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# Solved Examples on Integration

Prerequisite Topics:
Factoring
Partial Fractions
Exponents and Logarithms
Trigonometry
Differential Calculus

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Indicate the method(s) you used as applicable
Show all work

(1.) Evaluate $\displaystyle\int -7dx$

Power Rule

$\displaystyle\int -7dx \\[5ex] = \displaystyle\int -7 * 1 * dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \displaystyle\int -7x^0dx \\[5ex] = -7 \displaystyle\int x^0dx \\[5ex] = -7 * \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = -7 * \dfrac{x^1}{1} + C \\[5ex] = -7x + C$
(2.) Determine the antiderivative of $\cot x$

Integration by Algebraic Substitution

$\displaystyle\int \cot xdx \\[5ex] \cot x = \dfrac{\cos x}{\sin x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \sin x \\[3ex] \dfrac{dp}{dx} = \cos x \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{\cos x} \\[5ex] dx = \dfrac{dp}{\cos x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \cot xdx = \displaystyle\int \dfrac{\cos x}{\sin x} dx \\[5ex] = \displaystyle\int \dfrac{\cos x}{p} * \dfrac{dp}{\cos x} \\[5ex] = \displaystyle\int \dfrac{dp}{p} \\[5ex] = \ln p + C \\[3ex] = \ln \sin x + C$
(3.) JAMB Evaluate $\displaystyle\int_1^3 (x^2 - 1)dx$

$A.\:\: -6\dfrac{2}{3} \\[5ex] B.\:\: 6\dfrac{2}{3} \\[5ex] C.\:\: \dfrac{2}{3} \\[5ex] D.\:\: -\dfrac{2}{3} \\[5ex]$

Power Rule

$\displaystyle\int_1^3 (x^2 - 1)dx \\[5ex] \displaystyle (x^2 - 1)dx = \displaystyle\int x^2dx - \displaystyle\int 1dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \dfrac{x^{2 + 1}}{2 + 1} - \displaystyle\int x^0dx \\[5ex] = \dfrac{x^3}{3} - \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = \dfrac{x^3}{3} - x + C \\[5ex] \rightarrow \\[3ex] \displaystyle\int_1^3 (x^2 - 1)dx = \left[\dfrac{x^3}{3} - x\right]_1^3 \\[5ex] = \left(\dfrac{3^3}{3} - 3\right) - \left(\dfrac{1^3}{3} - 1\right) \\[5ex] = (9 - 3) - \left(\dfrac{1}{3} - \dfrac{3}{3}\right) \\[5ex] = 6 - \left(-\dfrac{2}{3}\right) \\[5ex] = 6 + \dfrac{2}{3} \\[5ex] = 6\dfrac{2}{3}$
(4.) JAMB Evaluate $\displaystyle\int \sin 3xdx$

$A.\:\: -\dfrac{1}{3}\cos 3x + c \\[5ex] B.\:\: \dfrac{1}{3}\cos 3x + c \\[5ex] C.\:\: \dfrac{2}{3}\cos 3x + c \\[5ex] C.\:\: -\dfrac{2}{3}\cos 3x + c \\[5ex]$

Integration by Algebraic Substitution

$\displaystyle\int \sin 3xdx \\[5ex] Let\:\: p = 3x \\[3ex] \dfrac{dp}{dx} = 3 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{3} \\[5ex] dx = \dfrac{dp}{3} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \sin 3xdx = \displaystyle\int \sin p * \dfrac{dp}{3} \\[5ex] = \dfrac{1}{3} * \displaystyle\int \sin pdp \\[5ex] = \dfrac{1}{3} * -\cos p + C \\[5ex] = \dfrac{1}{3} * - \cos 3x + C \\[3ex] = -\dfrac{1}{3}\cos 3x + c$
(5.) WASSCE:FM

(a) Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions

(b) Use the answer in (a) to evaluate $\displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3}dx$

$(a.) \\[3ex] \dfrac{x + 6}{(x + 1)^3} \\[5ex]$ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator

$\dfrac{x + 6}{(x + 1)^3} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} \\[5ex] = \dfrac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3} \\[5ex] = \dfrac{A[(x + 1)(x + 1)] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + x + x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + A + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + Bx + A + B + C}{(x + 1)^3} \\[5ex]$ Denominators are the same
Equate the numerators

$x + 6 = Ax^2 + 2Ax + Bx + A + B + C \\[3ex] Swap \\[3ex] Ax^2 + 2Ax + Bx + A + B + C = x + 6 \\[3ex] Ax^2 + x(2A + B) + A + B + C = 0x^2 + x + 6 \\[3ex] \implies \\[3ex] A = 0...eqn.(1) \\[3ex] 2A + B = 1...eqn.(2) \\[3ex] A + B + C = 6...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 2(0) + B = 1 \\[3ex] 0 + B = 1 \\[3ex] B = 1 - 0 \\[3ex] B = 1 \\[3ex] Substitute\;\;0\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 0 + 1 + C = 6 \\[3ex] 1 + C = 6 \\[3ex] C = 6 - 1 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{0}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \dfrac{x + 6}{(x + 1)^3} = 0 + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \therefore \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] = \dfrac{1(x + 1) + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 1 + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 6}{(x + 1)^3} \\[5ex] = LHS \\[3ex]$ Integration by Partial Fractions

