Find the area enclosed by the curve y=sinx and the x-axis between x=0 and x=22/7
To find the area enclosed by the curve y = sin(x) and the x-axis between x = 0 and x = 22/7, we can utilize integration. The area under a curve can be calculated by integrating the function over the given interval.
First, let's graph the function y = sin(x) and identify the region we want to find the area of.
The graph of y = sin(x) is a periodic curve that oscillates between -1 and 1. If we consider the interval x = 0 to x = 22/7 (which corresponds to one complete period), we can see that the curve is above the x-axis, forming a region bounded by the curve and the x-axis.
To find the area of this region, we need to integrate the function y = sin(x) over the interval x = 0 to x = 22/7. In other words, we need to find the definite integral ∫[0, 22/7] sin(x) dx.
To integrate sin(x), we can use the standard integration formula:
∫ sin(x) dx = -cos(x) + C,
where C is the constant of integration.
Applying this formula to our definite integral, we get:
∫[0, 22/7] sin(x) dx = -cos(x) ∣[0, 22/7]
= -cos(22/7) - (-cos(0))
Evaluating this expression, we find:
Area = -cos(22/7) - (-cos(0))
Since cos(0) = 1, and cos(22/7) is approximately -0.9999, we have:
Area ≈ -(-0.9999) - 1
≈ 0.9999 - 1
≈ -0.0001
Therefore, the area enclosed by the curve y = sin(x) and the x-axis between x = 0 and x = 22/7 is approximately -0.0001 square units.