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Express the integral in terms of the variable \(u\), but do not evaluate it. (a) \(\int_{1}^{3}(2 x-1)^{3} d x ; u=2 x-1\) (b) \(\int_{0}^{4} 3 x \sqrt{25-x^{2}} d x ; u=25-x^{2}\) (c) \(\int_{-1 / 2}^{1 / 2} \cos (\pi \theta) d \theta ; u=\pi \theta\) (d) \(\int_{0}^{1}(x+2)(x+1)^{5} d x ; u=x+1\)

Short Answer

Expert verified
The integrals are: (a) \(\frac{1}{2} \int_{1}^{5} u^{3} \, du\), (b) \(-\frac{3}{2} \int_{9}^{25} \sqrt{u} \, du\), (c) \(\frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos u \, du\), and (d) \(\int_{1}^{2} (u+1)u^{5} \, du\).

Step by step solution

01

Substitute for Part (a)

For the integral \( \int_{1}^{3}(2x-1)^{3} \, dx \) with \( u = 2x-1 \), calculate the derivative \( \frac{du}{dx} = 2 \). Thus, \( dx = \frac{du}{2} \). Change the limits of integration using \( x = 1 \rightarrow u = 2 \times 1 - 1 = 1 \) and \( x = 3 \rightarrow u = 2 \times 3 - 1 = 5 \). Hence, the integral becomes \( \int_{1}^{5} u^{3} \, \frac{du}{2} = \frac{1}{2} \int_{1}^{5} u^{3} \, du \).
02

Substitute for Part (b)

For \( \int_{0}^{4} 3x \sqrt{25-x^2} \, dx \) with \( u = 25-x^2 \), the derivative is \( \frac{du}{dx} = -2x \) implying \( xdx = \frac{-1}{2}du \). Change the limits using \( x=0 \rightarrow u=25 \) and \( x=4 \rightarrow u=9 \). The integral becomes \( \int_{25}^{9} 3 \cdot \sqrt{u} \cdot \frac{-1}{2} du = -\frac{3}{2} \int_{25}^{9} \sqrt{u} \, du \). Reverse integration limits to \( -\frac{3}{2} \int_{9}^{25} \sqrt{u} \, du \).
03

Substitute for Part (c)

For \( \int_{-1/2}^{1/2} \cos (\pi \theta) \, d\theta \) with \( u = \pi \theta \), \( \frac{du}{d\theta} = \pi \) so \( d\theta = \frac{du}{\pi} \). Change limits: \( \theta = -1/2 \rightarrow u = -\pi/2 \) and \( \theta = 1/2 \rightarrow u = \pi/2 \). The integral becomes \( \int_{-\pi/2}^{\pi/2} \cos u \, \frac{du}{\pi} = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos u \, du \).
04

Substitute for Part (d)

For \( \int_{0}^{1} (x+2)(x+1)^{5} \, dx \) with \( u = x+1 \), \( \frac{du}{dx} = 1 \) so \( dx = du \). Change limits: \( x=0 \rightarrow u=1 \) and \( x=1 \rightarrow u=2 \). Rewrite \( x+2 = u+1 \). The integral becomes \( \int_{1}^{2} (u+1)u^{5} \, du \).

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Key Concepts

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

Change of Variables
Integration by substitution, or change of variables, is a powerful technique in calculus used to simplify complex integrals. It involves replacing a difficult variable expression with a simpler one by introducing a new variable, usually denoted as \(u\). This method is particularly useful when dealing with intricate functions or compositions of functions.
Consider the integral \( \int (2x-1)^3 \, dx \). By letting \( u = 2x-1 \), we effectively simplify the polynomial inside the integral.
- Differentiate \( u = 2x - 1 \) to find \( \frac{du}{dx} = 2 \). This rearranges to \( dx = \frac{du}{2} \).- Substitute \( u \) and \( dx \) into the integral.
This transformation helps rewrite the original integral in terms of \( u \), leading to an easier integral \( \int u^3 \, \frac{du}{2} \). The substitution process transforms complex problems into ones that are more manageable.
Definite Integrals
Definite integrals calculate the area under a curve between two specified limits, often denoted as \(a\) and \(b\). These integrals provide a numerical result and are a foundational concept in calculus for determining total accumulation. When using the substitution method, the limits of integration also change to reflect the new variable.
For example, if you switch \( x \) to \( u \) via \( u = 2x - 1 \) in \( \int_{1}^{3} (2x-1)^3 \, dx \), calculate the new limits:- For \( x = 1 \), \( u = 2 \times 1 - 1 = 1 \).- For \( x = 3 \), \( u = 2 \times 3 - 1 = 5 \).
These calculations transform the integral to \( \frac{1}{2} \int_{1}^{5} u^3 \, du \), maintaining the correctness of the area's calculation under the transformed curve.
Limits of Integration
The limits of integration are the specified bounds between which the integration takes place. In definite integrals, these play a crucial role in the transformation process. When performing a substitution, adjusting these limits according to the new variable \( u \) is essential.
For the integral \( \int_{0}^{4} 3x \sqrt{25-x^2} \, dx \) with \( u = 25-x^2 \):- Originally, for \( x = 0 \), \( u = 25 - 0^2 = 25 \).- For \( x = 4 \), \( u = 25 - 4^2 = 9 \).
Notice the reversal of limits during integration, which requires altering the order or adjusting the sign of the integral. Correct limits ensure the substitution accurately reflects the original integral's outcome, particularly the geometric or physical interpretation of the problem.
Trigonometric Integrals
Trigonometric integrals often involve functions like sine and cosine, and substitution helps simplify them when integrating over specific intervals. This is common when the argument of the trigonometric function is not straightforward.
For instance, in \( \int_{-1/2}^{1/2} \cos(\pi \theta) \, d\theta \), use \( u = \pi \theta \):- With differentiation, \( \frac{du}{d\theta} = \pi \), giving \( d\theta = \frac{du}{\pi} \).- Limits need adjusting: for \( \theta = -1/2 \), \( u = \pi(-1/2) = -\pi/2 \); for \( \theta = 1/2 \), \( u = \pi(1/2) = \pi/2 \).
This substitution transforms the integral to \( \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos u \, du \), facilitating a simpler approach to finding the area or evaluating the integral. This method is particularly prevalent in solving problems involving waves and oscillations.

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Most popular questions from this chapter

(a) Give a geometric argument to show that \(\frac{1}{x+1}<\int_{x}^{x+1} \frac{1}{t} d t<\frac{1}{x}, \quad x>0\) (b) Use the result in part (a) to prove that $$ \frac{1}{x+1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}, \quad x>0 $$ (c) Use the result in part (b) to prove that $$ e^{x /(x+1)}<\left(1+\frac{1}{x}\right)^{x}0 $$ and hence that $$ \lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x}=e $$ (d) Use the result in part (b) to prove that $$ \left(1+\frac{1}{x}\right)^{x}0 $$

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$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=1}^{100}\left(2^{k+1}-2^{k}\right)$$

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