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Integral Calculus

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These are the solutions to the WASSCE Further Mathematics past questions on Integral Calculus.
The TI-84 Plus CE shall be used for applicable questions.
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Standard Integrals of Exponents

$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int ax^n dx = \dfrac{ax^{n + 1}}{n + 1} + C \\[7ex] (2.)\:\: \displaystyle\int (ax \pm b)^n dx = \dfrac{(ax \pm b)^{n + 1}}{a(n + 1)} + C \\[7ex] (3.)\:\: \displaystyle\int (-ax \pm b)^n dx = -\dfrac{(ax \pm b)^{n + 1}}{a(n + 1)} + C \\[5ex] $

Standard Integrals of Exponential Functions

$ \text{a, b, k, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[7ex] (2.)\:\: \displaystyle\int e^x dx = e^x + C \\[7ex] (3.)\:\: \displaystyle\int e^{kx} dx = \dfrac{e^{kx}}{k} + C \\[7ex] (4.)\:\: \displaystyle\int e^{-kx} dx = \dfrac{-e^{-kx}}{k} + C \\[7ex] (5.)\:\: \displaystyle\int e^{ax \pm b} dx = \dfrac{e^{ax \pm b}}{n} + C \\[7ex] (6.)\:\: \displaystyle\int k^{ax \pm b} dx = \dfrac{k^{ax \pm b}}{a\ln k} + C \\[7ex] (7.)\:\: \displaystyle\int xe^{ax} dx = \dfrac{e^{ax}(ax - 1)}{a^2} + C \\[7ex] (8.)\:\: \displaystyle\int x^ne^{ax} dx = \dfrac{x^ne^{ax}}{a} - \dfrac{n}{a}\displaystyle\int x^{n - 1}e^{ax} dx \\[7ex] (9.)\:\: \displaystyle\int \dfrac{dx}{1 + ke^{ax}} = x - \dfrac{\ln(1 + ke^{ax})}{a} + C \\[5ex] $

Standard Integrals of Logarithmic Functions

$ \text{n is a positive constant} \\[3ex] (1.)\:\: \displaystyle\int \ln x dx = x\ln x - x + C \\[7ex] (2.)\:\: \displaystyle\int (\ln x)^n dx = x(\ln x)^n - n\displaystyle\int (\ln x)^{n - 1}dx \\[7ex] (3.)\:\: \displaystyle\int \dfrac{dx}{x\ln x} = \ln|\ln x| + C \\[7ex] (4.)\:\: \displaystyle\int x^n\ln x dx = \dfrac{x^{n + 1}[\ln x(n + 1) - 1]}{(n + 1)^2} + C \\[5ex] $

Standard Integrals of Trigonometric Functions

$ (1.)\:\: \displaystyle\int \sin x dx = -cos x + C \\[7ex] (2.)\:\: \displaystyle\int cos x dx = sin x + C \\[7ex] (3.)\:\: \displaystyle\int \sec^2 x = \tan x + C \\[5ex] $

Standard Integrals of Hyperbolic Functions

$ (1.)\:\: \displaystyle\int \sinh x dx = \cosh x + C \\[7ex] (2.)\:\: \displaystyle\int \cosh x dx = \sinh x + C \\[5ex] $

Standard Integrals of Rational Functions

$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int \dfrac{1}{x} dx = \ln x + C \\[7ex] (2.)\:\: \displaystyle\int \dfrac{1}{ax \pm b} dx = \dfrac{\ln|ax \pm b|}{a} + C \\[7ex] (3.)\:\: \displaystyle\int \dfrac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}x + C \\[7ex] (4.)\:\: \displaystyle\int \dfrac{-1}{\sqrt{1 - x^2}} dx = \cos^{-1}x + C \\[7ex] (5.)\:\: \displaystyle\int \dfrac{1}{1 + x^2} dx = \tan^{-1}x + C \\[7ex] (6.)\:\: \displaystyle\int \dfrac{1}{\sqrt{x^2 + 1}} dx = \sinh^{-1}x + C \\[7ex] (7.)\:\: \displaystyle\int \dfrac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}x + C \\[7ex] (8.)\:\: \displaystyle\int \dfrac{1}{1 - x^2} dx = \tanh^{-1}x + C \\[5ex] $

Standard Integrals of Absolute Value Functions

$ \text{a, b, n are positive constants} \\[3ex] (1.)\:\: \displaystyle\int |x| dx = \dfrac{x|x|}{2} + C \\[7ex] (2.)\:\: \displaystyle\int |ax \pm b| dx = \dfrac{(ax \pm b)|ax \pm b|}{2a} + C \\[5ex] $

