How does one find the area under a semicircle ? (1 Viewer)

[Forbidden.]

Let's say I want to find the area under the semicircle of radius 2 units, defined by the definite integral:

Image Below

Using Junior Mathematics the area of a semicircle is 0.5 x pi x r² and if radius is 2 units then:

0.5 x pi x 2²
= 6.283185307 ...
= 6.3 units²(to 1 dp)

But using definite integral calculus:

Image Below

why ?

EDIT: btw guys, my 3U class only finished primitive functions and people like me want to jump ahead at things instead of starting integration next year, lol how slow we are.

 

[pLuvia]

You have an x variable in that square root, you can't just integrate like that. You have to use a trig substitution like x=sinu etc. IF you want to do it the long way

The easier way and acceptted way is by using the area of semi circle with radius 2

 

[Affinity]

mate you got your integral wrong. When you integrate f(g(x)) it's not F(g(x))/g'(x) in general, many students think that works because it does when g'(x) is a constant.

The correct integral of sqrt(4-x^2) is:
(x/2)sqrt(4-x^2) + 2arcsin(x/2)

 

[ianc]

for 2 unit you are not expected to integrate it - you are expected to be able to sketch it and realise that the area you are finding is that of a semicircle with radius 2. then you just use 0.5(pi)r^2

 

[Riviet]

BUT since f3nr15 is doing 3u, the trig substitution will need to be learnt.

 

[stariiskiez]

Let's say I want to find the area under the semicircle of radius 2 units, defined by the definite integral:

Image Below

Using Junior Mathematics the area of a semicircle is 0.5 x pi x r² and if radius is 2 units then:

0.5 x pi x 2²
= 6.283185307 ...
= 6.3 units²(to 1 dp)

But using definite integral calculus:

Image Below

why ?

EDIT: btw guys, my 3U class only finished primitive functions and people like me want to jump ahead at things instead of starting integration next year, lol how slow we are.

I think it's suppose to be 2 x intergral of (the equation) from 0 to 2... not from -2 to 2.

 

[Forbidden.]

I think it's suppose to be 2 x intergral of (the equation) from 0 to 2... not from -2 to 2.

If you mean replacing the -2 with 2, still doesn't work because a number squared will always be positive.

Or do you mean finding intergral between 0 and 2 then doubling it which I can't be bothered to work out later ? fkn definite integral calculus