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Chapter 6: Problem 72

Without evaluating integrals, prove that $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x=\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x$$.

Short Answer

Expert verified
Question: Prove that the following two integrals are equal without evaluating them: $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x = \int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x$$ Answer: We can prove that these integrals are equal by applying the property of definite integrals for derivatives and evaluating each expression. After performing these steps, we find that both integrals are equal to 0, proving their equality without having to evaluate them directly.

Step by step solution

01

Apply the property to both integrals

To find the value of each integral, we will need to apply the above-mentioned property to the integrals: $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x = 12 \sin \pi(2)^{2} - 12 \sin \pi(0)^{2}$$ and $$\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x = x^{10}(2-x)^{3} \Big|_0^2$$ Now, we just need to simplify each of these expressions.
02

Simplify the expressions for each integral

With the expressions obtained in Step 1, we will now substitute the limits in each integral and simplify: $$12 \sin \pi(2)^{2} - 12 \sin \pi(0)^{2} = 12 \sin(4\pi) - 12 \sin(0) = 0$$ Similarly, $$x^{10}(2-x)^{3} \Big|_0^2 = (2^{10}(2-2)^{3}) - (0^{10}(2-0)^{3}) = 0$$
03

Proving equality of both integrals

Now we have that: $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x = 0$$ and $$\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x = 0$$ As both these expressions are equal, we have proved that: $$\int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x = \int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x$$

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Key Concepts

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

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a central principle that connects differentiation and integration, two core operations in calculus. This theorem states that if a function is continuous on a closed interval \[a, b\] and F is an antiderivative of f on \[a, b\], then \[\int_a^b f(x)\,dx = F(b) - F(a)\].

This means that the definite integral of a function over an interval can be found by evaluating the difference of its antiderivative at the boundaries of the interval. In the textbook exercise, by using the fundamental theorem, the integrals are simplified to a matter of plugging in the upper and lower limits into the antiderivatives. The resulting conclusion, that both expressions lead to the same value, is grounded in this theorem.
Integration by Substitution
Integration by substitution, often known as u-substitution, is a method used to simplify integration problems. It's the inverse process of the chain rule of differentiation and is used when an integral contains some function and its derivative.

In practice, it involves choosing a new variable u to substitute a part of the original integrand, simplifying the integral in the process. For example, if you're given \[\int (2x)e^{x^2}\,dx\], you could let \[u = x^2\], and the integral becomes \[\int e^u\,du\], which is much simpler to evaluate. While this particular method wasn't used in the solution, it's a powerful tool in the calculus arsenal for tackling more complex integrals.
Definite Integrals
Definite integrals provide a way to calculate the accumulated quantity, such as area under a curve, within a certain interval \[a, b\]. Unlike an indefinite integral, which represents a general form of antiderivatives, a definite integral is a number that gives the net area under the graph of the function between the specified limits.

Using the fundamental theorem of calculus simplifies the process of evaluating definite integrals. As you've seen in the exercise, evaluating the definite integral from \[a\] to \[b\] of the derivative of a function results in a straightforward computation of differences at the boundaries, hence both integrals provided in the exercise equated to zero when evaluated.
Power Rule of Integration
The power rule for integration is a basic rule in integral calculus that can be used to find the integral of any power of x. The rule states that for any real number n, except for -1, the integral of \[x^n\] with respect to x is \[\frac{x^{n+1}}{n+1}\], plus the constant of integration in the indefinite case.

In the context of definite integrals, as shown in the textbook exercise, the power rule can be applied directly to evaluate the antiderivative at specific limits. The exercise simplifies to a form where it's evident that the final computation after applying the upper and lower limits results in zero, illustrating a practical application of the power rule.

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Most popular questions from this chapter

Suppose \(f(x)=\sqrt[3]{x}\) is to be approximated near \(x=8 .\) Find the linear approximation to \(f\) at 8 . Then complete the following table, showing the errors in various approximations. Use a calculator to obtain the exact values. The percent error is \(100 \mid\) approximation \(-\) exact \(|/|\) exact \(\mid .\) Comment on the behavior of the errors as \(x\) approaches 8 . $$ \begin{array}{|l|l|l|l|} \hline {}{} {x} & \text { Linear approx. } & \text { Exact value } & \text { Percent error } \\ \hline 8.1 & & & \\ \hline 8.01 & & & \\ \hline 8.001 & & & \\ \hline 8.0001 & & & \\ \hline 7.9999 & & & \\ \hline 7.999 & & & \\ \hline 7.99 & & & \\ \hline 7.9 & & & \\ \hline \end{array} $$

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