$(b.) \\[3ex] \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx \\[5ex] = \displaystyle\int_1^2 \left[\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}\right] dx \\[5ex] = \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] Let\;\; p = x + 1 \\[3ex] \dfrac{dp}{dx} = 1 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{1} = 1 \\[5ex] dx = dp \\[3ex] \displaystyle\int \dfrac{1}{(x + 1)^2} dx \\[5ex] = \displaystyle\int \dfrac{1}{p^2} dp \\[5ex] = \displaystyle\int p^{-2} dp \\[5ex] = \dfrac{p^{-2 + 1}}{-2 + 1} \\[5ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] = -\dfrac{1}{x + 1} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx \\[5ex] = \left[-\dfrac{1}{x + 1}\right]_1^2 \\[5ex] = -\dfrac{1}{2 + 1} - -\dfrac{1}{1 + 1} \\[5ex] = -\dfrac{1}{3} + \dfrac{1}{2} \\[5ex] = \dfrac{-2 + 3}{6} \\[5ex] = \dfrac{1}{6} \\[7ex] \displaystyle\int \dfrac{5}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{p^3} dp \\[5ex] = 5\displaystyle\int p^{-3} dp \\[5ex] = 5 * \dfrac{p^{-3 + 1}}{-3 + 1} \\[5ex] = 5 * \dfrac{p^{-2}}{-2} \\[5ex] = -\dfrac{5}{2} * p^{-2} \\[5ex] = -\dfrac{5}{2} * \dfrac{1}{p^2} \\[5ex] = -\dfrac{5}{2p^2} \\[5ex] = -\dfrac{5}{2(x + 1)^2} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \left[-\dfrac{5}{2(x + 1)^2}\right]_1^2 \\[5ex] = -\dfrac{5}{2(2 + 1)^2} - -\dfrac{5}{2(1 + 1)^2} \\[5ex] = -\dfrac{5}{2(3)^2} + \dfrac{5}{2(2)^2} \\[5ex] = -\dfrac{5}{2(9)} + \dfrac{5}{2(4)} \\[5ex] = -\dfrac{5}{18} + \dfrac{5}{8} \\[5ex] = \dfrac{-20 + 45}{72} \\[5ex] = \dfrac{25}{72} \\[7ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \dfrac{1}{6} + \dfrac{25}{72} \\[5ex] = \dfrac{12 + 25}{72} \\[5ex] = \dfrac{37}{72} \\[5ex] \therefore \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx = \dfrac{37}{72}$
(6.) ATAR
(a) Given that $\dfrac{2}{(x + 1)(x - 1)} = \dfrac{a}{x - 1} + \dfrac{b}{x + 1}$

determine the values for a and b

(b) Hence determine $\displaystyle\int \dfrac{1}{x^2 - 1}dx$

Numerator: cannot be simplified further

Form: Proper Fraction: Non-repeated Linear Factors at the Denominator

$\dfrac{2}{(x + 1)(x - 1)} \\[5ex] = \dfrac{a}{x - 1} + \dfrac{b}{x + 1} \\[5ex] = \dfrac{a(x + 1) + b(x - 1)}{(x - 1)(x + 1)} \\[5ex] = \dfrac{ax + a + bx - b}{(x - 1)(x + 1)} \\[5ex] = \dfrac{ax + bx + a - b}{(x - 1)(x + 1)} \\[5ex]$ Denominators are the same
Equate the numerators

$2 = ax + bx + a - b \\[3ex] ax + bx + a - b = 2 \\[3ex] ax + bx + a - b = 0x + 2 \\[3ex] \implies \\[3ex] a + b = 0 ...eqn.(1) \\[3ex] a - b = 2 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] 2a = 2 \\[3ex] a = \dfrac{2}{2} \\[5ex] a = 1 \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] 2b = -2 \\[3ex] b = -\dfrac{2}{2} \\[5ex] b = -1 \\[3ex] \therefore \dfrac{2}{(x + 1)(x - 1)} = \dfrac{1}{x - 1} + \dfrac{-1}{x + 1} \\[5ex] \dfrac{2}{(x - 1)(x + 1)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \\[5ex] \underline{Check} \\[3ex] \underline{RHS} \\[3ex] \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \\[5ex] = \dfrac{1(x + 1) - 1(x - 1)}{(x - 1)(x + 1)} \\[5ex] = \dfrac{x + 1 - x + 1}{(x - 1)(x + 1)} \\[5ex] = \dfrac{2}{(x - 1)(x + 1)} \\[5ex] = \dfrac{2}{(x + 1)(x - 1)} \\[5ex] = LHS \\[3ex]$ Integration by Partial Fractions

(b) Because of the word, "Hence"; we are expected to use the result we got from Part (a) to solve Part (b)
However, the partial fraction we did in (a) is slightly different from the integral in (b)
So, we need to make some modifications in order to solve it.

