Chapter 16: Problem 44
One value of \(k\) for which the area of the figure bounded by the curve \(y=8 x^{2}-x^{5}\), the straight lines \(x=1\) and \(x=k\) and the \(x\)-axis is equal to \(\frac{16}{3}\), is (A) \(-1\) (B) 3 (C) 2 (D) \(\sqrt[3]{8-\sqrt{17}}\)
Short Answer
Expert verified
The value of \(k\) is 2 (Option C).
Step by step solution
01
Understand the Problem
We need to find a value of \(k\) such that the area between the curve \(y=8x^2-x^5\), the lines \(x=1\) and \(x=k\), and the x-axis, is equal to \(\frac{16}{3}\).
02
Set up the Integral
The area under the curve from \(x=1\) to \(x=k\) can be found by integrating the function \(y=8x^2-x^5\) with respect to \(x\). The integral is given by:\[\int_{1}^{k} (8x^2 - x^5) \, dx = \frac{16}{3}\]
03
Calculate the Indefinite Integral
First, we calculate the indefinite integral of \(8x^2 - x^5\):\[\int (8x^2 - x^5) \, dx = \frac{8}{3}x^3 - \frac{1}{6}x^6 + C\]
04
Apply the Limits
Now, apply the definite integral from \(x=1\) to \(x=k\):\[\left[ \frac{8}{3}x^3 - \frac{1}{6}x^6 \right]_{1}^{k} = \frac{16}{3}\]Evaluating it gives:\[\left( \frac{8}{3}k^3 - \frac{1}{6}k^6 \right) - \left( \frac{8}{3}\cdot1^3 - \frac{1}{6}\cdot1^6 \right) = \frac{16}{3}\]
05
Simplify the Equation
Simplify the equation:\[\frac{8}{3}k^3 - \frac{1}{6}k^6 - \left( \frac{8}{3} - \frac{1}{6} \right) = \frac{16}{3}\]Calculate the constant part:\[\frac{8}{3} - \frac{1}{6} = \frac{16}{6} - \frac{1}{6} = \frac{15}{6} = \frac{5}{2}\]Thus,\[\frac{8}{3}k^3 - \frac{1}{6}k^6 - \frac{5}{2} = \frac{16}{3}\]
06
Rearrange and Solve for \(k\)
Rearrange the equation:\[\frac{8}{3}k^3 - \frac{1}{6}k^6 = \frac{16}{3} + \frac{5}{2}\]Convert \(\frac{5}{2}\) to a common denominator of 6:\[\frac{16}{3} = \frac{32}{6}, \frac{5}{2} = \frac{15}{6}\]So,\[\frac{8}{3}k^3 - \frac{1}{6}k^6 = \frac{32}{6} + \frac{15}{6} = \frac{47}{6}\]Now solve:\[\frac{1}{6}k^6 - \frac{8}{3}k^3 + \frac{47}{6} = 0\]Multiply the whole equation by 6:\[k^6 - 16k^3 + 47 = 0\]Test possible roots among the given options.
07
Quick Check with Given Options
Check the possible roots: Pick an option that satisfies the equation \(k^6 - 16k^3 + 47 = 0\) among options what matches.Calculating for \(k = 2\):\[2^6 - 16(2)^3 + 47 = 64 - 128 + 47 = 0\]Which confirms that \(k=2\) is a solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
Definite integrals are a foundational concept in calculus, specifically dealing with the calculation of areas under curves. When you see an integral with limits, such as \[\int_{a}^{b} f(x) \, dx\] it represents the area under the curve of the function \(f(x)\) from \(x = a\) to \(x = b\). This is called a definite integral and it provides a concrete numerical value.
- The limits \(a\) and \(b\) define the specific section of the curve you are interested in.
- The function \(f(x)\) describes the curve itself.
- The result is an exact value representing the net area, considering spaces above the x-axis as positive and below as negative.
Curve Area Calculation
Calculating the area bounded by a curve and the x-axis between two vertical lines is a common application of definite integrals. This method is useful for determining the total area between a polynomial function and the x-axis over a closed interval. In our example, the formula used is \[\int_{1}^{k} (8x^2 - x^5) \, dx = \frac{16}{3}\].
- To find the area under the curve, first derive the indefinite integral \(F(x)\), which gives you a general expression for the antiderivative.
- Apply the definite limits \(x = 1\) and \(x = k\) to evaluate the specific area.
- The task then becomes solving for \(k\) such that this area equals \(\frac{16}{3}\).
Polynomial Integration
Integrating polynomials involves finding the antiderivative of expressions such as \(ax^n\), which is a pivotal part of calculus. The general rule for integrating a polynomial function \(f(x) = ax^n\) is given by:\[ \int ax^n \, dx = \frac{a}{n+1}x^{n+1} + C \]where \(C\) is the integration constant for indefinite integrals.
- This applies to each term in a polynomial individually, meaning all terms are integrated one by one, and their antiderivatives are combined.
- In definite integrals, as seen with \(8x^2 - x^5\), we calculate each term’s antiderivative and evaluate it at the specified limits.
- The integration process for our example leads to: \(\int (8x^2 - x^5) \, dx = \frac{8}{3}x^3 - \frac{1}{6}x^6\).