Question

1x dx

Answer 1

Let the integral be f( x ),

f( x )= ∫ a b xdx

The integration of the function is,

∫ a b f( x )dx =( b−a ) lim n→∞ 1 n ( f( a )+f( a+h )+f( a+2h )+...+f( a+( n−1 )h ) )

Substitute the values,

a=a,b=b h= b−a n f( x )=x

And,

f( a+h )=a+h f( a+2h )=a+2h f( a+( n−1 )h )=a+( n−1 )h

Now, the integral becomes,

∫ a b xdx =( b−a ) lim n→∞ 1 n ( f( a )+f( a+h )+f( a+2h )+...+f( a+( n−1 )h ) ) =( b−a ) lim n→∞ 1 n ( ( a )+( a+h )+( a+2h )+...+( a+( n−1 )h ) ) =( b−a ) lim n→∞ 1 n ( ( na )+h( 1+2+...+( n−1 ) ) ) (1)

Simplify the above series,

1+2+3+...+n= n( n+1 ) 2 1+2+3+...+n−1= ( n−1 )( n−1+1 ) 2 = n( n−1 ) 2

Now, substitute the value of series in equation (1).

∫ a b xdx =( b−a ) lim n→∞ 1 n ( ( na )+ ( b−a ) n ( n( n−1 ) 2 ) ) [ ∵h= b−a n ] =( b−a ) lim n→∞ ( ( a )+ ( b−a ) n ( ( n−1 ) 2 ) ) =( b−a ) lim n→∞ ( ( a )+ ( b−a ) 2 ( n n − 1 n ) ) =( b−a )( a+ b−a 2 )

Further simplify,

∫ a b xdx = ( b−a )( b+a ) 2 = b 2 − a 2 2

Thus, the definite integral function f( x ) is b 2 − a 2 2 .