Aim: Integration by Substitution Course: Calculus Do Now: Aim: What is Integration by Substitution?

Aim: Integration by Substitution Course: Calculus 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

Aim: Integration by Substitution Course: Calculus Do Now Let u = 3x – 2x 2 General Power Rule nu n - 1 u’

Aim: Integration by Substitution Course: Calculus u - substitution & simplify substitute chain rule

Aim: Integration by Substitution Course: Calculus u – substitution in Integration From the definition of an antiderivative it follows that: 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

Aim: Integration by Substitution Course: Calculus Recognizing the Pattern inside function g(x) g(x)g(x)g(x)g(x) g’(x) g(x) = x 2 + 1g’(x) = 2x g(x)g(x) (g(x)) 2 g’(x)

Aim: Integration by Substitution Course: Calculus Model Problem g(x)g(x)g’ What is the inside function, u? cos(g(x))g’ 5x5x Check

Aim: Integration by Substitution Course: Calculus Multiplying/Dividing by a Constant g(x)g(x) What is the inside function?x factor of 2 is missing problem! k!k!

Aim: Integration by Substitution Course: Calculus Multiplying/Dividing by a Constant Check Constant Multiple Rule Integrate

Aim: Integration by Substitution Course: Calculus Recognizing the Pattern inside function g(x)g(x)g(x)g(x) g’(x) u = g(x) = x 2 + 1u’ = g’(x) = 2x g(x)g(x) (g(x)) 2 uu g(x)g(x)u u u’ u2u2 Change of Variable - u

Aim: Integration by Substitution Course: Calculus Change of Variables - u Calculate the differential Substitute in terms of u Constant Multiple Rule Antiderivative in terms of u Antiderivative in terms of x

Aim: Integration by Substitution Course: Calculus 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.

Aim: Integration by Substitution Course: Calculus Model Problem Substitute in terms of u Antiderivative in terms of u Antiderivative in terms of x

Aim: Integration by Substitution Course: Calculus Model Problem u = sin 3x Check

Aim: Integration by Substitution Course: Calculus General Power Rule for Integration u4u4 du u 5 /5 u = 3x – 1

Aim: Integration by Substitution Course: Calculus Model Problems u1u1 du u 2 /2 u = x 2 + x u 1/2 du u = x 3 – 2

Aim: Integration by Substitution Course: Calculus Model Problems u -2 du u -1 /(-1) u = 1 – 2x 2 u2u2 du u = cos x

Aim: Integration by Substitution Course: Calculus Change of Variables for Definite Integrals u = x du = 2x dx When x = 0, u = = 1 lower limitupper limit x = 1, u = = 2 determine new upper and lower limits of integration integration limits for x and u

Aim: Integration by Substitution Course: Calculus Model Problem When x = 1, u = 1 lower limitupper limit x = 5, u = 3 determine new upper and lower limits of integration

Aim: Integration by Substitution Course: Calculus Model Problem Area of region is 16/3 Before substitution Area of region is 16/3 After substitution =

Aim: Integration by Substitution Course: Calculus Even and Odd Functions To prove even, use f(x) = f(-x) and then substitute u = -x

Aim: Integration by Substitution Course: Calculus Model Problem = 0

Aim: Integration by Substitution Course: Calculus Multiplying/Dividing by a Constant

Aim: Integration by Substitution Course: Calculus Multiplying/Dividing by a Constant