Strand 5 — Functions & Calculus

Integration (Higher Level)

5th Year · 6th Year (Leaving Cert)

✓Calculate indefinite integrals of polynomial, trigonometric, exponential, and reciprocal functions.
✓Evaluate definite integrals using the Fundamental Theorem of Calculus.
✓Apply integration to find the area of a region bounded by a curve and the x-axis.
✓Determine the average value of a continuous function over a given interval.

Key concepts

Indefinite Integral

The indefinite integral, also known as the antiderivative, is the reverse process of differentiation. If we differentiate a function f(x) to get f'(x), then integrating f'(x) gives us f(x) plus a constant. It represents a family of functions.

∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C (for n ≠ -1) ∫k dx = kx + C ∫eᵃˣ dx = (1/a)eᵃˣ + C ∫(1/x) dx = ln|x| + C ∫sin(ax) dx = (-1/a)cos(ax) + C ∫cos(ax) dx = (1/a)sin(ax) + C ∫sec²(ax) dx = (1/a)tan(ax) + C

Constant of Integration (C)

When we differentiate a constant, the result is zero. Therefore, when finding an indefinite integral, there could have been any constant added to the original function. We represent this unknown constant with 'C'.

Definite Integral

A definite integral has upper and lower limits of integration. It evaluates to a numerical value and represents the net signed area between the function's graph and the x-axis over the specified interval. This is based on the Fundamental Theorem of Calculus.

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

Area Under a Curve

The area of the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is given by the definite integral. If the function is below the x-axis, the integral will yield a negative value, so we must take the absolute value of that portion of the integral to represent a positive area. If the curve crosses the x-axis, the area must be calculated in parts.

Area = ∫ₐᵇ f(x) dx (if f(x) ≥ 0 on [a, b]) If f(x) < 0, Area = |∫ₐᵇ f(x) dx| If f(x) crosses the x-axis at c, Area = ∫ₐᶜ f(x) dx + |∫ᶜᵇ f(x) dx|

Average Value of a Function

The average value of a continuous function f(x) over an interval [a, b] is the height of a rectangle with base (b-a) that has the same area as the region under the curve f(x) from a to b.

f_avg = (1/(b-a)) ∫ₐᵇ f(x) dx

Key facts to remember

1Integration is the reverse process of differentiation (antidifferentiation).
2Always include the constant of integration 'C' for indefinite integrals.
3The Fundamental Theorem of Calculus states that ∫ₐᵇ f(x) dx = F(b) - F(a), where F(x) is an antiderivative of f(x).
4When calculating area, any portion of the curve below the x-axis will yield a negative integral value; take its absolute value to represent a positive area.
5The average value of a function f(x) over [a, b] is given by f_avg = (1/(b-a)) ∫ₐᵇ f(x) dx.
6Remember the standard integrals: ∫xⁿ dx, ∫eᵃˣ dx, ∫(1/x) dx, ∫sin(ax) dx, ∫cos(ax) dx.
7For ∫(1/x) dx, the result is ln|x| + C, using the absolute value for the domain of the logarithm.

Worked examples

Example 1

Find ∫(3x² - 4x + 5 + e^(2x) + sin(3x)) dx

ISeparate the integral into individual terms: ∫3x² dx - ∫4x dx + ∫5 dx + ∫e^(2x) dx + ∫sin(3x) dx

IIApply the power rule and standard integral formulas:

III= 3(x³/3) - 4(x²/2) + 5x + (1/2)e^(2x) - (1/3)cos(3x) + C

IVSimplify the terms:

V= x³ - 2x² + 5x + (1/2)e^(2x) - (1/3)cos(3x) + C

Answer

x³ - 2x² + 5x + (1/2)e^(2x) - (1/3)cos(3x) + C

Remember to include the constant of integration, C, for indefinite integrals.

