To determine

To:

(i) Evaluate the integral,

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

Answer

(i) ∫π/62πcos⁡x  dx=-12

(ii) ∫π/62πcos⁡x  dx=∫π6π2cos⁡x  dx+∫π23π2cos⁡x  dx+∫3π22πcos⁡x  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=cos⁡x,  π6≤x≤2π

3) Calculation:

(i)

The function fx=cos⁡x is continuous on π/6, 2π

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

∫π/62πcos⁡x  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),

ddxsin⁡x=cos⁡x

From (2),

Fx=sin⁡x

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

∫π/62πcos⁡x  dx=F2π-F(π/6)

=sin⁡2π-sin⁡π6

=0-12

=-12

Therefore,

∫π/62πcos⁡x  dx=-12

(ii) To interpret integral as difference of areas

The curve y=cos⁡x,  π6≤x≤2π are given by,

From above graph, the given curve y=cos⁡x,  π6≤x≤2π bounded between x=π6 and x=2π

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

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

y=fx=cos⁡x bounded between x=π/6  and x=2π and

y=fx=cos⁡x 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/62πcos⁡x  dx=∫π/6π/2cos⁡x  dx+∫π/23π/2cos⁡x  dx+∫3π/22πcos⁡x  dx

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

∫π6π2cos⁡x  dx+∫π23π2cos⁡x  dx+∫3π22πcos⁡x  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),

ddxsin⁡x=cos⁡x

From (2),

Fx=sin⁡x

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

∫π6π2cos⁡x  dx+∫π23π2cos⁡x  dx+∫3π22πcos⁡x  dx=F2π-Fπ6

=sin⁡2π-sin⁡π/6

=0-12

=-12

∫π/62πcos⁡x  dx

Therefore,

∫π6π2cos⁡x  dx+∫π23π2cos⁡x  dx+∫3π22πcos⁡x  dx=-12

Conclusion:

Therefore, the required area is A=-12