Integration Help (1 Viewer)

[Drewk]

Find the area bounded by the curve y=sqrt(4-x^2), the x-axis and the y-axis in the first quadrant
Could some one plz do this through integration
i set limits 2 and 0, but it gives me like 5.3333
Please help

 

[Trebla]

Note that the curve is a semi-circle so you can just apply the formula for the area of a circle and divide it by 4

 

[Nws m8]

Note that te curve is a semi-circle so you can just apply the formula for the area of a circle and divide it by 4

+1

 

[Drewk]

Note that te curve is a semi-circle so you can just apply the formula for the area of a circle and divide it by 4

Yeah but why is it not working with integration??

 

[Trebla]

Yeah but why is it not working with integration??

It can be integrated directly using an Ext1 technique called substitution (e.g. let x = 2sin u)

 

[Drewk]

It can be integrated directly using an Ext1 technique called substitution (e.g. let x = 2sin u)

sorry if i seem a bit arrogant, but i just want to know if integrating this is wrong ?
Cause i think i am doing every thing correct but getting 5.333

 

[Nws m8]

sorry if i seem a bit arrogant, but i just want to know if integrating this is wrong ?
Cause i think i am doing every thing correct but getting 5.333

Treble is right man, you can't integrate that expression without substitution

 

[Trebla]

sorry if i seem a bit arrogant, but i just want to know if integrating this is wrong ?
Cause i think i am doing every thing correct but getting 5.333

The answer is pi. Show your working. This question cannot be done algebraically within the scope of the 2u course.

 

[Drewk]

The answer is pi. Show your working. This question cannot be done algebraically within the scope of the 2u course.

Limits are 2 and 0
Integrating sqrt(4-x^2)
= (2(4-x^2)^3/2)/3
Subbing in 2 i get 0, subbing in 0 i get 5.333

 

[Trebla]

Limits are 2 and 0
Integrating sqrt(4-x^2)
= (2(4-x^2)^3/2)/3
Subbing in 2 i get 0, subbing in 0 i get 5.333

Try differentiating your primitive, you will see that you do not get back the integrand. That whole power business only works if you have a power expression alone like xⁿ, not as a function of a another function.

 

[NewtownTutoring]

not as a function of a another function.

Unless that function happens to be linear, then just divide by the coefficient of whatever you're integrating with respect to.

 

[Kostie100]

Limits are 2 and 0
Integrating sqrt(4-x^2)
= (2(4-x^2)^3/2)/3
Subbing in 2 i get 0, subbing in 0 i get 5.333

You cannot use that rule as the function is not linear.

 

[iSplicer]

You cannot use that rule as the function is not linear.

^