Integral Calculus -4 - Mathematics MCQ

The equation of the curve is
p=f(r)
We know that tan 
and 
Hence 

Hence 

Given equations of cycloid are

Then 
and  ...(ii)
Hence  




So, 
Implies 
Hence From (iii), we ge

Limits for y -> 0,2 x = y²

To describe its first arch, θ varies from 0 to 2π
i.e., x varies from 0 to 2aπ

V = 4/5πa³

The area is present above the x-axis. Area above the x-axis is positive. The area is bounded by x-axis, curve y = f(x), straight lines x = a and x = b. Hence, area is found by integrating the curve with the lines as limits.

sin(4x)
=> sin(2x+2x)
=> sin(2x).cos(2x)+cos(2x).sin(2x)
=> 2[sin(x).cos(x)].cos(2x)+2[sin(x).cos(x)].cos(2x)
=> sin(4x)sin(x)=>2cos(x).cos(2x)+2cos(x).cos(2x)
Our integral is now reduced to 4cos(x).cos(2x)

I = ∫(0 to π) 4cos(x).cos(2x) 
I = ∫(0 to π) 4cos(x).(1−2sin2(x)) 
I = ∫(0 to π) 4cos(x)−8sin2(x).cos(x)) 
I =  ∫(0 to π)4cos(x) -  ∫8sin2(x).cos(x)) 
∫(0 to π) cos(x)=sin(x) 
∫(0 to π) sin²(x).cos(x))=sin3(x)3 
I = [sin(x)+8sin3(x)³+c ](0 to π)

Putting limits, hence we get answer 0.

Using the formula for even n we have

We have

Let us trace the curve roughly to get the limits of integration.
(i) The curve is symmetrical about x−axis.
(ii) It passes through the origin.The tangents at the origin are ay² = ax² or y=±x
∴ Origin is a node.
(iii) The curve has no asymptotes.
(iv)The curve meets the x−axis at (0,0) and (a,0).It meets the y−axis at (0,0) only.
From the equation of the curve, we have y=​ x(a−x)^1/2/(a)^1/2​​
For x>a,y is imaginary.Thus, no portion of the curve lies to the right of the line x=a. Also x→−∞,y→∞
∴ Area of the loop=2(Area of upper half of the loop)