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Chapter 7: Problem 5

What type of integrand is a good candidate for integration by parts?

Short Answer

Expert verified
A. Choosing u(x) and dv(x) randomly. B. Prioritizing functions as per the mnemonic LIATE. C. Ignoring patterns or simplifications. D. Differentiating and integrating the whole integrand. Answer: B. Prioritizing functions as per the mnemonic LIATE.

Step by step solution

01

Identify the form of the integrand

When determining if an integrand is a good candidate for integration by parts, the first step is to look at the integrand's form. An integrand is a candidate for integration by parts if it can be expressed as a product of two simpler functions, say, u(x) and v(x).
02

Choose u(x) and dv(x) smartly

When applying integration by parts, it is crucial to choose u(x) and dv(x) wisely. Generally, u(x) should be a function that becomes simpler when differentiated, whereas the function dv(x) must be the one that is easier to integrate. A common mnemonic to help in this choice is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), where you prioritize the choice of u(x) from left to right.
03

Apply integration by parts formula

Once u(x) and dv(x) are chosen, apply the integration by parts formula, which states: 鈭玼(x)v'(x)dx = u(x)v(x) - 鈭玽(x)u'(x)dx In the formula, differentiate u(x) to get du(x) and integrate dv(x) to get v(x). Then, substitute these values and solve the resulting integral if necessary. If the resulting integral is still complex, the integration by parts can be applied again.
04

Look for patterns or simplifications

When applying integration by parts multiple times, it might result in a pattern or simplification that makes the integral solvable. In such cases, keep applying the technique until the pattern becomes apparent.

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Most popular questions from this chapter

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 7.2 and an illustration.

Apply Simpson's Rule to the following integrals. It is easiest to obtain the Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(7 .\) Make \(a\) table similar to Table 7.8 showing the approximations and errors for \(n=4,8,16,\) and \(32 .\) The exact values of the integrals are given for computing the error. \(\int_{1}^{e} \ln x d x=1\)

Consider the family of functions \(f(x)=1 / x^{p},\) where \(p\) is a real number. For what values of \(p\) does the integral \(\int_{0}^{1} f(x) d x\) exist? What is its value?

\(\pi<\frac{22}{7}\) One of the earliest approximations to \(\pi\) is \(\frac{22}{7} .\) Verify that \(0<\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\frac{22}{7}-\pi .\) Why can you conclude that \(\pi<\frac{22}{7} ?\)
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