$\displaystyle\int \dfrac{1}{x^2 - 1}dx \\[5ex] \underline{Denominator} \\[3ex] x^2 - 1 = (x + 1)(x - 1) ...Difference\;\;of\;\;Two\;\;Squares \\[3ex] \displaystyle\int \dfrac{1}{x^2 - 1}dx \\[5ex] = \displaystyle\int \dfrac{1}{(x + 1)(x - 1)} dx \\[5ex] = \dfrac{1}{2} * \displaystyle\int \dfrac{2}{(x + 1)(x - 1)} dx \\[5ex] = \dfrac{1}{2} * \left[\displaystyle\int \left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\right) dx\right] \\[5ex] = \dfrac{1}{2} * \left[\displaystyle\int \dfrac{1}{x - 1}dx - \displaystyle\int \dfrac{1}{x + 1}dx\right] \\[5ex] = \dfrac{1}{2}[\ln|x - 1| - \ln|x + 1|] + C \\[5ex] = \dfrac{1}{2}\ln\dfrac{|x - 1|}{|x + 1|} + C$
(7.) WASSCE:FM Evaluate $\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx$

$A.\:\: 1\dfrac{1}{3} \\[5ex] B.\:\: 1\dfrac{1}{2} \\[5ex] C.\:\: 3 \\[3ex] D.\:\: 4\dfrac{1}{2} \\[5ex]$

Power Rule

$\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx \\[5ex] 3 - 3x^2 = 3(1 - x^2) \\[3ex] 1 - x^2 = 1^2 - x^2 = (1 + x)(1 - x)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] 3 - 3x^2 = 3(1 + x)(1 - x) = 3(x + 1)(1 - x) \\[3ex] \implies \\[3ex] \dfrac{3 - 3x^2}{x + 1} = \dfrac{3(x + 1)(1 - x)}{x + 1} = 3(1 - x) = 3 - 3x \\[5ex] \implies \\[3ex] \displaystyle\int (3 - 3x)dx = \displaystyle\int 3dx - \displaystyle\int 3xdx \\[5ex] \displaystyle\int 3dx = 3x \\[5ex] \displaystyle\int 3xdx = 3 * \dfrac{x^{1 + 1}}{1 + 1} = \dfrac{3x^2}{2} \\[5ex] \rightarrow \\[3ex] \displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx = \left[3x - \dfrac{3x^2}{2}\right]_0^1 \\[5ex] = \left[3(1) - \dfrac{3(1)^2}{2}\right] - \left[3(0) - \dfrac{3(0)^2}{2}\right] \\[5ex] = \left[3 - \dfrac{3(1)}{2}\right] - \left[0 - \dfrac{3(0)}{2}\right] \\[5ex] = \left(3 - \dfrac{3}{2}\right) - \left(0 - \dfrac{0}{2}\right) \\[5ex] = \left(\dfrac{6}{2} - \dfrac{3}{2}\right) - (0 - 0) \\[5ex] = \dfrac{6 - 3}{2} - 0 \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2}$
(8.) Evaluate $\displaystyle\int (3^p + 7^p)dp$

Standard Integral

$\displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[5ex] \displaystyle\int (3^p + 7^p)dp = \displaystyle\int 3^pdp + \displaystyle\int 7^pdp \\[5ex] \displaystyle\int 3^p dx = \dfrac{3^p}{\ln 3} \\[5ex] \displaystyle\int 7^p dx = \dfrac{7^p}{\ln 7} \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3^p + 7^p)dp = \dfrac{3^p}{\ln 3} + \dfrac{7^p}{\ln 7} + C$
(9.) Evaluate $\displaystyle\int \sqrt{2 + \sqrt{x}} dx$

Integration by Algebraic Substitution

$\displaystyle\int \sqrt{2 + \sqrt{x}} dx \\[3ex] Let\:\: p = 2 + \sqrt{x} \\[3ex] \dfrac{dp}{dx} = \dfrac{1}{2\sqrt{x}} \\[5ex] \dfrac{dx}{dp} = 2\sqrt{x} \\[5ex] dx = 2\sqrt{x}dp \\[3ex] \rightarrow \\[3ex] \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \displaystyle\int \sqrt{p} * 2\sqrt{x} dp \\[3ex] = 2\displaystyle\int \sqrt{p} * \sqrt{x} dp \\[3ex] p = 2 + \sqrt{x} \rightarrow \sqrt{x} = p - 2 \\[3ex] = 2 * \displaystyle\int \sqrt{p}(p - 2) dp \\[3ex] \displaystyle\int \sqrt{p}(p - 2) dp = \displaystyle\int p\sqrt{p}dp - \displaystyle\int 2\sqrt{p}dp \\[3ex] \displaystyle\int p\sqrt{p}dp = \displaystyle\int p^1 * p^{\dfrac{1}{2}}dp = \displaystyle\int p^{1 + \dfrac{1}{2}}dp = \displaystyle\int p^{\dfrac{3}{2}}dp \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{p^{\dfrac{3}{2} + 1}}{\dfrac{3}{2} + 1} \\[7ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} \div \dfrac{5}{2} \\[5ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} * \dfrac{2}{5} \\[5ex] p^{\dfrac{5}{2}} = (\sqrt{p})^5 = (\sqrt{p})^4 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^4 * \sqrt{p} = p^2\sqrt{p} \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{2p^2}{5} \\[5ex] \displaystyle\int 2\sqrt{p}dp = 2\displaystyle\int \sqrt{p}dp \\[3ex] \displaystyle\int \sqrt{p}dp = \dfrac{p^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \\[7ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} \div \dfrac{3}{2} \\[5ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} * \dfrac{2}{3} \\[5ex] p^{\dfrac{3}{2}} = (\sqrt{p})^3 = (\sqrt{p})^2 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^2 * \sqrt{p} = p\sqrt{p} \\[5ex] \displaystyle\int 2\sqrt{p}dp = \dfrac{2p\sqrt{p}}{3} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{p}(p - 2) dp = \dfrac{2p^2\sqrt{p}}{5} - \dfrac{2p\sqrt{p}}{3} \\[5ex] = 2p\sqrt{p}\left(\dfrac{p}{5} - \dfrac{1}{3}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p}{15} - \dfrac{5}{15}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] \rightarrow \\[3ex] 2 * \displaystyle\int \sqrt{p}(p - 2) dp = 2 * 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] = 4p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] Substitute\:\:back \\[3ex] = 4(2 + \sqrt{x})\sqrt{2 + \sqrt{x}}\left(\dfrac{3(2 + \sqrt{x}) - 5}{15}\right) \\[5ex] \dfrac{3(2 + \sqrt{x}) - 5}{15} = \dfrac{6 + 3\sqrt{x} - 5}{15} = \dfrac{1 + 3\sqrt{x}}{15} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{2 + \sqrt{x}} dx = 4(2 + \sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right)\left(\dfrac{1 + 3\sqrt{x}}{15}\right) \\[5ex] \therefore \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \dfrac{4}{15}(2 + \sqrt{x})(1 + 3\sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right) + C$
(10.) Find the integral of $\tan x$

