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Chapter 5: Problem 38

Evaluate the integrals by any method. $$ \int_{\pi^{2}}^{4 \pi^{2}} \frac{1}{\sqrt{x}} \sin \sqrt{x} d x $$

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Step by step solution

01

Identify the Substitution

To evaluate the integral \( \int_{\pi^{2}}^{4 \pi^{2}} \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx \), we notice that \( x \) appears inside both the square root and the sine function. Thus, we will use the substitution \( u = \sqrt{x} \).
02

Differentiate and Solve for dx

Differentiate the substitution \( u = \sqrt{x} \) to find \( du = \frac{1}{2\sqrt{x}} \, dx \), which means \( dx = 2u \, du \). Substituting \( \sqrt{x} = u \) and \( dx = 2u \, du \) into the integral modifies it into a simpler form.
03

Change the Limits of Integration

With the substitution \( u = \sqrt{x} \), change the limits of integration. When \( x = \pi^2 \), \( u = \sqrt{\pi^2} = \pi \). When \( x = 4\pi^2 \), \( u = \sqrt{4\pi^2} = 2\pi \). The new limits of integration become \( u = \pi \) to \( u = 2\pi \).
04

Substitute and Simplify the Integral

Substitute the expressions and limits into the integral: \[\int_{\pi}^{2\pi} \frac{1}{u} \sin u \cdot 2u \, du = \int_{\pi}^{2\pi} 2 \sin u \, du.\] The terms \( u \) cancel out, leaving the simplified integral \( 2 \int_{\pi}^{2\pi} \sin u \, du \).
05

Evaluate the Integral

The integral \( \int \sin u \, du = -\cos u + C \). Apply the limits of integration to the expression:\[2 ([-\cos u]_{\pi}^{2\pi}).\]
06

Compute the Definite Integral

Compute the expression:\[2 ([-\cos(2\pi) + \cos(\pi)]) = 2([ -1 - -1 ]) = 2(0) = 0.\] Hence, the evaluation of the integral gives \( 0 \).

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Key Concepts

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

The Substitution Method
The substitution method is a powerful technique used in calculus to simplify integrals, especially when dealing with complex expressions. In the given problem, we used substitution to transform the integral
  • We observed the expression has \( x \) inside both \( \sqrt{x} \) and \( \sin(\sqrt{x}) \).
  • This suggests a natural substitution: let \( u = \sqrt{x} \).
  • By differentiating, we derived \( du = \frac{1}{2\sqrt{x}} \, dx \), which can be rearranged to \( dx = 2u \, du \).
Using substitution, we change the variable of integration, simplifying the integral's form. Substitution is particularly useful when a function and its derivative are present in the integrand. Here, the transformation allows us to eliminate the square root and simplify the sine term. Always remember:
  • Identify an inner function within the integrand.
  • Express \( dx \) in terms of \( du \).
  • Change the limits of integration if dealing with definite integrals.
Definite Integrals
Definite integrals are used to calculate the net area under a curve within specific boundaries. Unlike indefinite integrals, which determine a whole family of antiderivative functions, definite integrals compute an exact value.
  • Our original integral had limits from \( \pi^2 \) to \( 4\pi^2 \).
  • After substitution, we needed to transform these limits to match the new variable \( u \).
  • This resulted in new limits \( u = \pi \) to \( u = 2\pi \).
The key steps include evaluating the antiderivative of the integrand, applying the integration limits, and performing subtraction. The definite integral's value, calculated using the Fundamental Theorem of Calculus, shows the function's accumulation over a specific interval, providing a concrete numerical result. In our case, calculating the area resulted in zero, indicating that the contributions on the interval cancel out.
Trigonometric Integrals
Trigonometric integrals involve functions like \( \sin(x) \) and \( \cos(x) \). These integrals are common in calculus, often requiring specific techniques for evaluation.
  • In our problem, the integral reduced to \( \int \sin u \, du \).
  • The antiderivative of \( \sin u \) is \( -\cos u \).
  • Trigonometric identities and substitutions can simplify such integrals.
Understanding the basic antiderivatives is crucial for solving trigonometric integrals smoothly. When integrating \( \sin u \), the result \( -\cos u \) indicates the general behavior of sine as it transitions smoothly between positive and negative values. Thus, these integrals often relate to geometric interpretations, like wave functions or periodic phenomena, highlighting their significance across various fields in math and physics.

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Most popular questions from this chapter

Prove the Mean-Value Theorem for Integrals (Theorem 5.6 .2 ) by applying the Mean-Value Theorem (4.8.2) to an antiderivative \(F\) for \(f .\)

Evaluate the integrals by any method. $$ \int_{0}^{e} \frac{d x}{2 x+e} $$

CASprograms have commands for working with most of the important nonelementary functions. Check your CAS documentantion for information about the error function erf(x) [see Formula (12)], and then complete the following. (a) Generate the graph of erf(x). (d) Use the graph to make a conjecture about the existence and location of any inflection points of erf \((x) .\) (e) Check your conjecture in part (d) using the second derivative of erf(x). (f) Use the graph to make a conjecture about the existence of horizontal asymptotes of erf(x). (g) Check your conjecture in part (f) by using the CAS to find the limits of erf(x) as \(x \rightarrow \pm \infty\). (b) Use the graph to make a conjecture about the existence and location of any relative maxima and minima of erf(x). (c) Check your conjecture in part (b) using the derivative of erf(x).

Let \(F(x)=\int_{0}^{x} \frac{t-3}{t^{2}+7} d t\) for \(-\infty< x <+\infty\) (a) Find the value of \(x\) where \(F\) attains its minimum value. (b) Find intervals over which \(F\) is only increasing or only decreasing. (c) Find open intervals over which \(F\) is only concave up or only concave down.

Express \(F(x)\) in a piecewise form that does not involve an integral. $$ F(x)=\int_{0}^{x} f(t) d t, \text { where } f(x)=\left\\{\begin{array}{ll}{x,} & {0 \leq x \leq 2} \\ {2,} & {x>2}\end{array}\right. $$
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