Integration by Substitution MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics Fall 2018
Objectives After this lesson we will be able to use the method of integration by substitution to find the integrals of more complicated functions.
Power Rule Recall the Power Rule for integration, x r dx = 1 r + 1 x r+1 + C if r 1.
Power Rule Recall the Power Rule for integration, x r dx = 1 r + 1 x r+1 + C if r 1. From the simple power rule we can develop a more generally applicable integration formula.
Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4.
Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x 3 + 2 is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x 3 + 2.
Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x 3 + 2 is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x 3 + 2. d [ (x 3 + 2) 5] = 5(x 3 + 2) 4 (3x 2 ) dx [ ] d 1 dx 5 (x 3 + 2) 5 = 3x 2 (x 3 + 2) 4
Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x 3 + 2 is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x 3 + 2. d [ (x 3 + 2) 5] = 5(x 3 + 2) 4 (3x 2 ) dx [ ] d 1 dx 5 (x 3 + 2) 5 = 3x 2 (x 3 + 2) 4 Therefore 3x 2 (x 3 + 2) 4 dx = 1 5 (x 3 + 2) 5 + C.
Substitution Consider the indefinite integral: (x 3 + 2) 4 3x 2 dx
Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du
Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du
Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du = 1 5 u5 + C
Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du = 1 5 u5 + C = 1 5 (x 3 + 2) 5 + C
General Power Rule Theorem If u is a differentiable function of x, then u n du dx dx = u n du = 1 n + 1 un+1 + C if n 1.
General Power Rule Theorem If u is a differentiable function of x, then u n du dx dx = u n du = 1 n + 1 un+1 + C if n 1. Remark: in some integrals we may be able to substitute the symbol u for a function and then carry out an elementary integration. This process is known as integration by substitution.
Integration with Function Notation k g (x) dx = k g(x) + C (g(x)) r g 1 (x) dx = r + 1 (g(x))r+1 + C (if r 1) g (x) dx = ln g(x) + C g(x) e g(x) g (x) dx = e g(x) + C.
Integration with u-substitution Suppose u is a differentiable function of x, then k du = ku + C u r 1 du = r + 1 ur+1 + C (if r 1) 1 du = ln u + C u e u du = e u + C.
Guidelines for Integration by Substitution 1. Let u be a function of x (usually part of a composition of functions in the integrand). 2. Solve for x and dx in terms of u and du. 3. Convert the integral in variable x into an integral in variable u. 4. After integrating, rewrite the antiderivative as a function of x. 5. Check the answer by differentiating. f (g(x)) g (x) dx = f (u) du = f (u) + C = f (g(x)) + C
Example (1 of 4) Evaluate the indefinite integral (3x 2 + 6)(x 3 + 6x) 2 dx.
Example (1 of 4) Evaluate the indefinite integral (3x 2 + 6)(x 3 + 6x) 2 dx. If we let u = x 3 + 6x then du = (3x 2 + 6) dx. (3x 2 + 6)(x 3 + 6x) 2 dx = u 2 du = 1 3 u3 + C = 1 ( 3 x 3 + 6x) + C 3
Example (2 of 4) Evaluate the indefinite integral 2(3x 4 + 1) 2 dx.
Example (2 of 4) Evaluate the indefinite integral 2(3x 4 + 1) 2 dx. If we let u = 3x 4 + 1 then du = 12x 3 dx which cannot be made to equal 2 dx, thus here we must expand the product and then integrate without substitution. ( 2(3x 4 + 1) 2 dx = 18x 8 + 12x 4 + 2) dx = 2x 9 + 12 5 x 5 + 2x + C
Example (3 of 4) Evaluate the indefinite integral x 2 e 2x 3 dx.
Example (3 of 4) Evaluate the indefinite integral x 2 e 2x 3 dx. If we let u = 2x 3 then du = 6x 2 dx 1 6 du = x 2 dx. Making the substitution we get x 2 e 2x 3 dx = 1 6 eu du = 1 6 eu + C = 1 6 e 2x 3 + C.
Example (4 of 4) Evaluate the indefinite integral (ln x) 3 dx. x
Example (4 of 4) Evaluate the indefinite integral (ln x) 3 dx. x If we let u = ln x then du = 1 x dx. Making the substitution we get (ln x) 3 dx = (ln x) 3 1 x x dx = u 3 du = 1 4 u4 + C = 1 4 (ln x)4 + C.
Propensity to Consume As the income level of a family increases, the family tends to save more and spend less than 100% of their income on necessities. The rate of change of consumption with respect to income is called the marginal propensity to consume. Example Suppose that a family of four with a total income of $25, 000 will spend 100% of their income on necessities. As the income of the family increases above $25, 000 the marginal propensity to consume is dq dx = 0.94 (x 24, 999) 0.06.
Example 1. Find the consumption Q as a function of income x. 2. Find the income saved and the income consumed if x = $75, 000.
Example 1. Find the consumption Q as a function of income x. Q(x) = 0.94 dx (x 24, 999) 0.06 = 0.94 u 0.06 du = u 0.94 + C Q(x) = (x 24, 999) 0.94 + C Q(25, 000) = 25, 000 = (25, 000 24, 999) 0.94 + C Q(x) = (x 24, 999) 0.94 + 24, 999 2. Find the income saved and the income consumed if x = $75, 000.
Example 1. Find the consumption Q as a function of income x. Q(x) = 0.94 dx (x 24, 999) 0.06 = 0.94 u 0.06 du = u 0.94 + C Q(x) = (x 24, 999) 0.94 + C Q(25, 000) = 25, 000 = (25, 000 24, 999) 0.94 + C Q(x) = (x 24, 999) 0.94 + 24, 999 2. Find the income saved and the income consumed if x = $75, 000. Q(75, 000) = (75, 000 24, 999) 0.94 + 24, 999 = 51, 123 consumed and 75, 000 51, 123 = 23, 877 saved.
Demand Function Suppose the quantity demanded x is a function of the price per unit p for a certain product. If dx dp = 300 (0.03p 1) 3 and x = 10, 000 when p = $105 find the demand function x = f (p).
Solution dx dp = 300 (0.03p 1) 3 If u = 0.03p 1 then 100 3 du = dp and 300 x = (0.03p 1) 3 dp 100 = 300 3 u 3 du = 10000 u 3 du f (p) = 5000u 2 5000 + C = (0.03p 1) 2 + C f (105) = 5000 (0.03(105) 1) 2 + C = 10000 f (p) = 5000 (0.03p 1) 2 + 8918.33