Proving Theorems on Definite Integrals at SGGSJ Government College, Paonta Sahib
Explore mathematical theorems on definite integrals at SGGSJ Government College, Paonta Sahib, with detailed proofs and explanations. Understand concepts such as definite integral properties and their applications in solving mathematical problems.
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SGGSJ GOVERNMENT COLLEGE PAONTA SAHIB
Theorem 1: On Definite Integral Theorem 2 : On Definite Integral Theorem 3 : On Definite Integral Theorem 4 : On Definite Integral Theorem 5 : On Definite Integral Theorem 6 : On Definite Integral Theorem 7 : On Definite Integral
b b dx = = f( z ) a Theorem 1: Prove that f(x) f(x) dx f( z ) dz dz a Proof . Let f ( x ) Now L.H.S = = f (x) R.H.S R.H.S = [ (z) ] { from eqn. ( = [ (z) ] { from eqn. (1 1) putting x=z, = = (b) (b) (a) (a) L.H.S = R.H.S L.H.S = R.H.S Hence the result Hence the result dx = = ( x ) f ( x ) dx ( x ) . ( . (1 1) ) b b dx = =[ [ (x) (x)] ] = (b) = (b) (a) (a) Now L.H.S f (x) dx a a b a ) putting x=z, dx dx = = dz dz } }
b a Theorem 2 : Prove that f(x) dx = = - - f(x) f(x) dx a f(x) dx b dx Proof. Let f(x) Then L .H .S = f(x) R.H.S = - - f(x) = [ b ] L..H..S = R.H.S Hence the result . f(x) dx dx = (x) = (x) b b Then L .H .S = dx = [ (x) ] = (b) = [ (x) ] = (b) (a) a (a) f(x) dx a a a b dx = = - - [ (x)] [ (x)] = = - - [ (a) = [ b ] [b] L..H..S = R.H.S Hence the result . [ (a) (b) ] [b] R.H.S = (b) ] f(x) dx b
c b b Theorem 3 : Prove that f(x) Proof . Let f(x) f(x) dx L.H.S. = f(x) R.H.S. = = f(x) = =[ (c) L..H..S = R.H.S Hence the result dx = = f(x) dx + + f(x) c f(x) dx f(x) dx f(x) dx dx a a Proof . Let dx = (x) = (x) b b [ (x)] = = (b) a b dx = = [ (x)] L.H.S. = f(x) dx a c (b) (a) (a) b dx + + f(x) dx = = [ (x)] + [ (x)] [ (x)] + [ (x)] (c)] = = (b) (b) (a) c a R.H.S. f(x) dx f(x) dx c c a (a) ] + + [ (b) [ (c) (a) ] L..H..S = R.H.S [ (b) (c)] (a) Hence the result
a a dx = = f(a Theorem 4 : Prove that f(x) Proof. Proof. Put x When x= When x= 0 0 , z = a ; When x= a , z = f(x) f(x) dx = = f(a f(a- -x) f(a f(a - -x) 0 f(a - -x) f(x) dx x) dx dx 0 0 Put x= =a a - -z z dx dx = = - - dz dz , z = a ; When x= a , z = 0 0 dx = =- - f(a f(a - -z) a a a 0 dz = = f(a z) dz f(a - -z) z) dz dz 0 0 a x) dx dx a a 0 dx = = f(a f(a - -x) x) dx dx x) dx 0
2a a a Theorem 5 : Prove that f ( x ) Proof. We have f ( x ) Let Put x= Put x= 2 2a a z , dx = = f(x) dx = = f ( x ) f ( x ) dx dx + + f( f(2 2a a - - x) dx + + f ( x ) f ( x ) dx f ( x ) dx f(x) dx x) dx ( (1 1) ) dx 0 0 0 dx Proof. We have f ( x ) dx 2a Let I = I = f ( x ) 1 f ( x ) dx z , dx dx= = dz dz dx 0 When x= a , z= a ; When x= When x= a , z= a ; When x= 2 2a , z= I I= = - - f( f(2 2a a - - z) a , z= 0 0 dz = = f( f(2 2a a z) 0 a a z) dz z) dz dz 1 0 a = = f( f(2 2a a - - x) f(x) from eqn. ( from eqn. (1 1) we get , f(x) x) dx dx = = f( f(2 2a a - - x) ) we get , dx = = f(x) f(x) dx 0 dx 0 2a a f(x) dx x) dx dx 0 0 2a a a dx + + f( f(2 2a a - - x) 0 f(x) dx 0 x) dx dx
Theorem 6 : Prove that (i) if f(2a-x) = f(x), then f(x) dx = 2 f(x) dx (ii) if f(2a-x) = -f(x), then f(x) dx =0 2a a 0 0 2a 0 a a 2a Proof. We have f(x) dx = f(x) dx + f(2a-x) dx ..(1) (i) if f(2a-x) = f(x), then from eqn.(1) f(x) dx = f(x) dx + f(x) dx f(x) dx = 2 f(x) dx 0 0 0 0 0 a a 2a 0 0 0 2a a (ii) if f(2a-x) = -f(x), then from eqn.(1) f(x) dx = f(x) dx - f(x) dx 2a a a 0 0 0
f(x) dx =0 na a Theorem 7 : If f(x) = f(a+x), then prove that f(x) dx = n f(x)dx 0 0 na a na 2a Proof. We have f(x) dx = f(x) dx + f(x) dx + + f(x) dx Let I = f(x)dx Put x= a+z dx dx = = dz dz When x = a, z = 0 When x= 2a, z= a I = I = f(a+z) dx = f(a+x) dx = f(x) dx [ f(a+x)= f(x) ] 0 a 0 2a n-1 1 a a a a 1 0 0 0
2a a f(x) dx = f(x) dx (2) Again let I = f(x) dx 2 0 0 3a 2a Put x= a+z, dx dx = = dz dz When x = When x = 2 2a, z = a a, z = a When x = When x = 3 3a, z = I = I = f(a+z) dz = f(a+x) dx = f(x) dx [ f(a+x) = f(x) ] 2 a a a, z = 2 2a a 2a 2a 2a a 3a a f(x) dx= f(x) dx [ from eqn. 2 ] 2a 0 and so on. from eqn.( from eqn.(1 1) we get ) we get
na a a a f(x) dx = f(x) dx + f(x) dx + f(x) dx + .. to n terms. 0 0 0 na a 0 f(x) dx = n f(x) dx 0 0
