Integration Question (Sum + Difference of Areas) (1 Viewer)

[ThomasMcCorquodale]

"Calculate the area bounded by x=y^(2/3), and x=y^(2) in the first quadrant."

The correct answer is 4/15 units^2 but I keep getting 5/12 units^2. Could someone please confirm the correct answer? Cheers!

 

[VBN2470]

HINT: Draw a sketch of both functions and determine their points of intersection, which will define your interval on which you are integrating over.

 

[ThomasMcCorquodale]

Cheers for the quick response! Just one other question.

Find the area of the region between the graphs of y = 2x - x^2 and 3y = x^2 - 4x + 6. The correct answer is 1/36 units^2.

I'm just wondering how to deal with the 2nd equation. I should be fine from there.

 

[VBN2470]

Again, you apply the same procedure of solving the two equations simultaneously to obtain points of intersection which will then determine your limits of integration. So you would to 3(2x - x^2) = x^2 - 4x + 6 --> x = 3/2 or x = 1; then integrate over your desired interval.

 

[ThomasMcCorquodale]

Thanks so much