<PRE><B>Question 32659</B>
do you mean integral 0 to 1 OF SQRT(X)
IF SO INTEGRAL SQRT.(X)=(2/3)*(X)^(3/2)..FROM THE FORMULA..
INTEGRAL (X)^N = (X^(N+1))/(N+1)
SO PUTTING THE LIMITS 0 TO 1 WE GET
(2/3)(1^(3/2)-0^(3/2))=2/3
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OK NOW THE QUESTION IS CLEAR.WE HAVE F(X)=SQRT(X).WE HAVE TO FIND THE
AREA UNDER THE CURVE BY APPROXIMATING AT HIGHER AND LOWER LIMITS OF
INTERVALS.SO MAKE A TBLE LIKE THIS...
TOTAL INTERVAL.....................0 TO 1
SUB DIVISIONS .....4 NOS....SO EACH SUB INTERVAL IS 0.25
AREA =SUB INTERVAL * F(X)=0.25*F(X)...AT LOWER /UPPER LIMITS OF INTERVALS.

    BEGINING OF INTERVAL                                         END OF ITERVAL
X       F(X)=X^0.5      AREA            X       F(X)=X^0.5      AREA
0.000   0.000   0.000           0.250   0.500   0.125
0.250   0.500   0.125           0.500   0.707   0.177
0.500   0.707   0.177           0.750   0.866   0.217
0.750   0.866   0.217           1.000   1.000   0.250

TOTAL           0.518           TOTAL           0.768

       AVERAGE AREA =  (0.518+0.768)/2 = 0.643

       ACTUAL AREA =           INTEGRAL SQRT.(X)…FROM 0 TO 1
                       =2/3=0.667                      AS SHOWN EARLIER.</PRE>