{"id":3477,"date":"2023-10-05T10:38:51","date_gmt":"2023-10-05T10:38:51","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3477"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-7-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-7-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 7 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

WEEK SEVEN
\n<\/strong>TOPIC: INTEGRATION
\n<\/strong>Integration: <\/strong>This is defined as anti- differentiation. Suppose, y = x3<\/sup> + 2x, the first derivative is
\n 3x2<\/sup> + 2. (dy\/dx = 3x2 <\/sup>+ 2)
\nthen the anti \u2013 derivative of 3x2<\/sup> + 2 = x3<\/sup> + 2x
\n Thus, integration is the reverse process of differentiation and denoted by the symbol \u222b.
\nIf dy\/dx = xn<\/sup><\/strong> , then \u222b dy\/dx = xn+1 <\/sup> + C
\n\t\t\t\t<\/strong>n + 1 (n \u2260 -1)
\n<\/strong>where c is the arbitrary constant.<\/p>\n

\u00a0INDEFINITE INTEGRAL CONCEPTS
\n<\/strong>General Concept
\n<\/strong>Example: Evaluate the following integrals: 1. \u222bx2<\/sup> dx 2. \u222b<\/strong>x5\/2<\/sup> dx 3.\u222b <\/strong>4\/ x5 dx 4.\u222b <\/strong>\u221ax8<\/sup> 5.\u222b <\/strong>(7x4<\/sup> + 2) dx 6.\u222b<\/strong>(x5<\/sup> + 2x4<\/sup> \u2013 x3<\/sup> + 6) dx
\nSolution;
\n1. x3 <\/sup>+ C 2.x5\/2 + 1 <\/sup>= 2x7\/2 <\/sup>+ c 3. \u222b <\/strong>4x-5<\/sup> dx = 4x-5+1<\/sup>= – 4x \u2013 4 <\/sup>+ C
\n 3 5\/2 + 1 7 -5 + 1
\n4. \u222b<\/strong> x4<\/sup> = x4+1 <\/sup>= x5<\/sup> + C 5. 7x4 +1 <\/sup>+ 2x0+1 <\/sup>=7x5<\/sup> + 2x + C
\n\t\t\t 4 + 1 5
\n6. x6<\/sup> + 2x5<\/sup> \u2013 x4<\/sup>+ 6x + C
\n 6 5 4
\nNB: Integral of a constant is not zero but the variable in the question.<\/p>\n

\u00a0Evaluation<\/strong>: Evaluate the following integrals; 1. \u222b (<\/strong>12x3<\/sup> \u2013 x6<\/sup> +1\/x2<\/sup>) dx 2. \u222b<\/strong>x2<\/sup>(3x2<\/sup> + 4x) dx<\/p>\n

\u00a0Trigonometric integral:
\n<\/em><\/strong>\"\"\/The trigonometric integrals can be summarized in the following table. Remember that this is the reverse process of differentiation.
\n F(x) \u222b<\/strong> f(x)dx
\n Sin x – cos x + c
\n Cos x sin x + c
\n Sec2<\/sup>x tan x + c
\n Cosec2<\/sup>x – cot x + c
\n Sec x tan x sec x + c
\nex<\/sup>ex<\/sup> + c
\n 1\/x ln x + c<\/p>\n

\u00a0Example: Evaluate each of the following integrals.
\n1. \u222b <\/strong>sin x \u2013 5 cos x)dx 2. \u222b (<\/strong>5sinx + 3x2<\/sup>)dx
\nSolution:
\n1. . \u222b sin x \u2013 5 cos x )dx = \u222bsin x dx – \u222b5 cos x dx
\n= -cos x \u2013 5 ( sin x ) + c
\n = – cos x \u2013 5sin x + c
\n2.\u222b (5sinx + 3x2 <\/sup>)dx = \u222b5 sin x dx + \u222b3×2 dx
\n = -5cos x + x3<\/sup> + c
\n3. \u222b e2x<\/sup> dx = e2x<\/sup>\/2 + c<\/p>\n

\u00a0Evaluation: Evaluate the integrals:
\n 1. \u222b<\/strong> (3 cos x + 2 sin x) dx 2. \u222b e2×2 + 5x<\/sup> dx<\/p>\n