Integration by Algebraic Substitution

$\displaystyle\int \tan xdx \\[5ex] \tan x = \dfrac{\sin x}{\cos x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \cos x \\[3ex] \dfrac{dp}{dx} = -\sin x \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{\sin x} \\[5ex] dx = -\dfrac{dp}{\sin x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \tan xdx = \displaystyle\int \dfrac{\sin x}{\cos x} dx \\[5ex] = \displaystyle\int \dfrac{\sin x}{p} * -\dfrac{dp}{\sin x} \\[5ex] = \displaystyle\int -\dfrac{dp}{p} \\[5ex] = -\ln p + C \\[3ex] = -\ln \cos x + C$
(11.) Evaluate $\displaystyle\int \dfrac{1}{\sqrt{9 - x^2}} dx$

Standard Integral: Trigonometric Substitution

$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C \\[5ex] For:\;\;\displaystyle\int \dfrac{1}{\sqrt{9 - x^2}} dx \\[5ex] Compare:\;\;a^2 - x^2 \;\;to\;\; 9 - x^2 \\[3ex] a^2 = 9 \\[3ex] a = \sqrt{9} \\[3ex] a = 3 \\[3ex] \implies \\[3ex] \displaystyle\int \dfrac{1}{\sqrt{9 - x^2}} dx \\[5ex] = \sin^{-1}\left(\dfrac{x}{3}\right) + C$
(12.) Evaluate $\displaystyle\int (v^{-2} + 8v^{-1})dv$

Power Rule

$\displaystyle\int (v^{-2} + 8v^{-1})dv \\[5ex] = \displaystyle\int v^{-2}dv + \displaystyle\int 8v^{-1}dv \\[5ex] = \dfrac{v^{-2 + 1}}{-2 + 1} + 8 * \displaystyle\int v^{-1}dv \\[5ex] = \dfrac{v^{-1}}{-1} + 8 * \ln v + C \\[5ex] = -v^{-1} + 8\ln v + C \\[5ex] = -\dfrac{1}{v} + 8\ln v + C \\[5ex] = 8\ln v - \dfrac{1}{v} + C$
(13.) MEHA Consider the function $g(x) = \dfrac{1}{2}x^3 + x^2 + 1$ and the shaded region shown below.

Calculate the area of the shaded region.

The shaded region is the area of the rectangle minus the area enclosed by the curve
 $-$ =

$1st:\;\;Area\;\;of\;\;the\;\;rectangle = 9 * 2 = 18\;square\;\;units \\[3ex] 2nd:\;\;Area\;\;enclosed\;\;by\;\;the\;\;curve \\[3ex] = \displaystyle\int_0^2 g(x) dx \\[3ex] = \displaystyle\int_0^2 \left(\dfrac{1}{2}x^3 + x^2 + 1\right) dx \\[5ex] = \left[\dfrac{1}{2} * \dfrac{x^4}{4} + \dfrac{x^3}{3} + x\right]_0^2 \\[5ex] = \left[\dfrac{x^4}{8} + \dfrac{x^3}{3} + x\right]_0^2 \\[5ex] = \left[\dfrac{2^4}{8} + \dfrac{2^3}{3} + 2\right] - \left[\dfrac{0^4}{8} + \dfrac{0^3}{3} + 0\right] \\[5ex] = \dfrac{16}{8} + \dfrac{8}{3} + \dfrac{2}{1} - 0 \\[5ex] = \dfrac{2}{1} + \dfrac{8}{3} + \dfrac{2}{1} \\[5ex] = \dfrac{6 + 8 + 6}{3} \\[5ex] = \dfrac{20}{3}\;square\;\;units \\[5ex] 3rd:\;\;Area\;\;of\;\;the\;\;shaded\;\;region \\[3ex] = 18 - \dfrac{20}{3} \\[5ex] = \dfrac{54 - 20}{3} \\[5ex] = \dfrac{34}{3}\;square\;\;units$
(14.) NSC Determine the following integrals:

