7. Find the area of the region bounded by the curves: y = sin x, the line y= 1 and y= axis to the right of the y-axis.

Question

Show complete solution

Answer 1

The curve meets the line y=1 at (π/2,1). Since we need to find the area between the curve and the y-axis the integrand xdy best applies. What we have is a set of infinitesimal rectangles of length x and width dy. Since y=sin(x), dy=cos(x)dx so we can replace dy with cos(x)dx and integrate xcos(x)dx by parts; or we could integrate sin^-1(y)dy. But my preference is to find the area between the curve and the x-axis using ydx, then subtracting the result from the area of the rectangle enclosing the curve, which has an area of π/2. So we have the integral interval [0,π/2] for x:

π/2-₀∫^π/2ydx=π/2-₀∫^π/2sin(x)dx=π/2+[cos(x)]₀^π/2=π/2+(0-1)=π/2-1=0.5708 square units approx.

To give you a more comprehensive answer we can calculate the other integrals and make sure they come up with the same result.

J=₀∫^π/2xcos(x)dx: let u=x and dv=cos(x)dx⇒du=dx and v=sin(x).

J=xsin(x)-∫sin(x)dx=xsin(x)+cos(x) evaluated for x∈[0,π/2]:

π/2-0+0-1=π/2-1, as before.

To integrate ₀∫¹sin^-1(y)dy we make the substitution y=sin(x), dy=cos(x)dx. Deja vu! The rest is history!

Hope you enjoyed the excursion!