Question

Prove directly from Definition 3.1 .1 that $$ \int_{a}^{b} x^{2} d x=\frac{b^{3}-a^{3}}{3} $$ Do not assume in advance that the integral exists. The proof of this is part of the problem. HINT: Let \(P=\left\\{x_{0}, x_{2}, \ldots, x_{n}\right\\}\) be an arbitrary partition of \([a, b] .\) Use the mean value theorem to show that $$ \frac{b^{3}-a^{3}}{3}=\sum_{j=1}^{n} d_{j}^{2}\left(x_{j}-x_{j-1}\right) $$ for some points \(d_{1}, \ldots, d_{n},\) where \(x_{j-1}

Short answer

Question: Prove directly from the definition of a Riemann integral that the integral of the function f(x) = x^2 from a to b is equal to (b^3 - a^3)/3.

Step-by-step solution

Define Riemann sums using an arbitrary partition P

We start by defining an arbitrary partition P of the interval [a,b]. Let's denote this partition P as: $$ P=\left\\{x_0, x_1, \ldots, x_n\right\\} $$ where \(a = x_0 < x_1 < \dots < x_n = b\). For each subinterval \([x_{j-1}, x_j]\), let \(d_j\) be a point between \(x_{j-1}\) and \(x_j\), that is, \(x_{j-1} < d_j < x_j\). The Riemann sum for the function f(x) = x^2 with respect to the partition P can be expressed as: $$ R(f,P) = \sum_{j=1}^n f(d_j)(x_j - x_{j-1}). $$

Apply the Mean Value Theorem

Now, we can apply the Mean Value Theorem to the function \(g(x) = x^3\) on the interval \([a, b]\). This states that there must exist a point \(c\) in \((a, b)\) such that: $$ \frac{g(b) - g(a)}{b-a} = g'(c) $$ Similarly, we apply the Mean Value Theorem to the function \(g(x) = x^3\) on the intervals \([x_{j-1}, x_j]\) for all \(j = 1, \ldots, n\). This means that for each j, there must exist a point \(d_j\) in \([x_{j-1}, x_{j}]\) such that: $$ \frac{g(x_j) - g(x_{j-1})}{x_j - x_{j-1}} = g'(d_j). $$ Since \(g'(x) = 3x^2\), we have: $$ \frac{(x_j)^3-(x_{j-1})^3}{x_j - x_{j-1}} = 3d_j^2 $$ or $$ (x_j)^3-(x_{j-1})^3 = 3d_j^2 (x_j - x_{j-1}). $$

Relate the sum to Riemann sums

We want to show that the sum in the equation above is equal to the Riemann sum calculated in Step 1. To do this, we sum over all j from 1 to n: $$ \sum_{j=1}^{n} (x_j)^3-(x_{j-1})^3 = \sum_{j=1}^{n} 3d_j^2 (x_j - x_{j-1}) $$ Observe that the left-hand side is a telescoping series, where all terms cancel out, except for the first and the last term: $$ b^3 - a^3 = \sum_{j=1}^{n} 3d_j^2 (x_j - x_{j-1}) $$ Now, we can divide both sides of the equation by 3: $$ \frac{b^3 - a^3}{3} = \sum_{j=1}^{n} d_j^2 (x_j - x_{j-1}) $$ On the right-hand side of the equation, we have: $$ \sum_{j=1}^{n} d_j^2 (x_j - x_{j-1}) = \sum_{j=1}^n f(d_j)(x_j - x_{j-1}) = R(f,P) $$ Thus, we have successfully related our sum to the Riemann sums for the function f(x) = x^2: $$ \int_{a}^{b} x^2 dx = \frac{b^3 - a^3}{3} $$

Key concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental result in calculus that provides a bridge between values of a function and its derivative. It states that for any given function that is continuous on a closed interval \[a, b\], and differentiable on the open interval \( (a, b) \), there exists at least one point \((c)\) in the interval such that:
\[\frac{f(b) - f(a)}{b - a} = f'(c)\]
This means that the average rate of change of the function over the interval is equal to the instantaneous rate of change at some point within the same interval.

• In simpler terms, if you travel from point A to B, there is at least one instance when your speed exactly matches your average speed for the entire journey.
• For our task, when g(x) = x^3, we use MVT to identify a point \[d_j\] within each subinterval \[x_{j-1}, x_j\] such that \[g'(d_j) = 3d_j^2 \].

Using this theorem, we connect the algebraic expression \[\frac{g(x_j) - g(x_{j-1})}{x_j - x_{j-1}} = 3d_j^2 \] to our partitioned segments, allowing us to establish equivalencies that make calculating integrals feasible.

Riemann Sums

Riemann Sums are a method for approximating the total value of a definite integral, often represented as the area under a curve. A Riemann Sum is computed by dividing an interval \[a, b\] into smaller subintervals.
For each subinterval \[x_{j-1}, x_j\], a sample point \[d_j\] is selected, and the function’s value at that point \[f(d_j)\] is multiplied by the width of the subinterval \[(x_j - x_{j-1})\].
The sum of these products gives an approximation of the integral:

• The formula for the Riemann Sum over n subintervals is:\[\sum_{j=1}^n f(d_j)(x_j - x_{j-1})\]
• This method allows us to approach the exact definite integral as the number of subintervals increases and each subinterval width approaches zero.

When applied to the function \((f(x) = x^2)\) in our solution, this sum takes the form of \[\sum_{j=1}^{n} d_j^2 (x_j - x_{j-1})\], offering a stepping stone to evaluate \[\int_{a}^{b} x^2 dx\].

Definite Integral

The concept of a Definite Integral is integral to understanding areas under curves and solving real-world problems involving total accumulations. The definite integral of a function \(f(x)\) over the interval \[a, b\] is expressed as:\[\int_{a}^{b} f(x) dx\]This represents the signed area between the function and the x-axis, from x = a to x = b.
Key points about definite integrals include:

• The process of computing a definite integral essentially involves summing the areas of thin rectangles beneath the curve, as approached through Riemann Sums.
• As the partition \[P\] becomes finer, meaning the widths of the subintervals decrease to zero, the Riemann Sum approaches the exact value of the definite integral.
• In our exercise, we move from Riemann Sums to establish that:\[\int_{a}^{b} x^2 dx = \frac{b^3 - a^3}{3}\]

In essence, the definite integral provides a precise way to find the net area under the curve, accounting for above and below axis segments, using the limit of Riemann Sums.