$(14.1) \;\;\; \displaystyle\int x(x^2 + 6x)dx \\[3ex] (14.2) \;\;\; \displaystyle\int \left(3^x + \dfrac{1}{x}\right)dx \\[5ex]$ (14.3) The sketch below represents the area bounded by the function g defined by $g(x) = 3x^2$ and the points where x = k and x = 4

$(14.1) \\[3ex] \boldsymbol{{Power\;\;Rule}} \\[3ex] \displaystyle\int x(x^2 + 6x)dx \\[3ex] = \displaystyle\int (x^3 + 6x^2)dx \\[3ex] = \dfrac{x^4}{4} + \dfrac{6x^3}{3} + C \\[5ex] = \dfrac{x^4}{4} + 2x^3 + C \\[5ex] (14.2) \\[3ex] \boldsymbol{{Special\;\;Integrals}} \\[3ex] \displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[5ex] \displaystyle\int 3^x dx + \displaystyle\int \dfrac{1}{x} dx \\[5ex] = \dfrac{3^x}{\ln 3} + \ln x + C \\[5ex] (14.3) \\[3ex] Area = \displaystyle\int_{x_1}^{x_2} g(x) dx \\[3ex] 56 = \displaystyle\int_{k}^{4} 3x^2 dx \\[3ex] 56 = \displaystyle\int_{k}^{4} 3x^2 dx \\[3ex] 56 = \left[\dfrac{3x^3}{3}\right]_k^4 \\[5ex] 56 = [x^3]_k^4 \\[3ex] 56 = 4^3 - k^3 \\[3ex] k^3 = 4^3 - 56 \\[3ex] k^3 = 64 - 56 \\[3ex] k^3 = 8 \\[3ex] k = \sqrt[3]{8} \\[3ex] k = 2$
(15.) Determine the indefinite integral $\displaystyle\int \dfrac{2dt}{\sqrt{16 - 36t^2}}$

Standard Integral: Trigonometric Substitution

$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C \\[5ex] For:\;\;\displaystyle\int \dfrac{2dt}{\sqrt{16 - 36t^2}} \\[5ex]$ (1.) The coefficient of $t^2$ is not unity. We need to make it unity (make it 1)
(2.) We need to simplify the integral to look like the standard integral

$\displaystyle\int \dfrac{2dt}{\sqrt{16 - 36t^2}} \\[5ex] = \displaystyle\int \dfrac{2dt}{\sqrt{36\left(\dfrac{16}{36} - t^2\right)}} \\[10ex] = \displaystyle\int \dfrac{2dt}{6\sqrt{\dfrac{4}{9} - t^2}} \\[10ex] = \dfrac{2}{6} \displaystyle\int \dfrac{dt}{\sqrt{\dfrac{2^2}{3^2} - t^2}} \\[10ex] = \dfrac{1}{3} \displaystyle\int \dfrac{dt}{\sqrt{\left(\dfrac{2}{3}\right)^2 - t^2}} \\[10ex] Compare:\;\;a^2 - x^2 \;\;to\;\;\left(\dfrac{2}{3}\right)^2 - t^2 \\[5ex] a = \dfrac{2}{3} \\[5ex] x = t \\[3ex] \implies \\[3ex] = \dfrac{1}{3}\sin^{-1}\left(t \div \dfrac{2}{3}\right) + C \\[5ex] = \dfrac{1}{3}\sin^{-1}\left(\dfrac{3t}{2}\right) + C$
(16.) KCSE Find the area enclosed by the curve $y = x^2 + 2x$ the straight lines x = 1, x = 3, and the x-axis

$x^2 + 2x = 0 \\[3ex] x(x + 2) = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x + 2 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = -2 \\[3ex]$ The zeros of the function...where the graph include areas above or below the x-axis are x = 0 and x = -2
But we are asked to calculate the area between x = 1 and x = 3
Because 0 and -2 are not included in that range, we skip those areas and only focus on the area between x = 1 and x = 3

$Area = \displaystyle\int_1^3 (x^2 + 2x) dx \\[5ex] Area = \left[\dfrac{x^3}{3} + \dfrac{2x^2}{2}\right]_1^3 \\[5ex] Area = \left[\dfrac{x^3}{3} + x^2\right]_1^3 \\[5ex] x = 3 \\[3ex] \dfrac{3^3}{3} + 3^2 \\[5ex] 9 + 9 \\[3ex] 18 \\[3ex] x = 1 \\[3ex] \dfrac{1^3}{3} + 1^2 \\[5ex] \dfrac{1}{3} + 1 \\[5ex] \dfrac{1 + 3}{3} \\[5ex] \dfrac{4}{3} \\[5ex] Area = 18 - \dfrac{4}{3} \\[5ex] Area = \dfrac{54 - 4}{3} \\[5ex] Area = \dfrac{50}{3}\;square\;\;units$
(17.) HSC Determine the following:

(a) $\displaystyle\int (2x^2 - x^3) dx$

(b) $\displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin (x)}{3 - \cos(x)} dx$