Other Standard Integrals

$ \underline{\text{Algebraic Substitution}} \\[3ex] (1.)\:\: \displaystyle\int f(x)f'(x) dx = \dfrac{f^2(x)}{2} + C \\[7ex] (2.)\:\: \displaystyle\int \dfrac{f'(x)}{f(x)} dx = \ln f(x) + C \\[7ex] (3.)\:\: \displaystyle\int \dfrac{-f'(x)}{f(x)} dx = -\ln f(x) + C \\[7ex] \underline{\text{Trigonometric Substitution}} \\[3ex] (4.)\;\; \displaystyle\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex] (5.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex] (6.)\;\; \displaystyle\int \sqrt{a^2 - x^2}dx = \dfrac{a^2}{2}\left[\sin^{-1}\left(\dfrac{x}{a}\right) + \dfrac{x\sqrt{a^2 - x^2}}{a^2}\right] + C \\[7ex] \underline{\text{Integration by Parts (Integration of Products)}} \\[3ex] (7.)\;\; \displaystyle\int vdu = uv - \displaystyle\int udv \\[7ex] \underline{\text{Hyperbolic Substitution}} \\[3ex] (8.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{a^2 + x^2}} = \sinh^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex] (9.)\;\; \displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\left(\dfrac{x}{a}\right) + C \\[7ex] \underline{\text{Integral Reflection Method}} \\[3ex] \text{Also known as Symmetric Substitution OR Functional Transformation in Integration} \\[3ex] \displaystyle\int_a^b \dfrac{f(x_1)}{f(x_1) + f(x_2)}dx \\[7ex] Let\;\; a + b = \text{some constant},\;c \\[3ex] \text{Set the substitution}:\;\; x = c - x \\[3ex] \implies \\[3ex] f(x) = \text{original function} \\[3ex] f(c - x) = \text{transformed function} \\[3ex] \text{Assume the original function and the transformed function complements each other} \\[3ex] f(x) + f(c - x) = 1 \\[3ex] Let\;\; I_1 = \displaystyle\int_a^b \dfrac{f(x_1)}{f(x_1) + f(x_2)}dx \quad\dots\text{integral of original function} \\[7ex] Let:\;\;I_2 = \displaystyle\int_a^b \dfrac{f(x_2)}{f(x_2) + f(x_1)}dx \quad\dots\text{integral of transformed function} \\[7ex] I_1 = I_2 = \text{the integral},\;I \\[4ex] I_1 + I_2 = 2I \\[3ex] $ Because we are integrating over the same limits and the variable, x is just being swapped with c − x, the integral remains the same as long as the transformation preserves the total area under the curve.

$ 2I = \displaystyle\int_a^b 1 dx = [x]_a^b = b - a \\[5ex] I = \dfrac{b - a}{2} $

Theorems

(1.) The Fundamental Theorem of Calculus:
Assume:
a closed interval: $[c, d]$
a continuous function, $f$ defined on the interval
an antiderivative of the function, $F$ defined on the interval
then:
$\displaystyle\int_c^d f(x) dx = F(d) - F(c)$

(2.)

Applications

(1.) Economics

$ (a.) \\[3ex] TC = \displaystyle\int MC(x) dx \\[5ex] where \\[3ex] x = \text{number of items} \\[3ex] MC = \text{Marginal Cost function} \\[3ex] TC = \text{Total Cost function} \\[3ex] \text{constant of integration is the fixed cost} \\[5ex] (b.) \\[3ex] TR = \displaystyle\int MR(x) dx \\[5ex] where \\[3ex] x = \text{number of items} \\[3ex] MR = \text{Marginal Revenue function} \\[3ex] TR = \text{Total Revenue function} \\[3ex] \text{constant of integration = 0 if no item is produced} \\[5ex] (c.) \\[3ex] NC = \displaystyle\int MPC(x) dx \\[5ex] where \\[3ex] x = \text{disposable national income} \\[3ex] MPC = \text{marginal propensity to consume} \\[3ex] NC = \text{national consumption function} \\[5ex] $ (2.) Average Value of a Function
(3.) Trapezium Rule (or Trapezoidal Rule)

