calchap8B_PorTcd2 - Integration by substitution Given the composition H(x = F g(x the chain rule for differentiation gives H(x = F g(x g(x The SFToC

calchap8B_PorTcd2 - Integration by substitution Given the...

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Unformatted text preview: Integration by substitution Given the composition H ( x ) = F ( g ( x ) ) , the chain rule for differentiation gives H ′ ( x ) = F ′ ( g ( x ) ) g ′ ( x ) . The SFToC applied to H ′ gives integraldisplay x a H ′ ( t ) dt = H ( x ) − H ( a ) or in terms of F and g integraldisplay x a F ′ ( g ( t ) ) g ′ ( t ) dt = F ( g ( x ) ) − F ( g ( a ) ) Now use SFToC to replace the difference of the F ’s by an integral to get integraldisplay x a F ′ ( g ( t ) ) g ′ ( t ) dt = integraldisplay g ( x ) g ( a ) dF du du – p. 1 Now replace F ′ by f in the last line above to get integraldisplay x a f ( g ( t ) ) g ′ ( t ) dt = integraldisplay g ( x ) g ( a ) f ( u ) du the usual form for integration by substitution for definite integrals. For anti-derivatives we have integraldisplay x f ( g ( t ) ) g ′ ( t ) dt = integraldisplay g ( x ) f ( u ) du or integraldisplay f ( g ( x ) ) g ′ ( x ) dx = integraldisplay f ( u ) du where u = g ( x ) . All of these formulas look as though we have put u = g ( t ) , or u = g ( x ) , and then said du = g ′ ( t ) dt or du = g ′ ( x ) dx . We exploit these appearances for mechanistic calculations when doing “integration by substitution”. – p. 2 Ex 5. Evaluate a) integraldisplay x 2 radicalbig 1 + x 3 dx b) integraldisplay 4 1 sin( π √ x ) √ x dx SOLUTION: a) If set u = 1 + x 3 then du dx = 3 x 2 or, abusing notation du = 3 x 2 dx so integraldisplay x 2 radicalbig 1 + x 3 dx = integraldisplay √ u du 3 = 1 3 u 3 / 2 3 / 2 + C by using result for anti-derivative of a power of u . Converting RHS back to x we get integraldisplay x 2 radicalbig 1 + x 3 dx = 2(1 + x 3 ) 3 / 2 9 + C – p. 3 b) Try u = √ x then du dx = 1 2 √ x or du = dx 2 √ x so that integraldisplay 4 1 sin( π √ x ) √ x dx = integraldisplay 2 1 sin( πu )2 du = 2 bracketleftbigg − cos( πu ) π bracketrightbigg 2 1 = − 2 π (cos 2 π − cos π ) = − 4 π General advice: use substitution to hide “ugly” functions. – p. 4 Integrals of functions with special properties Proposition 5. i) If f ( x ) is even on [ − a, a ] then integraltext a − a f ( t ) dt = 2 integraltext a f ( t ) dt ii) If f ( x ) is odd on [ − a, a ] then integraltext a − a f ( t ) dt = 0 iii) If f is periodic with period T then integraltext T + β β f ( t ) dt = integraltext T f ( t ) dt for all β ∈ R . PROOF: Will only prove part (iii); the others use similar ideas. integraldisplay β + T β f ( t ) dt = integraldisplay β f ( t ) dt + integraldisplay β + T f ( t ) dt use t = s- T in the first integral = integraldisplay T β + T f ( s − T ) ds + integraldisplay β + T f ( t ) dt = integraldisplay T β + T f ( s ) ds + integraldisplay β + T f ( t ) dt as f ( s- T ) = f ( s ) by periodicity = integraldisplay T f ( t ) dt....
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