Question

1. Find the area of the finite region enclosed by the curve y=2√x and the lines x=3 and x=0.
2. Calculate in square units the area of the finite region bounded by the line $\frac{x}{4} + \frac{y}{5} = 1$ , y-0 and x = b

Answer 1

9514 1404 393

Answer:

1. 4√3 ≈ 6.92820
2. 10

Step-by-step explanation:

1. The integral from 0 to 3 of 2x^(1/2) can be found using the power rule:

  area = 2(2/3x^(3/2)) for x = 3

  = 4/3(3√3)

  area = 4√3 . . . square units

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2. The line equation for y=0 is corrupted, so we assume the line equation for x is intended to be x=0. The area is a right triangle with side lengths 4 and 5, so the geometric area formula applies:

  A = 1/2bh = 1/2(4)(5) = 10

The area of the triangular region is 10 square units.

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Additional comment

If you really intend the right bound of the enclosed area to be x=b, then the area is the shape of a trapezoid. One base length is 5; the other is (5 -5/4b), and the height of the trapezoid is (b). The area is then ...

  A = 1/2(b1+b2)h

  A = 1/2(5 +5 -5/4b)(b) = 5b -5/8b^2