Answered: Evaluating an integral without the…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Evaluating an integral without the Fundamental Theorem of
Caleulus Evaluate f,7 In (1 + tan x) dx using the following
steps.
a. If f is integrable on [0, b], use substitution to show that
SöF(x) dx = [?(f(x) + f(b – x)) dv.
tan a + tan B
b. Use part (a) and the identity tan (a + B)
- tan a tan B
to evaluate In (1 + tan x) dr.

Expert Solution

Step 1

$w e h a v e t o e v a l u a t e \int_{0}^{\frac{π}{4}} \ln (1 + \tan x) d x u \sin g t h e f o l l o w i n g s t e p s : a) i f f i s i n t e g r a b l e o n [0, b], u s e s u b s t i t u t i o n t o s h o w t h a t \int_{0}^{b} f (x) d x = \int_{0}^{\frac{b}{2}} (f (x) + f (b - x)) d x b) u s e p a r t a) a n d i d e n t i f y \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} t o e v a l u a t e \int_{0}^{\frac{π}{4}} n (1 + \tan x) d x$

Step 2

$a) C o n s i d e r t h e i n t e g r a l \int_{0}^{b} f (x) d x s p l i t t h e i n t e r v a l o f i n t e g r a t i o n [0, b] i n t o t w o s u b i n t e r v a l s [0, \frac{b}{2}] a n d [\frac{b}{2}, b] \Rightarrow \int_{0}^{b} f (x) d x = \int_{0}^{\frac{b}{2}} f (x) d x + \int_{\frac{b}{2}}^{b} f (x) d x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) c o n s i d e r t h e i n t e g r a l \int_{\frac{b}{2}}^{b} f (x) d x l e t b - x = t a n d d i f f e r e n t i a t e t h e a b o v e e q u a t i o n w i t h r e s p e c t t o x d x = - d t C o n s i d e r t h e e x p r e s s i o n x = b - t w h e n x = \frac{b}{2} t h e n t = b - \frac{b}{2} = \frac{b}{2} w h e n x = b t h e t = b - b = 0 t h u s, \int_{\frac{b}{2}}^{b} f (x) d x = \int_{\frac{b}{2}}^{0} f (b - t) \times - d t = \int_{0}^{\frac{b}{2}} f (b - t) d t = \int_{0}^{\frac{b}{2}} f (b - x) d x n o w s u b s t i t u e t h e v a l u e o f i n t e g r a l \int_{\frac{b}{2}}^{b} f (x) d x i n e q . (1) \int_{0}^{b} f (x) d x = \int_{0}^{\frac{b}{2}} f (x) d x + \int_{0}^{\frac{b}{2}} f (b - x) d x = \int_{0}^{\frac{b}{2}} (f (x) + f (b - x)) d x$