Question

In Exercises \(31-36,\) find the area of the region bounded by the graphs of the equations. $$ y=3 x^{2}+1, \quad x=0, \quad x=2, \quad y=0 $$

Short answer

The area of the region bounded by the graphs of the equations is 10 square units.

Step-by-step solution

Understand the Region

The area that needs to be found is enclosed by the function \(y=3 x^{2}+1\) and the lines \(x=0\), \(x=2\), and \(y=0\). So, the area lies above the x-axis, below the curve, and between \(x=0\) and \(x=2\).

Set Up the Definite Integral

The definite integral that will give us the area is the integral from \(x=0\) to \(x=2\) of the function \(y=3 x^{2}+1\). The integral can therefore be written as: \[ \int_{0}^{2} (3x^{2}+1) \, dx \]

Calculate the Integral

Calculate the integral: \[ \int (3x^{2}+1) \, dx = x^{3} + x + C \]Then evaluate this at \(x=2\) and \(x=0\): \[ Area = (2^{3}+2)-(0+0) = 8+2-0 = 10 \]

Key concepts

Area Under the Curve

When studying calculus and specifically the concept of the definite integral, a fundamental application is to calculate the area under a curve.

The area under a curve, in a very intuitive sense, represents the total 'space' or 'region' that is bounded by the curve of a function and the x-axis between two points. In the exercise given, we are looking at the region under the function given by the equation y=3x^2+1, and above the x-axis (since the area below the x-axis would be considered negative). This area is enclosed by vertical lines at x=0 and x=2, and the x-axis itself acting as the bottom boundary (y=0).

By integrating the function with respect to x from 0 to 2, we can find the total area under the curve. The result of the integral gives us the exact area, which in the realm of real-life applications, could represent anything from physical space to economic surplus depending on the context of the problem.

Riemann Sum

The Riemann sum is a method for approximating the total value of a function over a certain interval, and it plays a critical role in the understanding of integration. The way it works is by 'slicing' the area under the curve into small rectangles (or sometimes trapezoids) and then adding up the area of all these shapes to get a total sum.

Each rectangle's area is calculated by multiplying the height of the function at that point by the width of the interval slice. The finer the slices, or partitions, the more accurate our approximation becomes. In the context of our definite integral calculation, the Riemann sum is the approach taken towards building up to the exact answer. The limit of these sums, as the width of the partitions approaches zero, is exactly what we define as the definite integral of the function.

So when we are calculating the definite integral from 0 to 2 for the function y=3x^2+1, imagine that we are actually adding up infinitely many infinitesimally thin rectangles under the curve between x=0 and x=2 to get to the exact area of 10.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a technique used to simplify the process of integration when dealing with more complex functions. This method is the reverse process of the chain rule in differentiation.

To apply this method, one often looks for an inner function (typically represented as u) that, when derived, appears elsewhere in the integral. Substituting u into the integral, along with its differential du, often transforms the integral into a simpler one that is easier to evaluate.

In our exercise, integration by substitution isn't necessary because the integral of 3x^2+1 is straightforward. However, if the curve equation was more complex, such as involving a nested expression like \(3\text{(x+1)}^2+1\), we could let u = x+1, then du = dx, and the integral would become simpler to solve after substitution. Understanding when and how to use u-substitution is a valuable skill in calculus, as it allows us to integrate a broader range of functions effectively.