Problem 27
Question
Use an integral to find the specified area. Under \(y=6 x^{3}-2\) for \(5 \leq x \leq 10\).
Step-by-Step Solution
Verified Answer
The area under the curve from \(x = 5\) to \(x = 10\) is \(14052.5\).
1Step 1: Set up the integral to find the area
To find the area under the curve given by the function \(y = 6x^3 - 2\) from \(x = 5\) to \(x = 10\), we need to compute the definite integral of the function over this interval. The integral setup is: \[\int_{5}^{10} (6x^3 - 2) \, dx\]
2Step 2: Integrate the function
To integrate the function \(6x^3 - 2\), we find the antiderivative. The antiderivative of \(6x^3\) is \(\frac{6x^4}{4} = \frac{3x^4}{2}\), and the antiderivative of \(-2\) is \(-2x\). Thus, the indefinite integral is: \[\int (6x^3 - 2) \, dx = \frac{3x^4}{2} - 2x + C\]
3Step 3: Evaluate the definite integral
Use the antiderivative to evaluate the definite integral from 5 to 10. Substitute \(x = 10\) and \(x = 5\) into the antiderivative and compute the difference:\[\left[ \frac{3(10)^4}{2} - 2(10) \right] - \left[ \frac{3(5)^4}{2} - 2(5) \right]\]Calculate each part: \[\left[ \frac{3 imes 10000}{2} - 20 \right] = 14980\]\[\left[ \frac{3 imes 625}{2} - 10 \right] = 927.5\]Subtract the two results:\[14980 - 927.5 = 14052.5\]
4Step 4: Conclude with the final area result
After evaluating the integral, conclude with the area under the curve on the specified interval. Thus, the area under the curve \(y = 6x^3 - 2\) from \(x = 5\) to \(x = 10\) is \(14052.5\).
Key Concepts
Area Under a CurveAntiderivativeEvaluating IntegralsCalculus Problems
Area Under a Curve
The concept of finding the area under a curve is a fundamental application of definite integrals in calculus. When we discuss the 'area under a curve,' we are referring to the space between the curve and the x-axis over a specified interval. In our example, the function is given as \(y = 6x^3 - 2\), and we want to find the area from \(x = 5\) to \(x = 10\).
To compute this area, we set up a definite integral for the function across the interval, which is written as \(\int_{5}^{10} (6x^3 - 2) \, dx\). This integral expression represents the total accumulated area between the curve and the x-axis bounded by \(x = 5\) and \(x = 10\). It is crucial to note that if the curve dips below the x-axis, this area would count as negative—integrals naturally consider both above and below the x-axis areas to give a complete understanding of net area.
To compute this area, we set up a definite integral for the function across the interval, which is written as \(\int_{5}^{10} (6x^3 - 2) \, dx\). This integral expression represents the total accumulated area between the curve and the x-axis bounded by \(x = 5\) and \(x = 10\). It is crucial to note that if the curve dips below the x-axis, this area would count as negative—integrals naturally consider both above and below the x-axis areas to give a complete understanding of net area.
Antiderivative
The antiderivative is often described as the 'reverse' of taking a derivative. It is the function whose derivative is the original function you started with. For our problem \(y = 6x^3 - 2\), we need to find the antiderivative to evaluate the integral.
The process involves integrating each term separately. The term \(6x^3\) becomes \(\frac{3x^4}{2}\) when integrated, as we apply the power rule which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\). The constant \(-2\) simply becomes \(-2x\) after integration. Thus, the antiderivative of the function is \(\frac{3x^4}{2} - 2x + C\), where \(C\) is the constant of integration, which we ignore for definite integrals.
Understanding how to compute the antiderivative allows one to handle diverse calculus problems, involving curve analysis and area estimation accurately.
The process involves integrating each term separately. The term \(6x^3\) becomes \(\frac{3x^4}{2}\) when integrated, as we apply the power rule which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\). The constant \(-2\) simply becomes \(-2x\) after integration. Thus, the antiderivative of the function is \(\frac{3x^4}{2} - 2x + C\), where \(C\) is the constant of integration, which we ignore for definite integrals.
Understanding how to compute the antiderivative allows one to handle diverse calculus problems, involving curve analysis and area estimation accurately.
Evaluating Integrals
Evaluating integrals, specifically definite integrals, is a necessary step after finding the antiderivative, especially to solve calculus problems involving area. To evaluate the definite integral such as \(\int_{5}^{10} (6x^3 - 2) \, dx\), we use the Fundamental Theorem of Calculus.
This theorem states that if you have found an antiderivative of a function \(f\), you can evaluate \(\int_{a}^{b} f(x) \, dx\) by taking \[F(b) - F(a)\], where \(F\) is the antiderivative.
In our case, inserting \(10\) and \(5\) into our antiderivative, \(\left( \frac{3(10)^4}{2} - 2(10) \right)\) and \(\left( \frac{3(5)^4}{2} - 2(5) \right)\), we find the difference to finally compute the area. This leads to a result of \(14052.5\). By calculating these values, students develop a robust understanding of integrating functions over specific intervals.
This theorem states that if you have found an antiderivative of a function \(f\), you can evaluate \(\int_{a}^{b} f(x) \, dx\) by taking \[F(b) - F(a)\], where \(F\) is the antiderivative.
In our case, inserting \(10\) and \(5\) into our antiderivative, \(\left( \frac{3(10)^4}{2} - 2(10) \right)\) and \(\left( \frac{3(5)^4}{2} - 2(5) \right)\), we find the difference to finally compute the area. This leads to a result of \(14052.5\). By calculating these values, students develop a robust understanding of integrating functions over specific intervals.
Calculus Problems
Calculus problems often involve using integrals to solve real-world problems, such as determining the area under a curve which is a frequent challenge in physics, engineering, and economics. Calculus problems help bridge abstract mathematical concepts with practical applications.
Understanding these problems requires grasping several interconnected concepts such as limits, derivatives, and integrals. The area problem we focused on illustrates this beautifully. Calculus empowers students to model and solve problems by employing integrals to analyze the properties of functions over intervals, which can describe velocity, growth patterns, economic profit, and more.
Solving calculus problems builds critical thinking and problem-solving skills, necessary for advanced studies and diverse professional fields. Mastery of such skills through exercises like these not only makes students proficient in mathematics but also equips them with the tools to tackle complex, applied problems.
Understanding these problems requires grasping several interconnected concepts such as limits, derivatives, and integrals. The area problem we focused on illustrates this beautifully. Calculus empowers students to model and solve problems by employing integrals to analyze the properties of functions over intervals, which can describe velocity, growth patterns, economic profit, and more.
Solving calculus problems builds critical thinking and problem-solving skills, necessary for advanced studies and diverse professional fields. Mastery of such skills through exercises like these not only makes students proficient in mathematics but also equips them with the tools to tackle complex, applied problems.
Other exercises in this chapter
Problem 26
Find the area between the graph of \(y=x^{2}-2\) and the \(x\) -axis, between \(x=0\) and \(x=3\).
View solution Problem 27
Use a calculator or computer to evaluate the integral. $$ \int_{1}^{3} \ln x d x $$
View solution Problem 28
Use a calculator or computer to evaluate the integral. $$ \int_{1.1}^{1.7} e^{t} \ln t d t $$
View solution Problem 28
Use an integral to find the specified area. Under \(y=2 \cos (t / 10)\) for \(1 \leq t \leq 2\)
View solution