under the curve and comparing it with the exact area, which gives an idea about the accuracy of the method for the given number of subdivisions, \(n=4\).

Step-by-step solution

Define the function, Interval and Midpoint Rule

The function is defined as \(f(x) = x^{2} - x^{3} \), and the interval is [0,1]. Divide the interval into \(n=4\) equal subintervals. Now, the Midpoint Rule is given by \[M_n = Δx [f(x_1) + f(x_2) ... + f(x_n)]\]. Here, \(Δx = \frac{(b-a)}{n}\) and \(x_i = a + \frac{1}{2}Δx + (i - 1)Δx \).

Find the Width, Midpoints and Values

First, calculate the width using \(Δx = \frac{(1-0)}{4}= 0.25 \). Then find the midpoints with \(x_i = 0 + \frac{1}{2}Δx + (i - 1)Δx \), so the midpoints can be calculated by \( x = 0 .125, 0.375, 0.625, 0.875 \). With these midpoints, we can find the function values \(f(x)\).

Calculate Approximate Area

Add the values and multiply the sum by the width, and you get the estimated value of the integral using the midpoint rule.

Find the exact Area

The integral over the interval \([0,1] \) of \(x^{2}-x^{3}\) yields the exact area under the curve. Compute this using the fundamental theorem of calculus.

Compare and Analyze

Compare the approximate area obtained in Step 3 with the exact area obtained in Step 4. This will show how accurate the midpoint rule is in this particular case for \( n=4 \).

Key concepts

Numerical Integration

When we talk about numerical integration, we're discussing methods to estimate the value of a definite integral, which is challenging to evaluate exactly. Instead of struggling with complex algebra, we use techniques like the Midpoint Rule, Trapezoidal Rule, or Simpson's Rule to approximate the area under the curve.

Imagine you're trying to fill a sandbox but you don't know the exact amount of sand needed. You could measure it out with a cup, estimating how many cups cover the bottom evenly. That's similar to how we use numerical integration - instead of cups, we use mathematical calculations to approximate the 'sand' that represents our area of interest. This method offers a practical approach to handle integrals of functions that might not have a simple antiderivative or when only data points are known, rather than a closed-form expression.

Definite Integrals

Definite integrals give us the total accumulation of a quantity as a function changes over a specific interval. To connect this with something tangible, think about filling a water trough. A definite integral would represent the total amount of water added to the trough as time passes, within a set period.

Formally, a definite integral is written as \( \int_{a}^{b} f(x) \,dx \) and calculates the net area under the curve of a function \( f(x) \) from \( a \) to \( b \) along the x-axis. The key here is 'net' area, which takes into account both above and below the x-axis. So, in essence, we're tallying up the continuous 'layers' of quantities across the interval to get our total.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is the Superman of calculus, connecting differentiation and integration into one powerful concept. It tells us that if we have a continuous function over an interval, the integral of that function corresponds to the net area under the curve between two points and also is the antiderivative evaluated at these points.

• Part 1 states that if \( F(x) \) is the antiderivative of \( f(x) \) over an interval \( [a, b] \) then \( \int_{a}^{b} f(x) \,dx = F(b) - F(a) \).
• Part 2 asserts that if you take the derivative of the integral of a function you get the original function back, kind of like hitting 'undo' on a change.

For our example, applying the theorem involves finding the antiderivative of \( f(x) \) and evaluating it at the endpoints of the interval to get the exact area under the curve.

Area Under a Curve

Visualizing the area under a curve is like looking at the shadow a line casts onto the x-axis. Intuitively, if a graph is above the x-axis, the area represents something positive, and if it's below, the area is negative.

A definite integral captures this 'shadow', which is the reason we use it to calculate the area under a curve. This can represent physical quantities such as distance, probability, or work depending on the context. In simple cases, we can find this area algebraically, but for more complex functions or when a function is best represented by a set of data points, numerical methods such as the Midpoint Rule are very useful tools – these allow us to approximate the area when an exact answer might be elusive or impractical to find.