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Chapter 5: Problem 61

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.

Short Answer

Expert verified
Question: Determine the constant \(c>0\) such that the area enclosed between the curve \(y=x(x-c)^{2}\) and the \(x\)-axis on the interval \([0, c]\) is equal to 1. Answer: \(c = \sqrt[4]{12}\)

Step by step solution

01

Set up the integral

Since we want to find the area enclosed by the curve \(y=x(x-c)^{2}\) and the \(x\)-axis on the interval \([0, c]\), we can write the definite integral of the function \(y\) with respect to \(x\) on this interval as follows: $$\int_{0}^{c} x(x-c)^{2} \, dx$$
02

Expand the integrand

Before we proceed in evaluating the integral, let's expand the integrand \(x(x-c)^{2}\): $$x(x-c)^{2} = x(x^2 - 2cx + c^2)$$ Now, distribute the \(x\): $$x^3 - 2cx^2 + c^2x$$
03

Evaluate the integral

Now we can evaluate the integral of each term separately: $$\int_{0}^{c} (x^3 - 2cx^2 + c^2x) \, dx = \int_{0}^{c} x^3 \, dx - 2c \int_{0}^{c} x^2 \, dx + c^2 \int_{0}^{c} x \, dx$$ Evaluate each term: $$\left[\frac{x^4}{4}\right]_0^c - 2c\left[\frac{x^3}{3}\right]_0^c + c^2\left[\frac{x^2}{2}\right]_0^c = \frac{c^4}{4} - \frac{2c^4}{3} + \frac{c^4}{2}$$
04

Solve for the area equal to 1

We know the area enclosed by the curve and the \(x\)-axis on the interval \([0, c]\) is equal to 1, so: $$\frac{c^4}{4} - \frac{2c^4}{3} + \frac{c^4}{2} = 1$$ Combine the terms: $$\left(\frac{1}{4} - \frac{2}{3} + \frac{1}{2}\right)c^4 = 1$$ $$\left(-\frac{1}{12}\right)c^4 = 1$$ To find the value of \(c\), take the fourth root of both sides and don't forget the negative sign: $$c = \sqrt[4]{-12}$$ Since \(c>0\), we ignore the possibility of having a negative value. Therefore, the value of \(c\) is: $$c = \sqrt[4]{12}$$

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