INTEGRALS

A Note In order to quicken this method, we can proceed as follows: After performing steps , and , there is no need of step . Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step. Let us illustrate this by examples. Example Evaluate .

Solution Put t = x + , then dt = x dx . t dt t = ( x + Hence, – ( ( ) – (– ) ( Alternatively , first we transform the integral and then evaluate the transformed integral with new limits. Let t = x + . Then dt = x dx .

Note that, when x = – , t = and when x = , t = Thus, as x varies from – to , t varies from to Therefore t dt – = ( Example Evaluate – INTEGRALS Solution Let t = tan – x , then . The new limits are, when x = , t = and when x = , . Thus, as x varies from to , t varies from to π . Therefore – t dt – EXERCISE .

Evaluate the integrals in Exercises to using substitution. . x + . φ φ φ .

. e dx Choose the correct answer in Exercises and . . The value of the integral is (A) (B) (C) (D) .

If f ( x ) = sin x t t dt , then f ′ ( x ) is (A) cos x + x sin x (B) x sin x (C) x cos x (D) sin x + x cos x . Some Properties of Definite Integrals We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily. P : f t dt P : .

In particular, a f x dx = P : P : f a P : f a (Note that P is a particular case of P ) P : ( f P : , if ( f f x and if f ( a – x ) = – f ( x ) P : a f x dx , if f is an even function, i.e., if f (– x ) = f ( x ). a f x dx , if f is an odd function, i.e., if f (– x ) = – f ( x ). We give the proofs of these properties one by one. Proof of P It follows directly by making the substitution x = t .

Proof of P Let F be anti derivative of f . Then, by the second fundamental theorem of calculus, we have F ( ) – F ( ) – [F ( ) F ( )] = − Here, we observe that, if a = b , then a f x dx = . Proof of P Let F be anti derivative of f . Then a f x dx = F( b ) – F( a ) ...

( ) a f x dx = F( c ) – F( a ) ... ( ) and c f x dx = F( b ) – F( c ) ... ( ) Adding ( ) and ( ), we get F( ) – F( ) This proves the property P . Proof of P Let t = a + b – x .

Then dt = – dx . When x = a , t = b and when x = b , t = a . Therefore a f x dx – ) b f a t dt INTEGRALS – ) a f a t dt (by P ) – ) a f a dx by P Proof of P Put t = a – x . Then dt = – dx .

When x = , t = a and when x = a , t = . Now proceed as in P . Proof of P Using P , we have . Let t = a – x in the second integral on the right hand side.

Then dt = – dx . When x = a , t = a and when x = a , t = . Also x = a – t . Therefore, the second integral becomes ( – ) a f t dt ( – ) a f t dt ( – ) a f Hence a f x dx ( f Proof of P Using P , we have ( f ...

( ) Now, if f ( a – x ) = f ( x ), then ( ) becomes a f x dx , and if f ( a – x ) = – f ( x ), then ( ) becomes a f x dx Proof of P Using P , we have a f x dx