Problem 61 - Calculus Early Transcendentals

Question

Find the value of \(c>0\) such that the region bounded by the cubic \(y=x(x-c)^{2}\) and the \(x\) -axis on the interval \([0, c]\) has area 1.

Step-by-Step Solution

Answer

Answer: \(c = \sqrt[4]{12}\)

Step 1: Write down the given function and interval

We are given the function \(y = x(x-c)^2\) and the interval \([0, c]\).

Step 2: Find the integral of the function

To find the area under the curve, we need to calculate the definite integral of the function over the given interval: \(\int_{0}^{c}x(x-c)^2dx\)

Step 3: Integrate the function

To integrate the function, we can use integration by parts. Let \(u=x\) and \(dv=(x-c)^2dx\). Then, \(du=dx\) and by integrating \(dv\), we get \(v= \frac{1}{3}(x-c)^3\). Now, apply the integration by parts formula: \(\int_{0}^{c}x(x-c)^2dx = \left[uv \right]_0^c - \int_{0}^{c}vdu\)

Step 4: Calculate the expressions

Substituting our expressions for \(u\), \(v\), \(du\), and the limits, we get: \(\left[\frac{1}{3}x(x-c)^3 \right]_0^c - \int_{0}^{c}\frac{1}{3}(x-c)^3dx\) Now, we evaluate the expression at the limits. At \(x=0\), the expression is 0. At \(x=c\), the expression becomes \(\frac{1}{3}c(c-c)^3=\frac{1}{3}(0)=0\). So, \(\left[\frac{1}{3}x(x-c)^3 \right]_0^c = 0\) Moving to the remaining part, we have \(- \int_{0}^{c}\frac{1}{3}(x-c)^3dx\) Now, make a substitution: let \(t = x-c \Rightarrow dt = dx\). When \(x=0, t=-c\), and when \(x = c, t=0\): \(- \int_{-c}^{0}\frac{1}{3}(t)^3 dt = \frac{-1}{3} \int_{-c}^0 t^3 dt\)

Step 5: Integrating the remaining part

Integrate the remaining part \(\frac{-1}{3} \int_{-c}^0 t^3 dt\): \(\frac{-1}{3} \left[\frac{1}{4}t^4 \right]_{-c}^0 = 0 - \left(-\frac{1}{12}c^4\right) = \frac{1}{12}c^4\) So the definite integral is: \(\int_{0}^{c}x(x-c)^2dx=\frac{1}{12}c^4\)

Step 6: Find the value of c such that the area is 1

Set the area equal to 1: \(\frac{1}{12}c^4 = 1\) Now, solve for \(c\): \(c^4 = 12\) \(c = \sqrt[4]{12}\) Therefore, the value of \(c\) such that the region bounded by the cubic \(y=x(x-c)^2\) and the x-axis on the interval \([0, c]\) has area 1 is \(c=\sqrt[4]{12}\).

Key Concepts

Area Under the Curve

When we talk about the 'area under the curve' in calculus, we're referring to finding the region bounded by a graph of a function and the x-axis. Calculating this area is often synonymous with performing definite integration over a specified interval.

The definite integral represents the accumulated total of the function's values over that interval. Think of it as a way to 'add up' small slices of the area under the graph, giving us a single numerical value that represents the size of that region.

For example, in the given exercise, to find the area of the region bounded by the curve represented by the function \(y=x(x-c)^{2}\) and the x-axis from \(x=0\) to \(x=c\), we set up the definite integral \(\textstyle\bigintsss_0^c x(x-c)^2 \text{\text{\text{\textcheckmark}}}x\). The function \(y=x(x-c)^{2}\) describes the height of the curve at any point \(x\), and integrating it from 0 to \(c\) sums up these heights to find the total area.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation and is used when we're dealing with integrals of products of functions. It's especially handy when our function is a multiplication of two simpler functions whose integral is not straightforward.

The formula for integration by parts is \( \textstyle\bigint u dv = uv - \textstyle\bigint v du \), where \(u\) and \(v\) are functions of \(x\). It's a strategic move: we pick \(u\) and \(dv\) such that \(du\) and \(v\) (the integral of \(dv\)) are easier to manage.

In the exercise, we applied this method to the function \(y = x(x-c)^2\). We chose \(u=x\) and \(dv=(x-c)^2dx\) because it simplifies the integration process. Calculating \(v\) from \(dv\) and substituting back into the integration by parts formula allowed us to solve the integral in a more manageable way. This elegant technique is crucial for solving complex calculus problems.

Calculus Problems

Calculus problems can range from straightforward to highly complex, often involving a series of steps and the application of various methods like differentiation, integration, and limit processes. The beauty of calculus lies in its ability to solve real-world problems, from computing the area of irregular shapes to determining rates of change in various scientific fields.

When tackling calculus problems, such as the one featuring the definite integral of \(x(x-c)^2\), it's important to follow a step-by-step approach. First, identify what is being asked – in our case, the area under the curve. Then, select the appropriate calculus technique, such as integration by parts if the function involves a product of terms. Bear in mind that some problems may even require multiple methods or substitutions, as we saw with our variable change \(t=x-c\).

Moreover, improving the comprehension of these problems involves getting your hands dirty with practice and understanding the fundamental concepts, not just memorizing formulas. For instance, recognizing how to decompose functions and apply integration techniques will build a solid foundation for solving an extensive range of calculus problems.