Pure mathematics

Integration: Techniques and Applications

Year 12 · Year 13

✓By the end of this lesson students will be able to integrate a range of standard functions.
✓By the end of this lesson students will be able to evaluate definite integrals and calculate areas under curves.
✓By the end of this lesson students will be able to apply integration by substitution and integration by parts.
✓By the end of this lesson students will be able to solve first-order differential equations by separating variables.

Key concepts

Indefinite Integration (Standard Integrals)

Integration is the reverse process of differentiation. An indefinite integral represents a family of functions, differing by a constant. This constant, 'c', is known as the constant of integration and must always be included for indefinite integrals.

∫ x^n dx = (x^(n+1))/(n+1) + c (for n ≠ -1) ∫ 1/x dx = ln|x| + c ∫ e^(kx) dx = (1/k)e^(kx) + c ∫ cos(kx) dx = (1/k)sin(kx) + c ∫ sin(kx) dx = -(1/k)cos(kx) + c ∫ sec²(kx) dx = (1/k)tan(kx) + c ∫ f'(x)/f(x) dx = ln|f(x)| + c ∫ f'(x)(f(x))^n dx = ((f(x))^(n+1))/(n+1) + c

Definite Integration & Areas

A definite integral has upper and lower limits of integration and evaluates to a numerical value. Geometrically, it represents the net signed area between the curve y = f(x) and the x-axis over the interval [a, b]. Areas below the x-axis contribute negatively to the definite integral.

∫_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a), where F(x) is an antiderivative of f(x).

Integration by Substitution

This technique simplifies integrals by transforming the variable of integration. It is particularly useful when the integrand contains a composite function and its derivative. The key is to choose a suitable substitution, u = g(x), then find du/dx and rewrite the integral entirely in terms of u and du.

∫ f(g(x))g'(x) dx = ∫ f(u) du, where u = g(x).

Integration by Parts

Integration by parts is used to integrate products of functions and is derived from the product rule for differentiation. The choice of which function to designate as 'u' and which as 'dv/dx' is crucial for simplifying the integral.

∫ u dv/dx dx = uv - ∫ v du/dx dx

Differential Equations (Separation of Variables)

A differential equation is an equation involving derivatives of an unknown function. Solving it means finding the function itself. The method of separation of variables is applicable to first-order differential equations of the form dy/dx = f(x)g(y), where terms involving y and dy can be separated from terms involving x and dx.

If dy/dx = f(x)g(y), then ∫ (1/g(y)) dy = ∫ f(x) dx.

Key facts to remember

1Always include the constant of integration, '+ c', for indefinite integrals.
2For definite integrals, the constant of integration cancels out.
3The integral of 1/x is ln|x|, not x^0/0.
4When using substitution for definite integrals, remember to change the limits of integration to be in terms of the new variable.
5For integration by parts, choose 'u' to be the function that simplifies when differentiated (e.g., polynomials, logarithms) and 'dv/dx' to be the function that is easily integrated (e.g., exponentials, trigonometric functions). The 'LIATE' rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) can help choose 'u'.
6For separable differential equations, rearrange the equation so all 'y' terms (and dy) are on one side, and all 'x' terms (and dx) are on the other, before integrating.
7Areas below the x-axis contribute negatively to a definite integral. To find the total area, integrate separately for positive and negative regions and take the absolute value of the negative parts.
8The integral of f'(x)/f(x) is ln|f(x)| + c.

Worked examples

Example 1

1. Evaluate the definite integral ∫_1^2 (x^2 + 1/x) dx.

IFirst, find the indefinite integral of each term:

II∫ x^2 dx = x^(2+1)/(2+1) = x^3/3

III∫ 1/x dx = ln|x|

IVSo, the indefinite integral is x^3/3 + ln|x| + c. For definite integrals, we do not need 'c' as it cancels out.

VNow, apply the limits:

VI[x^3/3 + ln|x|]_1^2 = (2^3/3 + ln|2|) - (1^3/3 + ln|1|)

VII= (8/3 + ln 2) - (1/3 + 0) (since ln 1 = 0)

VIII= 8/3 + ln 2 - 1/3

9= 7/3 + ln 2

Answer

7/3 + ln 2

Remember to use ln|x| for ∫ 1/x dx, especially if the domain includes negative values, though here x is positive.

