Calculus: Integration by Substitution and the General Power Rule

The General Power Rule for Integration

  • Technique Overview: Finding antiderivatives for more complicated functions often requires a technique called integration by substitution, which is based on the Chain Rule for differentiation.

  • Conceptual Foundation: When looking for a function whose derivative is 2x(x2+1)32x(x^2 + 1)^3, the key observation is the presence of the factor 2x2x. Because 2x2x is the exact derivative of (x2+1)(x^2 + 1), the integrand can be viewed in the form undudxu^n \frac{du}{dx}.

  • Defining the General Power Rule: The rule states that for an integrand containing a factor uu raised to a power nn, the derivative du/dxdu/dx must also be a factor of the integrand.

  • Substitution Notation: Letting u=x2+1u = x^2 + 1 and du=2x dxdu = 2x\,dx, the integral is rewritten:     ∫(x2+1)32x dx=∫u3 du=u44+C=(x2+1)44+C\int (x^2 + 1)^3 2x\,dx = \int u^3\,du = \frac{u^4}{4} + C = \frac{(x^2 + 1)^4}{4} + C

  • Role of the Derivative: The technique is effective specifically because the inner function's derivative is present in the integrand. If the derivative factor is missing or different, the General Power Rule cannot be applied directly without modification.

Adjusting the Integrand: Multiplying and Dividing by a Constant

  • Missing Constant Adjustment: In many cases, the derivative du/dxdu/dx is partially missing from the integrand. If the missing part is a constant factor, you can adjust the integrand to fit the General Power Rule.

  • Example 2: Integration with Constant Manipulation: Find ∫x(3−4x2)2 dx\int x(3 - 4x^2)^2 \,dx.

    • Step 1: Let u=3−4x2u = 3 - 4x^2.

    • Step 2: Calculate the derivative du/dx=−8xdu/dx = -8x. The integrand currently contains xx, but it needs −8x-8x.

    • Step 3: Multiply and divide the integrand by the constant −8-8. The dividing factor −1/8-1/8 is moved outside the integral sign.

    • Step 4: Integrate using the General Power Rule:         ∫x(3−4x2)2 dx=−18∫(3−4x2)2(−8x) dx\int x(3 - 4x^2)^2\,dx = -\frac{1}{8} \int (3 - 4x^2)^2(-8x)\,dx         (−18)[(3−4x2)33]+C( - \frac{1}{8} ) [ \frac{(3 - 4x^2)^3}{3} ] + C         −(3−4x2)324+C-\frac{(3 - 4x^2)^3}{24} + C

  • Rule for Variables: This strategy of multiplying and dividing only works for constants. You cannot multiply and divide by a variable to create the necessary du/dxdu/dx factor.

Limitations and Alternative Strategies

  • Failure of the General Power Rule (Example 4): Find ∫−8(3−4x2)2 dx\int -8(3 - 4x^2)^2 \,dx.

    • In this case, letting u=3−4x2u = 3 - 4x^2 requires du/dx=−8xdu/dx = -8x.

    • Although there is a −8-8 in the integrand, the variable xx is missing. Because you cannot factor a variable out of an integral sign, you cannot "create" the missing xx.

    • Solution Strategy: To solve this indefinite integral, expand the integrand: (3−4x2)2=9−24x2+16x4(3 - 4x^2)^2 = 9 - 24x^2 + 16x^4. Then distribute the −8-8 and use the Simple Power Rule on each term individually.

  • Moving Extra Constant Factors (Example 5): Find ∫7x2x3+1 dx\int 7x^2 \sqrt{x^3 + 1} \,dx.

    • Step 1: Identify u=x3+1u = x^3 + 1. The required derivative is du/dx=3x2du/dx = 3x^2.

    • Step 2: Notice the extra constant factor of 77. Move this factor outside the integral sign: 7∫x2(x3+1)1/2 dx7 \int x^2(x^3 + 1)^{1/2}\,dx.

    • Step 3: Multiply and divide by 33 to create the required derivative factor.         73∫(x3+1)1/2(3x2) dx\frac{7}{3} \int (x^3 + 1)^{1/2}(3x^2)\,dx

    • Step 4: Integrate:         73[(x3+1)3/23/2]+C=73[23(x3+1)3/2]+C=149(x3+1)3/2+C\frac{7}{3} [ \frac{(x^3 + 1)^{3/2}}{3/2} ] + C = \frac{7}{3} [ \frac{2}{3}(x^3 + 1)^{3/2} ] + C = \frac{14}{9}(x^3 + 1)^{3/2} + C

Integration by Substitution: Change of Variables

  • The Procedure: For more complex integrands where the General Power Rule is difficult to recognize, the substitution or "change of variables" procedure is used to completely rewrite the integral in terms of a new variable uu.

