Easy Integration (1 Viewer)

[untouchablecuz]

I just started integration yesterday and i was wondering how you would integrate the positive semi-circle function NOT by geometrical methods (area formula) e.g. using integration by substitution. Thanks in advance.

 

[kaz1]

Reverse chain rule?

 

[ianc]

okay you can't integrate it directly because it is not a function (use the straight line test)

however, you can observe that by symmetry (where the radius is a):

A = 2 (integral from 0 to a) sqrt[a^2 - x^2] dx

(note that section of the circle is a function)

you'll need to make the substitution x=cos@ or x=sin@, and then just integrate away.

hope this helps!!

 

[untouchablecuz]

...

a semi circle is a function (for every x value there is one y value)

and we havent learnt the reverse chain rule

 

[tacogym27101990]

okay you can't integrate it directly because it is not a function (use the straight line test)

however, you can observe that by symmetry (where the radius is a):

A = 2 (integral from 0 to a) sqrt[a^2 - x^2] dx

(note that section of the circle is a function)

you'll need to make the substitution x=cos@ or x=sin@, and then just integrate away.

hope this helps!!

actually you'll need to use the substitution x=acos@

 

[untouchablecuz]

One more thing, reading cambridge now, isn't the reverse chain rule essentially integration by substitution?

 

[Timothy.Siu]

One more thing, reading cambridge now, isn't the reverse chain rule essentially integration by substitution?

yeah

 

[Trebla]

One more thing, reading cambridge now, isn't the reverse chain rule essentially integration by substitution?

Yes, but integration by substitution removes complicated algebra. However, it also involves more steps, especially with a definite integral. If the primitive is really obvious like ∫xe^x² dx, then you can write the answer straight away without long-winded substitution (unless of course you are specifically instructed to use a substitution).

 

[tommykins]

okay you can't integrate it directly because it is not a function (use the straight line test)

however, you can observe that by symmetry (where the radius is a):

A = 2 (integral from 0 to a) sqrt[a^2 - x^2] dx

(note that section of the circle is a function)

you'll need to make the substitution x=cos@ or x=sin@, and then just integrate away.

hope this helps!!

or just use area of a quarter circle.

a = pir^2/4, r = a
A = pia^2*2/4 = pi.a^2/2

handy if you're able to spot it out during exams.

 

[ianc]

actually you'll need to use the substitution x=acos@

either x=acos@ or x=asin@ works fine.