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nexus9112 [7]
3 years ago
10

Find the area bounded by the curve y = x 1/2 + 2, the x-axis, and the lines x = 1 and x = 4 A. 16 B. 10 2/3 C. 7 1/2 D. 28 1/2

Mathematics
1 answer:
jeka943 years ago
6 0
To find the area of the curve subject to these constraints, we must take the integral of y = x ^ (1/2) + 2 from x=1 to x=4
Take the antiderivative: Remember that this what the original function would be if our derivative was x^(1/2) + 2 
antiderivative (x ^(1/2) + 2) =  (2/3) x^(3/2) + 2x
* To check that this is correct, take the derivative of our anti-derivative and make sure it equals x^(1/2) + 2

To find integral from 1 to 4:
Find anti-derivative at x=4, and subtract from the anti-derivative at x=1
2/3 * 4 ^ (3/2) + 2(4) - (2/3) *1 - 2*1
2/3 (8) + 8 - 2/3 - 2                               Collect like terms
2/3 (7) + 6                                             Express 6 in terms of 2/3
2/3 (7) + 2/3 (9)
2/3 (16) = 32/3 = 10 2/3          Answer is B

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Step-by-step explanation:

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3 years ago
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
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The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

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