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By using integration prove that the area of a circle of radius r units is πr2 square units

Answer
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Hint: Use the fact that the equation of a circle of radius r centred at the origin is x2+y2=r2. Observe that the curve is symmetrical in all the four quadrants. Hence find the area in the first quadrant, and hence the area of the circle will be four times the area in the first quadrant. For finding the area in the first curve quadrant express y in terms of x. Note that y>0 and hence take only the positive sign. Then use the fact that the area under the curve is given by abydx. Substitute suitable values of a and b and integrate and hence find the area.

Complete step-by-step answer:
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As is evident from the graph that the curve is symmetrical in the four quadrants. Hence, we will find the area in the first quadrant, and then the total area will be four times the area in the first quadrant.
Now, we have
x2+y2=r2
Subtracting x2 from both sides, we get
y2=r2x2
Hence, we have
y=±r2x2
Since in the first quadrant y0, we have
y=r2x2
Now consider the vertical strip CDEF.
Here CF = y and EF = dx.
Hence the area of the strip is ydx.
The total area in the first quadrant will be the sum of these vertical strips from A to B.
At point B, we have
y =0
Hence, we have
x2=r2x=±r
Since the abscissa of B is positive, we have
x=r
Hence, we have
The area in the first quadrant is 0rydx
Substituting the value of y, we get
The area in the first quadrant is 0rr2x2dx
Let I=0rr2x2dx
Finding the value of I:
Let x = r sint
Differentiating both sides with respect to t, we get
dxdt=rcostdx=rcostdt
When x = 0, we have rsint=0sint=0t=0
When x = r, we have
rsint=rsint=1t=π2
Hence, we have
I=0π2r2r2sin2trcostdt=r20π21sin2tcostdt
We know that 1sin2t=cos2t
Hence, we have
I=r20π2cos2tcostdt=r20π2|cost|cost
Since in the interval (0,π2) cost is positive, we have |cost|=cost
Hence, we have
I=r20π2cos2tdt (i)
We know that abf(x)dx=abf(a+bx)dx
Hence, we have
I=r20π2cos2(π2t)dt
We know that cos(π2x)=sinx
Hence, we have
I=r20π2sin2tdt (ii)
Adding equation (i) and equation (ii), we get
2I=r20π2(sin2t+cos2t)dt
We know that sin2t+cos2t=1
Hence, we have
2I=r20π21dt=r2(t|0π2)=r2(π20)=πr22
Dividing both sides by 2, we get
I=πr24
Hence the area in the first quadrant is πr24
Hence the area of the circle is 4×πr24=πr2
Q.E.D

Note:[1] We can directly solve I by using the fact that a2x2dx=x2a2x2+a22sin1xa+C
Hence, we have
I=0rr2x2=x2r2x2+r22sin1xr|0r=(r2r2r2+r22sin1rr)(02r202+r22sin10r)=(0+r22×π2)(0+0)=πr24
Which is the same as obtained above.

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