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This guide explores the technique of integration by substitution, including the application of the chain rule for differentiable functions. It provides step-by-step instructions on how to identify inner and outer functions, apply the general power rule, and change variables effectively. It includes model problems for better understanding, key strategies for definite integrals, and essential guidelines for making variable changes. This resource is ideal for students seeking to enhance their calculus skills and tackle more complex integrals with confidence.
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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