Question

Let \(B(c)\) denote the area of the region bounded from above by the graph of the curve \(y=x(1-x)\), from below by the \(x\) -axis, and from the right by the line \(x=c\). The domain of \(B\) is the interval \([0,1]\). (See Figure 14.) Given that \(B(1)=\frac{1}{6}\), (a) Find \(B(0)\) (b) Find \(B\left(\frac{1}{2}\right)\) (c) As best you can, sketch a graph of \(B(c)\).

Short answer

(a) 0, (b) \(\frac{1}{12}\), (c) Graph curves upward and levels at \(\frac{1}{6}\) at \(c=1\).

Step-by-step solution

Understanding the Problem

We need to find the area under the curve \(y = x(1-x)\) from \(x = 0\) to \(x = c\), where this area is denoted as \(B(c)\). The curve is a downward-opening parabola with roots at \(x = 0\) and \(x = 1\). Given that \(B(1) = \frac{1}{6}\), we know the total area under the parabola from 0 to 1 is \(\frac{1}{6}\).

Integrate the Function

To find \(B(c)\), we integrate \(y = x(1-x) = x - x^2\) with respect to \(x\) from 0 to \(c\). The integral is:\[\int_0^c (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^c\]

Evaluate the Integral at Known Values

We need to evaluate the definite integral obtained in Step 2 at the specific values of \(c\) asked in the problem.(a) For \(B(0)\):\[B(0) = \left[ \frac{0^2}{2} - \frac{0^3}{3} \right] = 0\](b) For \(B\left(\frac{1}{2}\right)\):\[B\left(\frac{1}{2}\right) = \left[ \frac{\left(\frac{1}{2}\right)^2}{2} - \frac{\left(\frac{1}{2}\right)^3}{3} \right] = \frac{1}{8} - \frac{1}{24} = \frac{1}{12}\]

Sketch the Graph

The function \(B(c)\), as defined, represents the area under the curve from \(x = 0\) to \(x = c\). Based on the integration results:- \(B(0) = 0\)- \(B\left(\frac{1}{2}\right) = \frac{1}{12}\)- \(B(1) = \frac{1}{6}\)The graph of \(B(c)\) is a curve starting at \(0\) (at \(c=0\)), increasing to \(\frac{1}{12}\) (for \(c = \frac{1}{2}\)), and peaking at \(\frac{1}{6}\) (for \(c=1\)). This indicates a concave down curve representing the cumulative area under the parabola from \(0\) to \(c\).

Key concepts

Definite Integral

In calculus, a definite integral is a concept that calculates the area between a function and the x-axis over a specified interval. It is written as \( \int_a^b f(x) \, dx \) and represents the total accumulation of values over that interval. The definite integral has limits of integration, noted as \(a\) and \(b\), which define where we start and stop measuring the area.

This method is used extensively for determining the area under a curve, solving real-world problems involving accumulation, and understanding changes in quantities over time. In the given exercise, the definite integral was used to find \(B(c)\), representing the area under the parabola defined by \(y = x(1-x)\) from \(x = 0\) to a variable endpoint \(c\).

To solve it, we integrated \(y = x(1-x) = x - x^2\). The integration gives us the function \( \frac{x^2}{2} - \frac{x^3}{3} \), which, once evaluated over the interval from \(0\) to \(c\), yields the expression \( \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^c \). This calculated value gives us the precise measurement of accumulated area under the curve for any given value of \(c\).

Area Under Curve

The area under a curve in a graph is a fundamental concept in calculus and physics. It represents the total sum of infinitesimally small rectangles under the line of the curve across a specified range on the x-axis. This measurement can provide valuable data related to distances, probabilities, and other mathematical and physical contexts.

In our exercise, we are measuring the area under the parabola \(y = x(1-x)\) to find \(B(c)\). Since this curve exists above the x-axis, the area involves a positive calculation of all points from the curve down to the x-axis between \(x = 0\) and \(x = c\). Understanding this visual and geometric perspective of areas as it applies to curves helps create a more tangible view of what an integral represents and why the outcome \(B(c)\) can be trusted as a "cumulative" area measure for different values of \(c\).

Measuring this area is akin to calculating how much space the parabola encloses from its curve to the x-axis, which is exactly what helps us understand the varying values at different maxima or endpoints in practical terms.

Parabola

The parabola is one of the most recognized shapes in both mathematics and nature, forming a symmetrical, arch-like curve. It can be represented by the quadratic function \(y = ax^2 + bx + c\). In this case, we are working with the specific downward-opening parabola \(y = x(1-x)\). The "downward-opening" portion indicates the "a" coefficient (here, it simplifies to \(-x^2\)) is negative, causing the curve to bend downwards.

The parabola has distinct features, such as a vertex, axis of symmetry, and roots or x-intercepts. For our exercise, the roots are at \(x = 0\) and \(x = 1\), which are automatically derived from setting \(y = 0\) in \(y = x(1-x)\) to find any points where the parabola crosses the x-axis.

Understanding the nature of this parabola helps us hugely in visualizing the "bounded region" under the curve between the x-axis and specific vertical cut-offs described in the problem scope. Recognizing these properties helps in grasping why the area can be calculated as a definite integral, and is intuitive to see where the values \(B(0)\), \(B\left(\frac{1}{2}\right)\), and \(B(1)\) are derived geometrically beneath the curved plot of \(y = x(1-x)\).