  • Relationship to the General Power Rule: If u=f(x)u = f(x), then du=f′(x) dxdu = f'(x)\,dx. The integral form is ∫g(u) du\int g(u)\,du.

  • Guidelines for Integration by Substitution:

    1. Identify u: Let uu be a function of xx (typically the part of the integrand inside a power, radical, or denominator).

    2. Rewrite Integral: Replace all instances of xx and dxdx with uu and dudu.

    3. Integrate: Find the resulting integral in terms of the variable uu.

    4. Resubstitute: Rewrite the antiderivative as a function of xx by replacing uu with the original function of xx.

    5. Check: Differentiate the final result to ensure it matches the original integrand.

  • Example 7: Integration by Substitution Implementation: Find ∫xx2−1 dx\int x\sqrt{x^2 - 1}\,dx.

    • Let u=x2−1u = x^2 - 1. Then du=2x dxdu = 2x\,dx, which implies 12 du=x dx\frac{1}{2}\,du = x\,dx.

    • Substitute: ∫u(12) du=12∫u1/2 du\int \sqrt{u} (\frac{1}{2}) \,du = \frac{1}{2} \int u^{1/2}\,du.

    • Integrate: 12[u3/23/2]+C=13u3/2+C\frac{1}{2} [ \frac{u^{3/2}}{3/2} ] + C = \frac{1}{3}u^{3/2} + C.

    • Resubstitute: 13(x2−1)3/2+C\frac{1}{3}(x^2 - 1)^{3/2} + C.

    • Verification: ddx[13(x2−1)3/2+C]=13(32)(x2−1)1/2(2x)=x(x2−1)1/2=xx2−1\frac{d}{dx} [ \frac{1}{3}(x^2 - 1)^{3/2} + C ] = \frac{1}{3}(\frac{3}{2})(x^2 - 1)^{1/2}(2x) = x(x^2 - 1)^{1/2} = x\sqrt{x^2 - 1}.

Extended Application: Propensity to Consume

  • Economic Context (2013 Data): In 20132013, the U.S. poverty level for a family of four was approximately 23,90023,900. Families at this level tend to consume 100%100\% of their income for necessities (food, clothing, shelter).

  • Consumption Behavior: As income increases, the average consumption percentage typically drops below 100%100\%.

    • Example: A family earning 30,00030,000 might save 600600 and consume 29,40029,400 (98%98\% of income).

  • Marginal Propensity to Consume: This is defined as the rate of change of consumption with respect to income.

  • Example 8: Analyzing Consumption: For a family of four in 2013, the marginal propensity to consume income xx (in dollars) is modeled by:     dQdx=0.98(x−23,899)−0.02\frac{dQ}{dx} = 0.98(x - 23,899)^{-0.02}     where QQ represents income consumed.

  • Finding the Consumption Model:

    • Step 1: Integrate the rate of change:         Q=∫0.98(x−23,899)−0.02 dxQ = \int 0.98(x - 23,899)^{-0.02}\,dx         Using substitution where u=x−23,899u = x - 23,899 and du=dxdu = dx:         Q=0.98∫u−0.02 du=0.98[u0.980.98]+C=(x−23,899)0.98+CQ = 0.98 \int u^{-0.02}\,du = 0.98 [ \frac{u^{0.98}}{0.98} ] + C = (x - 23,899)^{0.98} + C

    • Step 2: Solve for the integration constant CC using the initial condition. At the poverty level, income x=23,900x = 23,900 results in consumption Q=23,900Q = 23,900.         23,900=(23,900−23,899)0.98+C23,900 = (23,900 - 23,899)^{0.98} + C         23,900=10.98+C23,900 = 1^{0.98} + C         23,900=1+C  ⟹  C=23,89923,900 = 1 + C \implies C = 23,899

    • Step 3: Final Model: Q=(x−23,899)0.98+23,899Q = (x - 23,899)^{0.98} + 23,899.

  • Estimation for Specific Income: For a family with an income of x=37,000x = 37,000:     Q=(37,000−23,899)0.98+23,899Q = (37,000 - 23,899)^{0.98} + 23,899     Q≈10,838+23,899≈34,737Q \approx 10,838 + 23,899 \approx 34,737     The family consumes approximately 34,73734,737 out of their 37,00037,000 income.

  • Visualizing Consumption: Figure 5.4 displays the graph of QQ, showing the split between "Income consumed" and "Income saved" as income xx increases beyond the poverty point intersection at (23,900,23,900)(23,900, 23,900) to the estimated point (37,000,34,737)(37,000, 34,737).