(c) $\dfrac{d}{dy} \displaystyle\int_{-1}^y 3x^2 \cos(2x) dx$

$(a) \\[3ex] \displaystyle\int (2x^2 - x^3) dx \\[5ex] \boldsymbol{Power\;\;Rule} \\[3ex] = \dfrac{2x^3}{3} - \dfrac{x^4}{4} + C \\[5ex] (b) \\[3ex] \displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin (x)}{3 - \cos(x)} dx ...keep\;\;the\;\;limits\;\;of\;\;integration \\[5ex] \displaystyle\int \dfrac{\sin (x)}{3 - \cos(x)} dx \\[5ex] \boldsymbol{Algebraic\;\;Substitution} \\[3ex] Let\;\; p = 3 - \cos x \\[3ex] \dfrac{dp}{dx} = -(-\sin x) = \sin x \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{\sin x} \\[5ex] dx = \dfrac{dp}{\sin x} \\[5ex] \implies \\[3ex] \displaystyle\int \dfrac{\sin x}{p} * \dfrac{dp}{\sin x} \\[5ex] = \ln p \\[3ex] = \ln(3 - \cos x) \\[3ex] Bring\;\;the\;\;limits\;\;of\;\;integration \\[3ex] = [\ln(3 - \cos x)]_0^{\dfrac{\pi}{2}} \\[5ex] = \ln\left[3 - \cos\dfrac{\pi}{2}\right] - \ln[3 - \cos 0] \\[5ex] = \ln(3 - 0) - \ln(3 - 1) \\[3ex] = \ln 3 - \ln 2 \\[3ex] = \ln\left(\dfrac{3}{2}\right) \\[5ex] (c) \\[3ex] \dfrac{d}{dy} \displaystyle\int_{-1}^y 3x^2 \cos(2x) dx \\[5ex] \boldsymbol{Derivatives\;\;of\;\;Integrals} \\[3ex] \dfrac{d}{dy} \displaystyle\int_{a(y)}^{b(y)} f(x) dx = f[b(y)] * b'(y) - f[a(y)] * a'(y) \\[5ex] Compare:\;\; \implies \\[3ex] a(y) = -1 \\[3ex] b(y) = y \\[3ex] b'(y) = 1 \\[3ex] a'(y) = 0 \\[3ex] f(x) = 3x^2 \cos(2x) \\[3ex] f[b(y)] = f(y) = 3y^2 \cos(2y) \\[3ex] f[a(y)] = f(-1) = 3(-1)^2 \cos(2 * -1) = 3(1)\cos(-2) = 3\cos(-2) \\[3ex] \implies \\[3ex] = [3y^2\cos(2y) * 1] - [3\cos(-2) * 0] \\[3ex] = 3y^2\cos(2y) - 0 \\[3ex] = 3y^2\cos(2y)$
(18.) Integrate $xe^x$ with respect to x

Integration by Parts

$\displaystyle\int udv = uv - \displaystyle\int vdu \\[3ex] \displaystyle\int xe^x dx \\[3ex] Let\;\;u = x \qquad\qquad\qquad dv = e^x dx \\[3ex] \dfrac{du}{dx} = 1 \qquad\qquad\qquad\quad v = \displaystyle\int e^x dx \\[5ex] du = dx \qquad\qquad\qquad\quad v = e^x \\[3ex] = x(e^x) - \displaystyle\int e^x dx \\[5ex] = xe^x - e^x + C \\[3ex] = e^x(x - 1) + C$
(19.) Using the substitution $x = 3\tan\theta$, evaluate the indefinite integral $\displaystyle\int \dfrac{46dx}{x^2\sqrt{x^2 + 9}}$

Integration by Trigonometric Substitution

$x = 3\tan\theta \\[3ex] \dfrac{dx}{d\theta} = 3\sec^2\theta \\[5ex] dx = 3\sec^2\theta d\theta \\[3ex] x^2 = (3\tan\theta)^2 = 3^2\tan^2\theta = 9\tan^2\theta \\[3ex] x^2 + 9 = 9\tan^2\theta + 9 = 9(\tan^2\theta + 1) = 9\sec^2\theta \\[3ex] \sqrt{x^2 + 9} = \sqrt{9\sec^2\theta} = 3\sec\theta \\[3ex] \implies \\[3ex] \displaystyle\int \dfrac{46dx}{x^2\sqrt{x^2 + 9}} \\[5ex] = 46 \displaystyle\int \dfrac{3\sec^2\theta d\theta}{9\tan^2\theta(3\sec\theta)} \\[5ex] = \dfrac{46}{9} \displaystyle\int \dfrac{\sec\theta d\theta}{\tan^2\theta} \\[5ex]$ Let's keep $\dfrac{46}{9}$ for now and focus on integrating $\dfrac{\sec\theta}{\tan^2\theta}$ wrt $\theta$

$\displaystyle\int \dfrac{\sec\theta d\theta}{\tan^2\theta} \\[5ex] = \displaystyle\int (\sec\theta \div \tan^2\theta) d\theta \\[5ex] = \displaystyle\int \left(\dfrac{1}{\cos\theta} \div \dfrac{\sin^2\theta}{\cos^2\theta}\right) d\theta \\[5ex] = \displaystyle\int \left(\dfrac{1}{\cos\theta} * \dfrac{\cos^2\theta}{\sin^2\theta}\right) d\theta \\[5ex] = \displaystyle\int \dfrac{\cos\theta}{\sin^2\theta} d\theta \\[5ex]$ Integration by Algebraic Substitution

