#### To determine

**To:**

(i) Evaluate the integral,

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

#### Answer

(i) ∫π/62πcosx dx=-12

(ii) ∫π/62πcosx dx=∫π6π2cosx dx+∫π23π2cosx dx+∫3π22πcosx dx

#### 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, π6≤x≤2π

**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, π6≤x≤2π are given by,

From above graph, the given curve y=cosx, π6≤x≤2π 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/62πcosx dx=∫π/6π/2cosx dx+∫π/23π/2cosx dx+∫3π/22πcosx dx

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

∫π6π2cosx dx+∫π23π2cosx dx+∫3π22πcosx dx

=Fπ2-Fπ6+F3π2-Fπ2+F2π-F3π2

=F2π-Fπ6…(2)

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) (power rule of antiderivative),

ddxsinx=cosx

From (2),

Fx=sinx

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

∫π6π2cosx dx+∫π23π2cosx dx+∫3π22πcosx dx=F2π-Fπ6

=sin2π-sinπ/6

=0-12

=-12

∫π/62πcosx dx

Therefore,

∫π6π2cosx dx+∫π23π2cosx dx+∫3π22πcosx dx=-12

**Conclusion:**

Therefore, the required area is A=-12