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

#### ThomasMcCorquodale

##### New Member
"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

##### Well-Known Member
$\bg_white The integral would be given as \int_{0}^{1} x^{\frac{1}{2}}-x^{\frac{3}{2}} dx = \frac{4}{15}. Alternatively, you could do \int_{0}^{1} y^{\frac{2}{3}}-y^{2} dx = \frac{4}{15}$

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

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#### ThomasMcCorquodale

##### New Member
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

##### Well-Known Member
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.

Thanks so much