   Chapter 4.3, Problem 48E

Chapter
Section
Textbook Problem

# Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch. ∫ π / 6 2 π cos x   d x

To determine

To:

(i) Evaluate the integral,

(ii) Interpret the integral as a difference of areas.Illustrate with sketch.

Explanation

1) Concept:

i) Fundamental theorem of Calculus, Part 2

If f is continuous on a,b, then

abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, a function F such that F'=f

2) Given:

y=cosx,  π6x2π

3) Calculation:

(i)

The function fx=cosx is continuous on π/6, 2π

By using concept i) (Fundamental theorem of Calculus, Part 2),

π/62πcosx  dx=F2π-Fπ/6(1)

Where F is antiderivative of f, that is, a function F such that F'(x)=f(x) means

ddxFx=fx(2)

By using concept ii) (sine rule of antiderivative),

ddxsinx=cosx

From (2),

Fx=sinx

Substitute F(x) in (1) at x=π/6  and  x=2π,

π/62πcosx  dx=F2π-F(π/6)

=sin2π-sinπ6

=0-12

=-12

Therefore,

π/62πcosx  dx=-12

(ii) To interpret integral as difference of areas

The curve y=cosx,  π6x2π are given by,

From above graph, the given curve y=cosx,  π6x2π bounded between x=π6 and x=2π

The region enclosed by the curve y=cosx,  x=π6, x=2π is shown bythe black region

From the above graph, the curve y=fx=cosx bounded between x=π/6 and x=2π is continuous on π/6, 2π, divide the given region into three parts,

y=fx=cosx bounded between x=π/6  and x=2π and

y=fx=cosx bounded between x=π/6 and x=2π, is continuous on π/6,π/2, π/2,3π/2 andon 3π/2, 2π

Therefore,

π/62

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