Solved Examples on Integration

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 $

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 $

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} $

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 $

$ (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} $

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 $

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} $

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 $

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 $

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 $

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 $

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 $

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 $

$ (14.1) \\[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 $

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 $

$ 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 $

$ (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) $

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 $

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 $

$ 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 $

$ 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 $

$ (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.