$Let\;\;p = \sin\theta \\[3ex] \dfrac{dp}{d\theta} = \cos\theta \\[5ex] \dfrac{d\theta}{dp} = \dfrac{1}{\cos\theta} \\[5ex] d\theta = \dfrac{dp}{\cos\theta} \\[5ex] \implies \\[3ex] \displaystyle\int \dfrac{\cos\theta}{\sin^2\theta} d\theta = \displaystyle\int \dfrac{\cos\theta}{p^2} * \dfrac{dp}{\cos\theta} \\[5ex] = \displaystyle\int p^{-2}dp \\[3ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] Substitute\;\;back\;\;for\;\;p \\[3ex] = -\dfrac{1}{\sin\theta} \\[5ex] Find\;\;\sin\theta\;\;in\;\;terms\;\;of\;\;x \\[3ex] Recall: \\[3ex] x = 3\tan\theta \\[3ex] 3\tan\theta = x \\[3ex] \tan\theta = \dfrac{x}{3} \\[5ex] \tan^2\theta = \dfrac{x^2}{3^2} = \dfrac{x^2}{9} \\[5ex] \tan^2\theta + 1 = \dfrac{x^2}{9} + \dfrac{9}{9} = \dfrac{x^2 + 9}{9} \\[5ex] \sec^2\theta = \tan^2\theta + 1 ...Pythagorean\;\;Identity \\[3ex] \sec^2\theta = \dfrac{x^2 + 9}{9} \\[5ex] \cos^2\theta = \dfrac{1}{\sec^2\theta} = \dfrac{9}{x^2 + 9} \\[5ex] \sin^2\theta = 1 - \cos^2\theta ...Pythagorean\;\;Identity \\[3ex] \sin^2\theta = 1 - \dfrac{9}{x^2 + 9} \\[5ex] \sin^2\theta = \dfrac{x^2 + 9}{x^2 + 9} - \dfrac{9}{x^2 + 9} \\[5ex] \sin^2\theta = \dfrac{x^2 + 9 - 9}{x^2 + 9} \\[5ex] \sin\theta = \sqrt{\sin^2\theta} = \sqrt{\dfrac{x^2}{x^2 + 9}} \\[5ex] \sin\theta = \dfrac{x}{\sqrt{x^2 + 9}} \\[5ex]$ Get back $-\dfrac{1}{\sin\theta}$

$-\dfrac{1}{\sin\theta} \\[5ex] = -1 \div \dfrac{x}{\sqrt{x^2 + 9}} \\[5ex] = -1 * \dfrac{\sqrt{x^2 + 9}}{x} \\[5ex] = -\dfrac{\sqrt{x^2 + 9}}{x} \\[5ex]$ Get back $\dfrac{46}{9}$

$\dfrac{46}{9} * -\dfrac{\sqrt{x^2 + 9}}{x} \\[5ex] = -\dfrac{46\sqrt{x^2 + 9}}{9x} \\[5ex] \therefore \displaystyle\int \dfrac{46dx}{x^2\sqrt{x^2 + 9}} = -\dfrac{46\sqrt{x^2 + 9}}{9x} + C$
(20.) MEHA The gradient function of the curve f(x) is given by $f'(x) = x(3x - 2)$
Find the equation of the curve given that the x-intercept of f(x) is 2

$f'(x) = x(3x - 2) \\[3ex] f(x) = \displaystyle \int f'(x) dx \\[5ex] f(x) = \displaystyle\int x(3x - 2) dx \\[5ex] f(x) = \displaystyle\int (3x^2 - 2x)dx \\[5ex] f(x) = \dfrac{3x^3}{3} - \dfrac{2x^2}{2} + C \\[5ex] f(x) = x^3 - x^2 + C \\[3ex] y = x^3 - x^2 + C \\[3ex]...y = f(x) \\[3ex] Point = x-intercept = (x, 0) = (2, 0) \\[3ex] \implies x = 2,\;\;y = 0 \\[3ex] 0 = 2^3 - 2^2 + C \\[3ex] 0 = 8 - 4 + C \\[3ex] 0 = 4 + C \\[3ex] 4 + C = 0 \\[3ex] C = -4 \\[3ex] y = x^3 - x^2 - 4 \\[3ex] \therefore f(x) = x^3 - x^2 - 4$

(21.) WASSCE:FM Find the area enclosed by the curves $y = x^2 - 3x + 2$ and $y = -x^2 + 3x + 2$