Example 2

2. Evaluate the indefinite integral ∫ x√(x^2 + 1) dx using substitution.

ILet u = x^2 + 1.

IIDifferentiate u with respect to x: du/dx = 2x.

IIIRearrange to find dx in terms of du: dx = du/(2x).

IVSubstitute u and dx into the integral:

V∫ x√u (du/(2x))

VICancel out x:

VII∫ (1/2)√u du

VIIIRewrite √u as u^(1/2):

9∫ (1/2)u^(1/2) du

10Integrate with respect to u:

11(1/2) * (u^(1/2 + 1))/(1/2 + 1) + c

12= (1/2) * (u^(3/2))/(3/2) + c

13= (1/2) * (2/3)u^(3/2) + c

14= (1/3)u^(3/2) + c

15Substitute back u = x^2 + 1:

16= (1/3)(x^2 + 1)^(3/2) + c

Answer

(1/3)(x^2 + 1)^(3/2) + c

For definite integrals using substitution, remember to change the limits of integration to be in terms of u.

Example 3

3. (a) Evaluate ∫ x e^x dx using integration by parts. (b) Solve the differential equation dy/dx = x e^x y, given that y = 1 when x = 0.

I(a) For ∫ x e^x dx, use integration by parts: ∫ u dv/dx dx = uv - ∫ v du/dx dx.

IIChoose u = x (because its derivative simplifies) and dv/dx = e^x.

IIIThen du/dx = 1 and v = ∫ e^x dx = e^x.

IVSubstitute these into the formula:

V∫ x e^x dx = x * e^x - ∫ e^x * 1 dx

VI= x e^x - ∫ e^x dx

VII= x e^x - e^x + c

VIII= e^x(x - 1) + c

9(b) The differential equation is dy/dx = x e^x y.

10This is a separable differential equation. Separate the variables:

111/y dy = x e^x dx

12Integrate both sides:

13∫ 1/y dy = ∫ x e^x dx

14From part (a), we know ∫ x e^x dx = e^x(x - 1) + C_1.

15∫ 1/y dy = ln|y| + C_2.

16So, ln|y| = e^x(x - 1) + C (where C = C_1 - C_2 is a single constant).

17Now, use the initial condition y = 1 when x = 0 to find C:

18ln|1| = e^0(0 - 1) + C

190 = 1(-1) + C

200 = -1 + C

21C = 1.

22Substitute C back into the equation:

23ln|y| = e^x(x - 1) + 1

24To express y explicitly, exponentiate both sides:

25|y| = e^(e^x(x - 1) + 1)

26Since y=1 when x=0, y is positive, so |y|=y.

27y = e^(e^x(x - 1) + 1)

Answer

(a) e^x(x - 1) + c (b) y = e^(e^x(x - 1) + 1)

When solving differential equations, remember to combine the constants of integration from both sides into a single constant, usually on the side with the independent variable.

Common mistakes

✗Forgetting to add '+ c' for indefinite integrals.
✗Incorrectly applying the power rule for 1/x (i.e., treating it as x^-1 and trying to use x^(n+1)/(n+1)).
✗Not changing the limits of integration when using substitution for definite integrals.
✗Incorrectly choosing 'u' and 'dv/dx' for integration by parts, leading to a more complex integral.
✗Algebraic errors when separating variables in differential equations or when evaluating definite integrals.
✗Forgetting the absolute value in ln|x| or ln|f(x)|.
✗Treating 'dy/dx' as a single inseparable term rather than a ratio of differentials that can be manipulated.

Exam tips

★After finding an indefinite integral, differentiate your answer to check if you get back the original integrand.
★For area problems, sketch the graph of the function to identify any regions below the x-axis, as these will require careful handling (e.g., taking absolute values) to find the total area.
★Clearly show all steps for integration by substitution and by parts, including the definition of 'u' (and du/dx) or 'u', 'dv/dx', 'du/dx', 'v'.
★When solving differential equations, substitute the initial conditions *after* integrating to find the constant of integration, and then express the solution in the required form (e.g., y in terms of x).

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