SOLUTION: This is from integrational calculus. Find the upper and lower sum of equation y=sqrt(x) on intrevals [0,1] thank you.

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Question 32659: This is from integrational calculus.
Find the upper and lower sum of equation y=sqrt(x) on intrevals [0,1]
thank you.
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
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.