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

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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.

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.**

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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