/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 37

Evaluate the following integrals or state that they diverge. $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}$$. Answer: 2

Step by step solution

01

Identify the substitution

We notice that the integral involves a square root in the denominator and the expression inside the square root is linear (\(x-1\)). We can perform a substitution to simplify the integral. Let's let \(u = x-1\). Then, \(d u = d x\) and the limits of integration will change as well.
02

Change limits of integration

When \(x = 1\), \(u = 1 - 1 = 0\). When \(x = 2\), \(u = 2 - 1 = 1\). So the new limits of integration are from \(0\) to \(1\).
03

Rewrite the integral using the substitution

Now that we have our substitution \(u = x-1\) and the new limits of integration, we can rewrite the integral as: $$\int_{0}^{1} \frac{d u}{\sqrt{u}}$$
04

Evaluate the integral

We can rewrite the integral in terms of powers of \(u\): $$\int_{0}^{1} u^{-\frac{1}{2}} d u$$ Now let's find the antiderivative. To do this, we'll add \(1\) to the power and divide by the new power: $$\frac{u^{\frac{1}{2}}}{\frac{1}{2}}$$ Let's evaluate this antiderivative at the limits of integration: $$\left[\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right]_0^1$$ Now substitute the limits of integration: $$\frac{1^{\frac{1}{2}}}{\frac{1}{2}} - \frac{0^{\frac{1}{2}}}{\frac{1}{2}}$$ Finally, simplify the expression: $$2(1)-2(0) = 2$$ The final result of the definite integral is: $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}=2$$.

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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 \(\tan \theta \leq 0\) ). $$\begin{aligned} &\text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}=\\\ &\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{aligned}$$

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is an integer. $$\int x(a x+b)^{n} d x(\text { Use } u=a x+b$$

Two integration approaches Evaluate \(\int \cos (\ln x) d x\) two different ways: a. Use tables after first using the substitution \(u=\ln x.\) b. Use integration by parts twice to verify your answer to part (a).

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?

A remarkable integral It is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4}\) for all real numbers \(m .\) a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2\) \(1,3 / 2,\) and \(2,\) and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m.\)
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