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Calculate the area of the region enclosed by the x-axis and one arch of the curve y =...

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beatrice0866 | Student, College Freshman | (Level 1) Honors

Posted October 25, 2010 at 4:44 PM via web

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Calculate the area of the region enclosed by the x-axis and one arch of the curve y = sin x.

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted October 25, 2010 at 7:29 PM (Answer #1)

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y = sinx. To find the area under the curve and x axis for one arc of the function.

We know that sinx  has an arc   y = sinx above axis for x in (0 , pi).

So the area A under the arc of the curve y = sinx is got by integrating y = sinx fro x = 0 to x= pi.

A = Integral sinx dx from x = 0 to x = pi.

A= {(-cosx) at x= pi}- {-cosx)at x= 0}

A ={ - (-1) - (-1)} , as cos 0 = 1 and cos(pi) = -1

A= 1+1

A = 2.

Therefore the area undeer the curve y = sinx and x axis  in the interval (0,pi) is 2.

 

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william1941 | College Teacher | (Level 3) Valedictorian

Posted October 25, 2010 at 4:46 PM (Answer #2)

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We see that the curve given is the sine curve and it intersects the x-axis at x = 0, x = pi, x = 2*pi and so on. To find the area of the region enclosed by the x- axis and one arch of the curve we have to find the definite integral of the curve between 0 and pi.

Now Int [sin x] = - cos x.

Between the values x = 0 and x = pi, the integral is equal to –cos pi – (- cos 0) = - (-1) + 1 = 1 + 1 = 2.

Therefore the required area of the region between the x-axis and one arch of the curve y = sin x is equal to 2.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted October 25, 2010 at 5:16 PM (Answer #3)

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The area bounded by the curve y = sin x and x axis is the definite integral of the given function.

Since the boundary lines are not given, we'll suppose that x = -pi/2 and x = pi/2.

Int sin x dx = -cos x

We'll apply Leibniz Newton formula:

Int sin x dx = F(b) - F(a)

a = -pi/2 and b = pi/2Since the sine function is odd, we'll consider the integral as:

Int sin x dx = 2 Int sin x dx from x = 0 to x = pi/2.

2Int sin x dx = 2[-cos (pi/2) + cos 0]

2Int sin x dx = 2(0 + 1)

2Int sin x dx = 2

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