$ Given: \displaystyle\int_{c}^{d} f(x) dx \\[7ex] (a.) \\[3ex] \text{Definite Integral} = \dfrac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + ... + 2f(x_{n - 1}) + f(x_n)] \\[7ex] \text{Definite Integral} = \dfrac{\Delta x}{2} [f(x_0) + f(x_n) + 2(f(x_1) + f(x_2) + f(x_3) + ... + f(x_{n - 1}))] \\[7ex] where: \\[3ex] c = \text{lower limit of integration} \\[3ex] d = \text{upper limit of integration} \\[3ex] \Delta x = \text{width of the interval} \\[3ex] x_0, x_1, x_2, x_3, ..., x_{n - 1}, x_n \;\;\text{are the nodes (also referred to as ordinates)} \\[3ex] f(x_0) \;\;and\;\; f(x_n) \;\;\text{are the function values at the endpoints} \\[3ex] f(c), \; f(d), \; f(x_3), ..., f(x_{n - 1}) \;\;\text{are the function values at the intermediate points} \\[3ex] (b.) \\[3ex] \Delta x = \dfrac{d - c}{n} \\[5ex] \Delta x = \dfrac{d - c}{r - 1} \\[5ex] (c.) \\[3ex] n = r - 1 \\[3ex] where: \\[3ex] n = \text{number of intervals} \\[3ex] r = \text{number of ordinates} \\[5ex] $

(1.) Using the Trapezium Rule with seven ordinates, evaluate, correct to three decimal places, $ \displaystyle\int_{2.4}^{3.6} \dfrac{1}{\sqrt{x^2 - 2}} dx \\[5ex] $

Because we are asked to round the final answer to three decimal places, it is okay to round intermediate values to at least six decimal places.

$ \dfrac{1}{\sqrt{x^2 - 2}} \\[5ex] r = \text{number of ordinates (nodes)} = 7 \\[3ex] n = \text{number of intervals} = 7 - 1 = 6 \\[5ex] \underline{\text{Interval Width}} \\[3ex] \Delta x = \dfrac{d - c}{n} \\[5ex] where: \\[3ex] c = \text{lower limit of integration} = 2.4 \\[3ex] d = \text{upper limit of integration} = 3.6 \\[3ex] \Delta x = \text{width of the interval} \\[3ex] \Delta x = \dfrac{3.6 - 2.4}{6} \\[5ex] = \dfrac{1.2}{6} \\[5ex] = 0.2 \\[5ex] \underline{\text{Nodes (Ordinates)}} \\[3ex] \text{The nodes are:} \\[3ex] x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7 \\[3ex] = 2.4, 2.4 + 0.2, 2.4 + 2(0.2), 2.4 + 3(0.2), 2.4 + 4(0.2), 2.4 + 5(0.2), 2.4 + 6(0.2) \\[3ex] = 2.4, 2.6, 2.8, 3, 3.2, 3.4, 3.6 \\[5ex] \underline{\text{Function Values}} \\[3ex] f(x_0), f(x_1), f(x_2), f(x_3), f(x_4), f(x_5), f(x_6), f(x_7) \\[3ex] f(2.4) = \dfrac{1}{\sqrt{(2.4)^2 - 2}} \approx 0.515711 \\[5ex] f(2.6) = \dfrac{1}{\sqrt{(2.6)^2 - 2}} \approx 0.458349 \\[5ex] f(2.8) = \dfrac{1}{\sqrt{(2.8)^2 - 2}} \approx 0.413803 \\[5ex] f(3) = \dfrac{1}{\sqrt{(3)^2 - 2}} \approx 0.377964 \\[5ex] f(3.2) = \dfrac{1}{\sqrt{(3.2)^2 - 2}} \approx 0.348367 \\[5ex] f(3.4) = \dfrac{1}{\sqrt{(3.4)^2 - 2}} \approx 0.323423 \\[5ex] f(3.6) = \dfrac{1}{\sqrt{(3.6)^2 - 2}} \approx 0.302061 \\[7ex] \underline{\text{Trapezium Rule}} \\[3ex] \text{Definite Integral} = \dfrac{\Delta x}{2} [f(x_0) + f(x_n) + 2(f(x_1) + f(x_2) + f(x_3) + ... + f(x_{n - 1}))] \\[5ex] = \dfrac{0.2}{2} [0.515711 + 0.302031 + 2(0.458349 + 0.413803 + 0.377964 + 0.348367 + 0.323423)] \\[5ex] = 0.1(4.661554) \\[3ex] = 0.4661554 \\[3ex] \approx 0.466 \quad\dots\text{to 3 decimal places} $

Directly (by Calculator): This is the exact answer.
Trapezium Rule gives an approximate.

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Intermediate Steps:
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