\u00a0INTEGRATION BY ALGEBRAIC SUBSTITUTION
\n<\/strong>Sometimes integral are not given in the standard form, such integral are then reduced to standard form format before evaluation by algebraic substitution.
\nSuppose, an integral is given in the form \u222b f(ax + b)n<\/sup> dx<\/strong>
\n\t\tThen, the algebraic substitution is to represent the function in the bracket by any letter.
\nLet u = (ax + b) du\/dx = a, dx = du\/ a
\n<\/strong> \u222bun<\/sup> dx = \u222b un<\/sup> du\/a
\n<\/strong> = 1\/a \u222b un<\/sup> du<\/strong>
\n\t\tExample:
\nEvaluate the following integrals.
\n\"\"\/1. \u222b <\/strong>(2x2<\/sup> \u2013 5x )4<\/sup> dx 2. \u222b 7 dx 3. \u222b xcos 2x2<\/sup> dx 4. \u222b x2<\/sup> \u221a(x3<\/sup> + 5)dx
\n (5x \u2013 4 )5
\n<\/sup>Solution:
\n<\/strong>1. <\/strong>.\u222b <\/strong>(2x2<\/sup> \u2013 5x )4<\/sup> dx let u = 2x2<\/sup> \u2013 5x , du\/dx = 4x \u2013 5 , dx = du\/4x -5
\n . \u222b <\/strong>u 4 du\/ 4x -5 = u5<\/sup><\/strong>+ c
\n 5(4x \u2013 5)
\n= (2x2<\/sup> \u2013 5x )5<\/sup>+ c
\n\t\t\t20x \u2013 25
\n2. \u222b 7 dx let u = ( 5x \u2013 4) , du\/dx = 5, dx = du\/5
\n (5x \u2013 4 )5
\n<\/sup> \u222b 7 u -5<\/sup> du\/5 = 7u-4<\/sup>+ c
\n 5 x – 4
\n= 7(5x \u2013 4 )-4<\/sup>
\n\t\t -20
\n 3. \u222b x cos 2x2<\/sup> dx let u = 2x2<\/sup>, du\/dx = 4x, dx = du\/4x
\nthen; \u222bx cos 2x2<\/sup> dx = \u00a0\u00a0\u00a0\u00a0\u222bx cos u du\/4x = 1 x ( sin u ) = 1 sin 2x2<\/sup> + c
\n\t\t\t 4 4
\n 4. \u222b x2<\/sup> \u221a(x3<\/sup> + 5)dx let u = x3<\/sup> + 5, du\/dx = 3x2<\/sup>, dx = du\/3x2<\/sup><\/p>\n

\u00a0\u222b x2<\/sup> \u221a(x3<\/sup> + 5)dx = \u222bx2<\/sup> u1\/2 <\/sup>du\/3x2<\/sup> = 1 x u3\/2 <\/sup>= 2 (x3<\/sup> + 5)3\/2<\/sup>+ c
\n\t\t\t 3\/2 x 3 9<\/p>\n

\u00a0 Evaluation:
\n<\/strong> Evaluate the following integrals:
\n 1. \u222b(5x \u2013 7 )7\/2<\/sup>dx 2. \u222bcos 9x dx 3. \u222b xcos 2x dx.<\/p>\n

\u00a0 INTEGRATION BY PARTS
\n<\/strong>This technique is uniquely useful in evaluating integrals that are not in the standard form. Such integrals cannot be solved by algebraic substitution.
\nFrom the product rule of differentiation, it can be generalized thus;
\n \u222b vdu = uv – \u222b udv.<\/p>\n

\u00a0Example:<\/strong> Evaluate the following integral by parts.<\/p>\n

    \n
  1. \n
    \u222b 2x sin x dx 2. \u222b e2x<\/sup>cos 2x dx\n<\/div>\n

    solution:
    \n 1. \u222b 2x sin x dx , let v = 2x, dv\/dx = 2, dv = 2dx
    \n \u222bdu = sin x dx
    \n u = -cos x
    \n \u222b vdu = uv – \u222b udv.
    \n \u222b x2<\/sup> sin x = – x2<\/sup>cos x – \u222b- cos x x 2 dx
    \n = – x2<\/sup>cosx + 2\u222b cos x
    \n = – x2<\/sup>cos x + 2 sin x + c
    \n2. \u222b x2<\/sup>ex<\/sup> dx , let v = x2<\/sup>, dv\/dx = 2x, dv = 2xdx
    \ndu = ex<\/sup> u = ex<\/sup> dx
    \n\u222b vdu = uv – \u222budv
    \n<\/strong>
    \n\u00a0 \u222b x2<\/sup>ex<\/sup> = ex<\/sup> x2<\/sup> – \u222b ex <\/sup>2xdx
    \n = x2<\/sup> ex<\/sup>– 2\u222bex<\/sup> x dx
    \nthe integral part in the RHS will have to be evaluated using integration by parts;
    \nthus, v = x, dv\/dx = 1, dv = dx , du =ex<\/sup>, u = ex<\/sup>
    \n\t\t\t\t\u222b vdu = uv – \u222budv
    \n<\/strong> \u222b x ex<\/sup> = ex <\/sup>. x – \u222bex<\/sup>dx
    \n = ex<\/sup>.x – ex
    \n<\/sup>finally, \u222bx2<\/sup>ex<\/sup> = x2<\/sup> ex<\/sup> – 2(ex<\/sup>.x – ex<\/sup>)
    \n = x2<\/sup>ex<\/sup> \u2013 2xex<\/sup> + 2ex<\/sup> + c<\/p>\n

