Q. 245.0( 2 Votes )

Evaluate <span la

Answer :

Know that, limit of sums is the sum of limits of the function.

We have,


Let f(x) = x2– x …(i)


Here,


a = 1


b = 4


nh = b – a = 4 – 1 = 3 …(*)


We know limit of sums is given as,



Put a = 1 as deduced above.



Put x = 1 in (i),


f(1) = 1 – 1


Put x = 1 + h in (i),


f(1 + h) = (1 + h)2 – (1 + h)


Repeat the same for other values of x. We get,











, nh = 3








OR


We have two equations:


One of them is the equation of curve, say y1.


y = √(5 – x2)


Squaring on both sides,


y2 = (√(5 – x2))2


y2 = 5 – x2


x2 + y2 = 5


It is a circle.


The other equation is of a line,


y = |x – 1|


Take y = x – 1 and y = 1 – x …(ii)


Let us find intersection point of these two figures by substituting y = x – 1 in y = √(5 – x2). We get,


x – 1 = √(5 – x2)


Squaring both sides,


(x – 1)2 = [√(5 – x2)]2


x2 + 1 – 2x = 5 – x2


x2 + x2 – 2x + 1 – 5 = 0


2x2 – 2x – 4 = 0


x2 – x – 2 = 0


x2 – 2x + x – 2 = 0


x(x – 2) + (x – 2) = 0


(x + 1)(x – 2) = 0


(x + 1) = 0 or (x – 2) = 0


x = -1 or x = 2


Put x = -1 in y = x – 1,


y = -1 – 1


y = -2


Point of intersection of circle x2 + y2 = 5 and line y = x – 1 is (-1, -2).


Put x = 2 in y = x – 1,


y = 2 – 1


y = 1


Point of intersection between x2 + y2 = 5 and line y = 1 – x is (2, 1).



So, the area of the region is given by













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