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IBSL Maths Definite Integration © Christine Crisp Definite Integration e.g.1 1 2 3 x 2 2 dx is a definite integral The numbers on the integral sign are called the limits of integration Definite Integration Evaluating the Definite Integral The definite integration results in a value. e.g.1 1 2 3 x 2 2 dx 1 3x 3 2x 31 Find the indefinite integral but omit C Definite Integration Evaluating the Definite Integral The definite integration results in a value. e.g.1 1 2 3 x 2 2 dx 1 x 3 2x 2 Draw square brackets and hang the limits on the end Definite Integration Evaluating the Definite Integral The definite integration results in a value. e.g.1 1 2 3 x 2 2 dx 1 12 Replace x with • • x 3 2x 2 3 ( 2 ) 2(2) the top limit the bottom limit Definite Integration Evaluating the Definite Integral The definite integration results in a value. e.g.1 1 2 3 x 2 2 dx 1 12 x 3 2x 2 3 ( 2 ) Subtract and evaluate 2(2) Definite Integration Evaluating the Definite Integral The definite integration results in a value. e.g.1 1 2 3 x 2 2 dx 12 1 So, 2 1 x 3 2x 2 3 84 15 2 3 ( 2 ) 3 x 2 dx 15 2(2) Definite Integration SUMMARY The method for evaluating the definite integral is: Find the indefinite integral but omit C Draw square brackets and hang the limits on the end Replace x with • • the top limit the bottom limit Subtract and evaluate Definite Integration Evaluating the Definite Integral 1 1 e.g. 2 Find x 7x x 7 x 12 dx 12 x 2 3 1 1 Solution: x 2 7 x 12 dx 2 3 2 Indefinite integral but no C 1 1 Definite Integration Evaluating the Definite Integral 1 1 e.g. 2 Find x 7x x 7 x 12 dx 12 x 2 3 1 1 Solution: x 2 7 x 12 dx 2 3 2 1 1 3 2 1 7 ( 1 ) 7 ( 1 ) Substitute for x : 12 12( 1) top minus limit 2 3 2 3 limit bottom 1 7 1 7 12 12 Simplify 3 2 3 2 Definite Integration Evaluating the Definite Integral 1 7 12 3 2 1 7 3 2 12 In this example, if we can’t use a calculator, we can We must be 1 7 1 7 save time by collecting 12terms from 12both brackets. very careful 3 2 2 24 3 2 24 3 3 2 with the signs 7 7 0 2 2 Definite Integration Exercises 2 1 1. Find 1 2 3 x 4x 3 x 4 x 1 dx x 21 31 1 2 3 2 ( 2) 3 2( 2) 2 2 1 2 1 2 14 2 12 1 2 6 x 2 x 2. Find 6 x 2 x 3 dx 3 x 21 31 2 2 2 2 2 3 2 2( 2) 3 1( 2) 2 3( 2) 2( 2) 3 ( 2) 2 3( 2) 16 4 6 16 4 6 6 ( 14) 20