Question

Find the area under \(y=3 x^{2}\) from \(x=0\) to \(x=2\) using the limit of a sum.

Short answer

Answer: The area under the curve is 2 square units.

Step-by-step solution

Divide the interval into subintervals

Divide the given interval \([0, 2]\) into \(n\) equally spaced subintervals, each of width \(\Delta x\). To find the width of each subinterval, subtract the lower limit from the upper limit and divide by the number of subintervals, i.e. \(\Delta x = \frac{2 - 0}{n} = \frac{2}{n}\)

Define the height of the rectangles

Since the height of each rectangle is based on the value of the function \(f(x) = 3x^2\) at the right endpoint of each subinterval, define \(f(x_i) = 3x_i^2\) where \(x_i\) is the right endpoint of the \(i\)-th subinterval. Since the width of each subinterval is \(\Delta x\) and \(x=0\) is the left endpoint of the first subinterval, then \(x_i = 0 + i\Delta x = \frac{2i}{n}\)

Find the sum of the areas of the rectangles

For each subinterval, find the area of the rectangle with base \(\Delta x\) and height \(f(x_i)\), and sum all these areas: \(A_n = \sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n (3(\frac{2i}{n})^2) \frac{2}{n}\)

Simplify the sum of the areas

Simplify the sum by combining like terms and applying summation properties: \(A_n = \sum_{i=1}^n (3(\frac{2i}{n})^2) \frac{2}{n} = 12\frac{1}{n^3} \sum_{i=1}^n i^2\)

Evaluate the series

Using the formula for the sum of the first \(n\) squares, we have: \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\)

Combine the series and the area expression

Substitute the series expression back into the area expression and simplify: \(A_n = 12\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}\) \(A_n = 2\frac{(n+1)(2n+1)}{n^2}\)

Take the limit

As the number of subintervals (\(n\)) approaches infinity, the sum of the areas of the rectangles approaches the actual area under the curve. Therefore, take the limit as \(n\) approaches infinity: \(A = \lim_{n\to\infty} A_n = \lim_{n\to\infty} 2\frac{(n+1)(2n+1)}{n^2}\)

Evaluate the limit

To evaluate the limit, divide each term in the numerator and denominator by the highest power of \(n\) (which is \(n^2\)): \(A = \lim_{n\to\infty} \frac{(1+\frac{1}{n})(2+\frac{1}{n})}{1} = (1)(2) = 2\) Finally, the area under \(y=3x^2\) from \(x=0\) to \(x=2\) is 2 square units.

Key concepts

Riemann sum

In calculus, a Riemann sum is a certain kind of approximation of an integral by a finite sum. It is named after the German mathematician Bernhard Riemann. One very common application is to estimate the area under a curve in the XY-plane.
To find this area, the region under the curve is divided into simple shapes, typically rectangles (although trapezoids and other shapes can be used as well). The Riemann sum is simply the sum of the area of these shapes.
To arrive at a Riemann sum, one must:

• Divide the interval into smaller subintervals
• Choose a sample point in each subinterval
• Calculate the value of the function at each sample point
• Multiply this value by the width of the subinterval
• Add up all these products

The more subintervals you choose, the closer the Riemann sum will approximate the true area.

Definite integral

A definite integral is a form of an integral that calculates the net area under a curve between two specified points. It gives a result called the integral value, which represents the accumulated sum of areas -- just like the sum of regions under a curve between two points on the x-axis. The process of finding this value is called integration.
The definite integral is denoted as
\begin{align*}\int_{a}^{b} f(x) dx,\tag{1}\right.ewline\end{align*}where \(a\) and \(b\) are the lower and upper bounds of the interval, respectively, and \(f(x)\) is the function being integrated. In the context of area, if the function \(f(x)\) is above the x-axis, the integral gives the area between the function and the axis; if the function dips below the axis, the integral gives the signed area, which can be negative.

Limits in calculus

The concept of limits is fundamental in calculus and is used to define both derivatives and integrals. A limit describes the value that a function or sequence approaches as the input or index approaches some value.
In the context of the Riemann sum and definite integrals, limits play a crucial role. As the number of rectangles increases indefinitely, making the width \(\Delta x\) infinitely small, the Riemann sum approaches the exact area under the curve.
This process is described algebraically using a limit. For example, to compute the area under \(y=3x^2\) from \(x=0\) to \(x=2\), the limit as the number of rectangles \(n\) approaches infinity is calculated to find the exact area. That is, \(A = \lim_{n\to\infty} A_n\). This concept ensures that we get an accurate measure of area, not just an approximation.

Squaring function

The squaring function is a simple yet vital polynomial function in mathematics, particularly in calculus. It is written as \(f(x) = x^2\), and graphically, it is a curve known as a parabola that opens upwards. The squaring function is relevant because it's often one of the basic forms you'll encounter when finding the area under a curve.
In our exercise, we dealt with \(y=3x^2\), which is a scaled squaring function, where every value of \(x\) gives a \(y\) value that is three times the square of \(x\). When finding the area under such a curve from \(x=0\) to \(x=2\), we're essentially summing up infinitely many squares of \(x\) values, each multiplied by 3, to get the total 'area under the curve' – which, in this case, is a definite integral.