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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 77. (page 452)

Area under a Curve The area under the graph of $y = \frac{1}{1 + x^{2}}$ and above the $x$ -axis between $x = a a n d x = b$ is given by
$\tan^{- 1} b - \tan^{- 1} a$
(a) Find the exact area under the graph of $y = \frac{1}{1 + x^{2}}$ and above the $x$ -axis between $x = 0 a n d x = \sqrt{3} .$
(b) Find the exact area under the graph of $y = \frac{1}{1 + x^{2}}$ and above the $x$ -axis between $x = - \frac{\sqrt{3}}{3} a n d x = 1$ .

Short Answer

Expert verified

The exact Area of the curve is:

(a) \frac{蟺}{3} (b) \frac{5 蟺}{12} .

Step by step solution

01

Step 1. Given Information.

The given function is

$\tan^{- 1} b - \tan^{- 1} a$ .

02

Part (a) Step 1. Calculation of area.

The given values are $a = 0, b = \sqrt{3}$

Then, we get

$\begin{array}{rcl} \tan^{- 1} b - \tan^{- 1} a & = & \tan^{- 1} (\sqrt{3}) - \tan^{- 1} (0) \\ = & \frac{蟺}{3} - 0 \\ = & \frac{蟺}{3} . \end{array}$

03

Part (b) Step 1. Calculation of area.

The given values are $b = 1, a = - \frac{\sqrt{3}}{3}$ .

The values we get,

$\begin{array}{rcl} \tan^{- 1} b - \tan^{- 1} a & = & - \tan^{- 1} (- \frac{\sqrt{3}}{3}) + \tan^{- 1} (1) \\ = & - (- \frac{蟺}{6}) - \frac{蟺}{4} \\ = & \frac{蟺}{6} + \frac{蟺}{4} \\ = & \frac{5 蟺}{12} . \end{array}$ .

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