Problem 59 - Calculus Single Variable

Question

Let \(\Delta x=(b-a) / 2 .\) Use the Mean Value Theorem to find points \(s_{1}\) in \([a,(a+b) / 2]\) and \(s_{2}\) in \([(a+b) / 2, b]\) such that the Riemann sum $$ \mathcal{R}\left(f,\left\\{s_{1}, s_{2}\right\\}\right)=f\left(s_{1}\right) \cdot \Delta x+f\left(s_{2}\right) \cdot \Delta x $$ is equal to the Riemann integral of \(f\) over \([a, b]\). $$ f(x)=4 x^{3} $$

Step-by-Step Solution

Answer

\(s_1 = (0.03125)^{1/3}, s_2 = (0.46875)^{1/3}\).

Step 1: Identify the Integral and Subintervals

To solve this problem, start by identifying the Riemann integral of the function over the interval \([a, b]\), which is \([0, 1]\) since typically, a standard example starts with these default values unless specified otherwise. The integral of \(f(x) = 4x^3\) over \([a, b]\) will need to be found.

Step 2: Calculate the Riemann Integral

Calculate the integral of \(f(x) = 4x^3\) over \([0, 1]\). This involves finding \[ \int_{0}^{1} 4x^3 \, dx \].Use the power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Therefore: \[ \int_{0}^{1} 4x^3 \, dx = \left[ \frac{4x^4}{4} \right]_0^1 = \left[ x^4 \right]_0^1 = 1^4 - 0^4 = 1. \]

Step 3: Define \(\Delta x\) and Subintervals

Define \(\Delta x\) as \((b-a)/2\). Hence, \(\Delta x = (1-0)/2 = 0.5\). The subintervals are \([0, 0.5]\) and \([0.5, 1]\).

Step 4: Apply the Mean Value Theorem

The Mean Value Theorem for integrals states that there exists \(s_1\) in \([0, 0.5]\) such that \( \int_{0}^{0.5} 4x^3 \, dx = 4(s_1)^3 \times 0.5 \). Similarly, there exists \(s_2\) in \([0.5, 1]\) such that \( \int_{0.5}^{1} 4x^3 \, dx = 4(s_2)^3 \times 0.5 \).

Step 5: Solve for \(s_1\)

Calculate \( \int_{0}^{0.5} 4x^3 \, dx = \left[ x^4 \right]_0^{0.5} = (0.5)^4 - 0^4 = 0.0625 \). Equate to \(4(s_1)^3 \times 0.5 = 0.0625\), solve for \(s_1\):\[ 4(s_1)^3 \times 0.5 = 0.0625 \rightarrow 2(s_1)^3 = 0.0625 \rightarrow (s_1)^3 = 0.03125 \rightarrow s_1 = (0.03125)^{1/3}. \]

Step 6: Solve for \(s_2\)

Calculate \( \int_{0.5}^{1} 4x^3 \, dx = \left[ x^4 \right]_{0.5}^{1} = 1 - 0.0625 = 0.9375 \). Equate to \(4(s_2)^3 \times 0.5 = 0.9375\), solve for \(s_2\):\[ 4(s_2)^3 \times 0.5 = 0.9375 \rightarrow 2(s_2)^3 = 0.9375 \rightarrow (s_2)^3 = 0.46875 \rightarrow s_2 = (0.46875)^{1/3}. \]

Step 7: Conclusion

Thus, the points \(s_1\) and \(s_2\) exist that satisfy the conditions of the Riemann sum being equal to the integral value. Calculate the exact values for \(s_1\) and \(s_2\) for a precise answer.

Key Concepts

Riemann Sum

The Riemann Sum is a method for approximating the total area under a curve, or the integral of a function, on a closed interval. It breaks the interval into subintervals of equal or varying sizes based on \(\Delta x\), the width of each subinterval.
It's crucial to understanding how integrals can be approximated before modern computational tools were available.

• In each subinterval, a function's value is evaluated at specific points called sample points.
• The sum of these point evaluations, multiplied by the respective subinterval widths (\(\Delta x\)), gives the Riemann Sum.

In this exercise, we use the Riemann Sum formula \(\mathcal{R}\left(f,\left\{s_{1}, s_{2}\right\}\right)=f\left(s_{1}\right) \cdot \Delta x+f\left(s_{2}\right) \cdot \Delta x\) over an interval, with specific points \(s_1\) in \[0, 0.5\]\ and \(s_2\) in \[0.5, 1\]\.

Integral Calculus

Integral Calculus is the branch of mathematics concerned with the concept of integrals. It derives from the need to find the cumulative sum or area under curves defined by functions.
There are two fundamental parts:

• The definite integral, representing the exact area under a curve within set limits, and
• The indefinite integral, representing a family of functions.

To compute the integral of a function like \(f(x) = 4x^3\) over the interval \([0, 1]\), you use integral calculus to find the precise area.
This requires applying integration rules, such as the Power Rule, to evaluate the definite integral accurately.

Riemann Integral

The Riemann Integral is a formal definition of the definite integral of a function. It's used to find the exact area under a curve by using Riemann sums, as previously discussed. This is particularly helpful for functions that are continuous over a given interval.
For the function \(f(x) = 4x^3\) on \([0, 1]\), the Riemann Integral is the limit of the Riemann sums as the number of subintervals approaches infinity.
In practice, this integral is evaluated by finding the antiderivative of the function and calculating the difference at the interval bounds:

• Calculate the antiderivative using integration rules
• Apply the Fundamental Theorem of Calculus, \[ ext{finding} \ \ F(b)-F(a)\] \, where \(F\) is the antiderivative.

Power Rule for Integration

The Power Rule for Integration is a fundamental method used to find the integral of power functions. This rule is effective when dealing with monomials \(x^n\) where \(n\) is any real number except \(-1\).
The power rule states:\[ \ \int x^n \, dx = \frac{x^{n+1}}{n+1}+C \]

• Here, \(C\) represents the constant of integration for indefinite integrals.
• For definite integrals, the constant \(C\) cancels out as you evaluate the integral at upper and lower bounds.

When we integrate \(4x^3\), use the Power Rule to get \[ \frac{4x^4}{4} \], which simplifies to \(x^4\).
Evaluating this between 0 and 1 results in the value of the definite integral \((1^4 - 0^4 = 1)\), signifying the total area under the curve.