$Points\;\;of\;\;intersection\;\;of\;\;the\;\;two\;\;curves \\[3ex] y = y \\[3ex] \implies \\[3ex] x^2 - 3x + 2 = -x^2 + 3x + 2 \\[3ex] x^2 + x^2 - 3x - 3x + 2 - 2 = 0 \\[3ex] 2x^2 - 6x = 0 \\[3ex] 2x(x - 3) = 0 \\[3ex] 2x = 0 \;\;\;OR\;\;\; x - 3 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = 3 \\[3ex] \therefore limits\;\;of\;\;integration\;\;are:\;\; x = 0 \;\;\;AND\;\;\; x = 3 \\[3ex] first\;\;function:\;\; y = x^2 - 3x + 2 \\[3ex] second\;\;function:\;\; y = -x^2 + 3x + 2 \\[3ex] first\;\;function - second\;\;function \\[3ex] = x^2 - 3x + 2 - (-x^2 + 3x + 2) \\[3ex] = x^2 - 3x + 2 + x^2 - 3x - 2 \\[3ex] = 2x^2 - 6x \\[3ex] Area = \displaystyle\int_{lower\;\;limit}^{upper\;\;limit} (first\;\;function - second\;\;function) dx \\[5ex] = \displaystyle\int_{0}^{3} (2x^2 - 6x) dx \\[5ex] = \left[\dfrac{2x^3}{3} - \dfrac{6x^2}{2}\right]_0^3 \\[5ex] = \left[\dfrac{2x^3}{3} - 3x^2\right]_0^3 \\[5ex] x = 3 \\[3ex] \dfrac{2(3)^3}{3} - 3(3)^2 \\[5ex] 18 - 27 \\[3ex] -9 \\[3ex] x = 0 \\[3ex] \dfrac{2(0)^3}{3} - 3(0)^2 \\[5ex] 0 - 0 \\[3ex] 0 \\[3ex] \implies \\[3ex] Area = -9 - 0 \\[3ex] Area = -9 \\[3ex] But:\;\;Area\;\;cannot\;\;be\;\;negative \\[3ex] \therefore Area = 9\;square\;\;units$
(22.) HSC (a) Determine the area between the parabola $y = x^2 - x + 3$ and the straight line $y = x + 3$

(b) The area between the parabola $y = x^2 - x - 2$ and the straight line $y = x - 2$ is the same as the area determined in part (a).
Explain why this is the case.

$(a) \\[3ex] \underline{Points\;\;of\;\;Intersection} \\[3ex] Parabola:\;\;y = x^2 - x + 3 \\[3ex] Straight\;\;Line:\;\; y = x + 3 \\[3ex] y = y \\[3ex] \implies \\[3ex] x^2 - x + 3 = x + 3 \\[3ex] x^2 - x + 3 - x - 3 = 0 \\[3ex] x^2 - 2x = 0 \\[3ex] x(x - 2) = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x - 2 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = 2 \\[3ex] \therefore limits\;\;of\;\;integration\;\;are:\;\; x = 0 \;\;\;AND\;\;\; x = 2 \\[3ex] \underline{Area\;\;enclosed\;\;by\;\;the\;\;Parabola\;\;and\;\;the\;\;Straight\;\;Line} \\[3ex] first\;\;function:\;\; y = x^2 - x + 3 \\[3ex] second\;\;function:\;\; y = x + 3 \\[3ex] first\;\;function - second\;\;function \\[3ex] = x^2 - x + 3 - (x + 3) \\[3ex] = x^2 - x + 3 - x - 3 \\[3ex] = x^2 - 2x \\[3ex] Area = \displaystyle\int_{lower\;\;limit}^{upper\;\;limit} (first\;\;function - second\;\;function) dx \\[5ex] = \displaystyle\int_{0}^{2} (x^2 - 2x) dx \\[5ex] = \left[\dfrac{x^3}{3} - \dfrac{2x^2}{2}\right]_0^2 \\[5ex] = \left[\dfrac{x^3}{3} - x^2\right]_0^2 \\[5ex] = \left[\dfrac{2^3}{3} - 2^2\right] - \left[\dfrac{0^3}{3} - 0^2\right] \\[5ex] = \left[\dfrac{8}{3} - \dfrac{4}{1}\right] - [0 - 0] \\[5ex] = \dfrac{8 - 12}{3} \\[5ex] = -\dfrac{4}{3} \\[5ex] But:\;\;Area\;\;cannot\;\;be\;\;negative \\[3ex] \therefore Area = \dfrac{4}{3}\;square\;\;units \\[5ex] (b) \\[3ex] \underline{Points\;\;of\;\;Intersection} \\[3ex] Parabola:\;\;y = x^2 - x - 2 \\[3ex] Straight\;\;Line:\;\; y = x - 2 \\[3ex] y = y \\[3ex] \implies \\[3ex] x^2 - x - 2 = x - 2 \\[3ex] x^2 - x - 2 - x + 2 = 0 \\[3ex] x^2 - 2x = 0 \\[3ex] x(x - 2) = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x - 2 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = 2 \\[3ex]$ These are the same points of intersection as part (a)
Also:

$\underline{Area\;\;enclosed\;\;by\;\;the\;\;Parabola\;\;and\;\;the\;\;Straight\;\;Line} \\[3ex] first\;\;function:\;\; y = x^2 - x - 2 \\[3ex] second\;\;function:\;\; y = x - 2 \\[3ex] first\;\;function - second\;\;function \\[3ex] = x^2 - x - 2 - (x - 2) \\[3ex] = x^2 - x - 2 - x + 2 \\[3ex] = x^2 - 2x \\[3ex]$ The area enclosed is the same as part (a)
Because of the same points of intersection and the same areas enclosed, the area between the parabola and the straight line respectively for parts (a) and (b) are the same.
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