1. Find the area of the finite region enclosed by the curve y=2√x and the lines x=3 and x=0.
2 answers:
9514 1404 393
Answer:
- 4√3 ≈ 6.92820
- 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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<em>Additional comment</em>
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
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Answer:
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Step-by-step explanation:
If we connect a line from point A to point C we create a triangle, and we can use pythagorean theorem to find this distance. So the line from point A to point C will be our hypotenuse and the other two distances will be out side lengths.
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61=
=AC or 7.8102=AC