# The area enclosed between the curves y = x^3 and y = √x is in square units ?

##
A 3/5

B 5/12

C 12/5

D 5/4

E 5/3

A 3/5

B 5/12

C 12/5

D 5/4

E 5/3

I found:

Have a look at the diagram and the area described by the two curves:

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To find the area enclosed between the curves (y = x^3) and (y = \sqrt{x}), we need to determine the points of intersection of these two curves and then integrate the absolute difference between them over that interval.

To find the points of intersection, set (x^3 = \sqrt{x}) and solve for (x): [x^3 = \sqrt{x}] [x^6 = x] [x^6 - x = 0] [x(x^5 - 1) = 0]

So, (x = 0) or (x^5 - 1 = 0).

Solving (x^5 - 1 = 0) gives us (x = 1).

Now, we integrate the absolute difference of the curves from (x = 0) to (x = 1): [A = \int_{0}^{1} |x^3 - \sqrt{x}| , dx]

[A = \int_{0}^{1} (x^3 - \sqrt{x}) , dx]

[A = \left[\frac{x^4}{4} - \frac{2x^{3/2}}{3}\right]_{0}^{1}]

[A = \left(\frac{1}{4} - \frac{2}{3}\right) - \left(0 - 0\right)]

[A = \frac{1}{4} - \frac{2}{3}]

[A = \frac{3}{12} - \frac{8}{12}]

[A = -\frac{5}{12}]

Since area cannot be negative, we take the absolute value:

[A = \frac{5}{12}]

So, the area enclosed between the curves (y = x^3) and (y = \sqrt{x}) is ( \frac{5}{12} ) square units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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