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Aim: What is Integration by Substitution?. Do Now:. Chain Rule. If y = f ( u ) is a differentiable function of u and u = g ( x ) is a differentiable function of x , then y = f ( g ( x )) is a differentiable function of x and. or, equivalently,.
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Chain Rule If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and or, equivalently, Think of the composite function as having 2 parts – an inner part and an outer part. outer inner
Let u = 3x – 2x2 General Power Rule u’ n un - 1 Do Now
u - substitution chain rule substitute & simplify
Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then u – substitution in Integration From the definition of an antiderivative it follows that:
inside function g(x) g(x) (g(x))2 Recognizing the Pattern g(x) g(x) g’(x) g’(x) g(x) = x2 + 1 g’(x) = 2x
g’ g(x) cos(g(x)) g’ Model Problem What is the inside function, u? 5x Check
g(x) k! Multiplying/Dividing by a Constant What is the inside function? x2 + 1 problem! factor of 2 is missing
Multiplying/Dividing by a Constant Constant Multiple Rule Integrate Check
inside function Change of Variable - u Recognizing the Pattern g(x) u g(x) u g(x) u g(x) u g’(x) u’ u2 (g(x))2 u = g(x) = x2 + 1 u’ = g’(x) = 2x
Constant Multiple Rule Change of Variables - u Calculate the differential Substitute in terms of u Antiderivative in terms of u Antiderivative in terms of x
Guidelines for Making Change of Variables 1. Choose a substitution u = g(x). Usually it is best to choose the inner part of a composite function, such as a quantity raised to a power. 2. Compute du = g’(x) dx 3. Rewrite the integral in terms of the variable u. 4. Evaluate the resulting integral in terms of u. 5. Replace u by g(x) to obtain an antiderivative in terms of x. 6. Check your answer by differentiating.
Model Problem Substitute in terms of u Antiderivative in terms of u Antiderivative in terms of x
Model Problem u = sin 3x Check
u4 du u5/5 General Power Rule for Integration u= 3x – 1
u1/2 u1 du du u2/2 Model Problems u= x2 + x u= x3 – 2
u2 u-2 du du u-1/(-1) Model Problems u= 1 – 2x2 u= cos x
determine new upper and lower limits of integration lower limit upper limit When x = 0, u = 02 + 1 = 1 x = 1, u = 12 + 1 = 2 integration limits for x and u Change of Variables for Definite Integrals du= 2x dx u= x2 + 1
lower limit upper limit When x = 1, u = 1 x = 5, u = 3 Model Problem determine new upper and lower limits of integration
Model Problem Before substitution After substitution = Area of region is 16/3 Area of region is 16/3
Even and Odd Functions To prove even, use f(x) = f(-x) and then substitute u = -x
Model Problem = 0