Integration question, help needed asap ?

Question

Determine the area enclosed by the following curves and the x-axis.

y = sinθ between θ = -π/2 and θ = π

Please could you show your working, much appreciated.

Answer 1

Yeah.... I'm not that good at maths and you seem to be skiving off with your homework. No can do. Besides what is sin and that 0 sign? Don't answer. Just illustrating my point.

Answer 2

y = sinθ  over [-π/2, π]

note: when you have to find out an area, your integrand must be positive; the area itself can assume positive values only;

therefore, instead of finding merely the  definite integral of sinθ over [-π/2, π], you have rather to evaluate:

∫ [-π/2, π] | sinθ | dθ =

how do you get rid of the abs value? Simply, examining the sign of sinθ over [-π/2, π], that is:

sinθ ≤ 0  over [-π/2, 0]

sinθ ≥ 0  over [0, π]

(looking at sine function graph would greatly help...!)

hence you have to split your integral into:

∫ [-π/2, 0] (- sinθ) dθ + ∫ [0, π] sinθ dθ

this expression is just the same as:

∫ [-π/2, π] | sinθ | dθ =

in that | sinθ | means:

- sinθ if sinθ < 0;  

sinθ if sinθ ≥ 0

thus you have to evaluate:

∫ [-π/2, 0] (- sinθ) dθ + ∫ [0, π] sinθ dθ

the antiderivative of sinθ is - cosθ, therefore you get:

∫ [-π/2, 0] (- sinθ) dθ = [cosθ][-π/2, 0] = cos(0) - cos(-π/2) = 1 - 0 = 1

∫ [0, π] sinθ dθ = [- cosθ][0, π]  = - cos(π) - [- cos(0)] = - (-1) + 1 = 1 + 1 = 2

thus the required area is:

A = ∫ [-π/2, 0] (- sinθ) dθ + ∫ [0, π] sinθ dθ = 1 + 2 = 3

I hope it helps...

Bye!

Answer 3

Nice try little buddy - now get on with your homework ;)