Question

Use the region in the first quadrant bounded by √x, y=2 and the y - axis to determine the area of the region. Evaluate the integral.

A. 50.265

B. 4/3

C. 16

D. 8

E. 8π

F. 20/3

G. 8/3

E/ -16/3

Answer 1

G. 8/3

Explanation:

You want the area between y=2 and y=√x.

Bounds

The square root curve is only defined for x ≥ 0. It will have a value of 2 or less for ...

√x ≤ 2

x ≤ 4 . . . . square both sides

So, the integral has bounds of 0 and 4.

Integral

The integral is ...

`\displaystyle \int_0^4{(2-x^(1)/(2))}\,dx=\left[2x-(2)/(3)x^(3)/(2)\right]_0^4=8-(2)/(3)(√(4))^3=\boxed{(8)/(3)}`

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Additional comment

You will notice that this is 1/3 of the area of the rectangle that is 4 units wide and 2 units high. That means the area inside a parabola is 2/3 of the area of the enclosing rectangle. This is a useful relation to keep in the back of your mind.