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Chapter 11: Problem 70

Verify by direct computation that \(\int_{0}^{3}\left(1+x^{3}\right) d x\) \(\quad=\int_{0}^{1}\left(1+x^{3}\right) d x+\int_{1}^{2}\left(1+x^{3}\right) d x+\int_{2}^{3}\left(1+x^{3}\right) d x\) hence showing that Property 5 may be extended.

Short Answer

Expert verified
We have verified that \(\int_{0}^{3}\left(1+x^{3}\right) dx = \int_{0}^{1}\left(1+x^{3}\right) dx+\int_{1}^{2}\left(1+x^{3}\right) dx+\int_{2}^{3}\left(1+x^{3}\right) dx\). This demonstrates that Property 5 can be extended to split a given integral into smaller sub-intervals.

Step by step solution

01

Find the antiderivative of 1+x^3

To compute the integral, we first need to find the antiderivative of the function 1+x^3. The antiderivative is found by integrating the function with respect to x: \[F(x) = \int(1+x^3) \,dx = x + \frac{x^4}{4} + C\]
02

Compute the integral on the interval [0, 3]

Substitute the limits of integration into the antiderivative and compute the definite integral: \[\int_{0}^{3}(1+x^3) \,dx = F(3) - F(0) = \left(3 + \frac{3^4}{4} \right) - \left(0 + \frac{0^4}{4} \right) = 3 + \frac{81}{4}\]
03

Compute the integrals on the intervals [0, 1], [1, 2], and [2, 3]

Now we compute the integrals on the smaller intervals: \[\int_{0}^{1}(1+x^3) \,dx = F(1) - F(0) = \left(1 + \frac{1^4}{4}\right) - \left(0 + \frac{0^4}{4}\right) = 1+\frac{1}{4}\] \[\int_{1}^{2}(1+x^3) \,dx = F(2) - F(1) = \left(2 + \frac{2^4}{4}\right) - \left(1 + \frac{1^4}{4}\right) = 1+\frac{15}{4}\] \[\int_{2}^{3}(1+x^3) \,dx = F(3) - F(2) = \left(3 + \frac{3^4}{4}\right) - \left(2 + \frac{2^4}{4}\right) = 1+\frac{65}{4}\]
04

Verify the equality

Sum the three smaller integrals to check if they are equal to the integral on the interval [0, 3]: \[ \left(1+\frac{1}{4}\right) + \left(1+\frac{15}{4}\right) + \left(1+\frac{65}{4}\right)= 3 + \left(\frac{1}{4} +\frac{15}{4} + \frac{65}{4}\right) = 3 + \frac{81}{4}\] We can now see that both sides of the equality are the same: \[\int_{0}^{3}\left(1+x^{3}\right) dx = \int_{0}^{1}\left(1+x^{3}\right) dx+\int_{1}^{2}\left(1+x^{3}\right) dx+\int_{2}^{3}\left(1+x^{3}\right) dx\] Hence, we have verified by direct computation that Property 5 may be extended.

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Key Concepts

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

Antiderivative Calculation
Understanding antiderivatives is crucial for mastering the concept of integration. An antiderivative of a function, often denoted by a capital letter (like 'F' for the original function 'f'), is a function whose derivative is the original function. In essence, calculating an antiderivative is the reverse process of differentiation.

For the function given in the exercise, \(1+x^3\), we find its antiderivative by integrating it with respect to 'x'. This results in the antiderivative \(F(x) = x + \frac{x^4}{4} + C\), where 'C' represents the constant of integration. When dealing with definite integrals, as in our problem, the constant 'C' cancels out when evaluating the antiderivative at the upper and lower limits, which simplifies the process.

Understanding that the antiderivative represents the accumulated area under the curve of the original function from a starting point, which in the context of a definite integral, could be the lower limit, helps demystify the process. Each piece of the integral, when computed, represents a section of the total area under the curve.
Integral Properties
Integration is governed by a set of properties that can simplify the calculation process and enhance our understanding of definite integrals. One such property, often referenced as the 'additivity' property, indicates that the integral of a function over an interval can be split into the sum of integrals over subintervals.

This property is reflective of the fact that the area under a curve between two points can be seen as the sum of areas over smaller segments between those points. As demonstrated in the step-by-step solution, the integral from \(0\) to \(3\) for the function \(1+x^3\) can be broken down into sub-intervals [0, 1], [1, 2], and [2, 3], and the sum of the integrals over these intervals equals the integral over the original interval. Such properties are not only mathematically appealing but also practical, offering a method to solve more complex integrals by breaking them down into simpler, manageable parts.
Integration by Intervals
Dividing the interval of integration can be particularly useful when facing complex integrals or when attempting to integrate numerically. The division into smaller intervals translates the problem into calculating the area under the curve in segments, which is the fundamental principle of integral calculus.

In the exercise at hand, the original problem is approached by dividing the interval from \(0\) to \(3\) into three smaller intervals. By computing the area under the curve for each segment and adding them together, we reach the same result as we would by integrating over the entire interval. This method not only verifies the integral properties but also provides a practical example of how integration can be tackled piece by piece, a concept that extends far beyond textbook exercises and into real-world applications where functions may be defined piecewise or behave differently across various segments.

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