Example 2

Evaluate ∫₁² (x + 1/x) dx

IRewrite the integrand with negative exponents where appropriate: ∫₁² (x + x⁻¹) dx

IIFind the antiderivative of each term:

III= [x²/2 + ln|x|]₁²

IVApply the Fundamental Theorem of Calculus (F(b) - F(a)):

V= ((2)²/2 + ln|2|) - ((1)²/2 + ln|1|)

VISimplify the expression:

VII= (4/2 + ln 2) - (1/2 + 0)

VIII= (2 + ln 2) - 1/2

9= 3/2 + ln 2

Answer

3/2 + ln 2

Recall that ln|1| = 0. Always evaluate F(b) - F(a) carefully.

Example 3

Find the area of the region bounded by the curve y = x² - 4, the x-axis, from x = 0 to x = 3.

IIdentify x-intercepts to determine if the curve crosses the x-axis within the interval [0, 3]. Set y = 0: x² - 4 = 0 => (x-2)(x+2) = 0 => x = 2 or x = -2.

IIThe relevant intercept within [0, 3] is x = 2. This means the curve is below the x-axis for x ∈ [0, 2) and above for x ∈ (2, 3].

IIISet up the integral for area, taking the absolute value for the portion below the x-axis:

IVArea = |∫₀² (x² - 4) dx| + ∫₂³ (x² - 4) dx

VFind the indefinite integral: ∫(x² - 4) dx = x³/3 - 4x

VIEvaluate the first definite integral: [x³/3 - 4x]₀² = ((2)³/3 - 4(2)) - ((0)³/3 - 4(0)) = (8/3 - 8) - 0 = 8/3 - 24/3 = -16/3

VIIEvaluate the second definite integral: [x³/3 - 4x]₂³ = ((3)³/3 - 4(3)) - ((2)³/3 - 4(2)) = (27/3 - 12) - (-16/3) = (9 - 12) - (-16/3) = -3 + 16/3 = -9/3 + 16/3 = 7/3

VIIICalculate the total area:

9Area = |-16/3| + 7/3 = 16/3 + 7/3 = 23/3

Answer

23/3 square units

Area must always be a positive value. If the integral yields a negative result, it means the region is below the x-axis, so take its absolute value.

Example 4

Find the average value of the function f(x) = x³ over the interval [1, 3].

IRecall the formula for the average value of a function: f_avg = (1/(b-a)) ∫ₐᵇ f(x) dx

IIIdentify a = 1, b = 3, and f(x) = x³.

IIISubstitute these values into the formula:

IVf_avg = (1/(3-1)) ∫₁³ x³ dx

Vf_avg = (1/2) ∫₁³ x³ dx

VIFind the antiderivative of x³: x⁴/4

VIIEvaluate the definite integral:

VIIIf_avg = (1/2) [x⁴/4]₁³

9f_avg = (1/2) ((3)⁴/4 - (1)⁴/4)

10f_avg = (1/2) (81/4 - 1/4)

11f_avg = (1/2) (80/4)

12f_avg = (1/2) (20)

13f_avg = 10

Answer

The average value of a function is a single numerical value, representing a 'mean height' of the function over the interval.

Common mistakes

✗Forgetting to include the constant of integration 'C' in indefinite integrals.
✗Incorrectly applying the limits of integration in definite integrals (e.g., F(a) - F(b) instead of F(b) - F(a)).
✗Not handling areas below the x-axis correctly, leading to negative area values or incorrect total area.
✗Making algebraic or arithmetic errors when evaluating the antiderivative at the limits of integration.
✗Confusing differentiation rules with integration rules (e.g., integrating e^(ax) as ae^(ax) instead of (1/a)e^(ax)).

Exam tips

★Always check your answer for indefinite integrals by differentiating it to ensure you get the original integrand.
★For area problems, sketch the graph of the function to visually identify any sections that lie below the x-axis and require separate integration.
★Be meticulous with arithmetic and substitution when evaluating definite integrals, especially with fractions and negative signs.
★Practise the standard integral formulas for polynomial, exponential, and trigonometric functions until they are second nature, as these are frequently tested.

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