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Chapter 5: Problem 5

Find the area under the curve \(y=f(x)\) over the stated interval. $$ f(x)=x^{3} ;[2,3] $$

Short Answer

Expert verified
The area is \(\frac{65}{4}\) square units.

Step by step solution

01

Identify the function and interval

The function given is \(f(x) = x^3\) and the interval over which we need to find the area is \([2, 3]\).
02

Set up the integral for the area

To find the area under the curve \(f(x) = x^3\) from \(x = 2\) to \(x = 3\), we need to evaluate the definite integral \(\int_{2}^{3} x^3 \, dx\).
03

Compute the integral

To find \(\int x^3 \, dx\), we use the power rule for integration. The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\). Therefore, \(\int x^3 \, dx = \frac{x^4}{4} + C\).
04

Evaluate the indefinite integral using limits

Substitute the upper and lower limits into the evaluated indefinite integral: \(\left. \frac{x^4}{4} \right|_2^3 = \left( \frac{3^4}{4} \right) - \left( \frac{2^4}{4} \right)\).
05

Calculate the definite integral

Calculate \(\frac{3^4}{4} - \frac{2^4}{4}\). This simplifies to \(\frac{81}{4} - \frac{16}{4} = \frac{65}{4}\).
06

Conclusion

The area under the curve \(y = x^3\) from \(x=2\) to \(x=3\) is \(\frac{65}{4}\) square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite integral
A definite integral is a concept used to find the total area under a curve within a specific interval. This means calculating the area between the graph of a function and the x-axis, from one point to another. In our exercise, we explored the definite integral
  • Function: \(f(x) = x^3\)
  • Interval: [2, 3]
To set this up, we use the integral notation \[\int_{2}^{3} x^3 \, dx\]This symbol tells us we're finding the accumulation of areas from x = 2 to x = 3, ensuring that we consider slices of this function across that range. The result gives us a precise value for the area under the curve over the specified interval.
Power rule for integration
The power rule for integration is a nifty tool that helps simplify integrating power functions like those found in polynomial expressions. It is particularly useful for functions of the form \(x^n\). This rule states \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \(C\) is the constant of integration, although for definite integrals, this constant eventually cancels out. Making use of the power rule in our example:
  • Function: \(f(x)=x^3\)
We increased the exponent by one to get \(x^4\), then divided by the new exponent to get \[\int x^3 \, dx = \frac{x^4}{4} + C\]After figuring this out, we proceed to evaluate the integral with specific limits.
Evaluating integrals
To convert an indefinite integral into a definite one, we substitute the upper and lower limits of the interval into the antiderivative. This effectively evaluates the integral. For our case:
  • Upper Limit: \(x=3\)
  • Lower Limit: \(x=2\)
Since \[\left. \frac{x^4}{4} \right|_2^3\]was derived from the power rule, we simply substitute and subtract: \[ \frac{3^4}{4} - \frac{2^4}{4} = \frac{81}{4} - \frac{16}{4} \]This calculation gives us our final answer, \[\frac{65}{4}\]which represents the area under the curve. Evaluating each part ensures accurate results for finding the area beneath a curve, perfect for understanding both pure math and practical situations involving integration.

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Most popular questions from this chapter

Evaluate the integrals by any method. $$ \int_{0}^{\pi / 6} \tan 2 \theta d \theta $$

Sketch the region described and find its area. The region under the curve \(y=3 \sin x\) and over the interval \([0,2 \pi / 3] .\)

Evaluate the integrals by any method. $$ \int_{0}^{\pi / 4} 4 \sin x \cos x d x $$

(a) Use a graphing utility to generate the graph of $$ f(x)=\frac{1}{100}(x+2)(x+1)(x-3)(x-5) $$ and use the graph to make a conjecture about the sign of the integral $$ \int_{-2}^{5} f(x) d x $$ (b) Check your conjecture by evaluating the integral.

Explain why the Fundamental Theorem of Calculus may be applied without modification to definite integrals in which the lower limit of integration is greater than or equal to the upper limit of integration.
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