    \u00a0Evaluation:
    \n<\/strong>Evaluate 1. \u222bx2<\/sup>cos x dx 2. \u222bx3<\/sup> e-x<\/sup> dx<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0INTEGRATION BY PARTIAL FRACTION
    \n<\/strong>S<\/strong>ometimes rational functions are not expressed in the proper standard form; such function can be evaluated by transforming them into standard form through partial fractions. The knowledge of partial fractions is needed here to evaluate the functions.<\/p>\n

    \u00a0Example; Integrate each of the following with respect to x;
    \n1 2x + 32 x + 8
    \n (2x + 1) (x \u2013 1) ( x2<\/sup> + 3x + 2)
    \nsolution:
    \n1. resolve into partial fraction; 2x + 3 = A + B
    \n (2x + 1)(x \u2013 1) (2x + 1) (x \u2013 1)
    \n 2x + 3 = A(x-1) + B(2x + 1)
    \nwhen x = 1, 2(1) + 3 = B(2 + 1)
    \n 5 = 3B, B= 5\/3
    \nwhen x = -1\/2, 2(-1\/2) + 3 = A(-1\/2 \u2013 1 )
    \n 2 = – 3\/2 A A = – 4\/ 3
    \n; 2x + 3 = \u222b -4 + \u222b 5 dx<\/strong>
    \n\t\t\t\t\t (2x + 1)(x \u2013 1) 3( 2x + 1) 3(x \u2013 1)
    \n = -4 ln (2x + 1) + 5 ln (x \u2013 1)
    \n 2 x 3 3
    \n = – 2\/3 ln (2x + 1) + 5\/3 ln ( x \u2013 1) + c
    \n2.x + 8 = x + 8 = A + B
    \n (x2<\/sup> + 3x + 2) (x + 1)(x + 2) x + 1 x + 2
    \n x + 8 = A(x+2) + B(x+1)
    \nwhen x = -2,
    \n – 2 + 8 = B(-2+1)
    \n 6 = -B, B = – 6
    \nwhen, x = -1, – 1 + 8 = A(-1 + 2)
    \n 7 = A.
    \nthus, x + 8 =\u222b 7 + \u222b – 6
    \n x + 1 x + 2
    \n = 7ln (x+ 1) \u2013 6 ln(x+2) + c
    \nEvaluation
    \n<\/strong> Integrate by partial fraction.
    \n 1. 4x + 3 2. 1
    \n\t\t\t\t (x \u2013 3)(x+2) (x2<\/sup>+ 3x + 2)<\/p>\n

    \u00a0GENERAL EVALUATION\/REVISIONAL QUESTIONS
    \n<\/em><\/strong>1.Find the derivatives of the following with respect to x;
    \n(a)\u00a0\u00a0\u00a0\u00a0y = (15 + 5x)(1 + 2x) (b)\u00a0\u00a0\u00a0\u00a0y = (1 + 2x)12<\/sup> (c) y = 3x2<\/sup> (3 \u2013 2x + 4x2<\/sup>) \u00bd<\/sup>
    \n\t\t\t\t2. Given that the gradient function of a curve is 8x \u2013 2, find the equation of the curve at point (2, 4)
    \n3. Find \u222b(x2<\/sup> + 1)(x3 <\/sup>– 2)dx
    \n4. Find \u222bx2<\/sup> e2x<\/sup>dx <\/p>\n

    \u00a0Reading Assignment:<\/strong> Read Integration, Page 31 \u2013 46 Further Mathematics project III.<\/p>\n

    \u00a0WEEKEND ASSINGMENT
    \n<\/strong>1. Evaluate \u222b (x5<\/sup> + 3)dx . A. x6<\/sup>\/6 + 3x + c B. x5<\/sup>\/6 + c C. x6<\/sup>\/6 + c
    \n 2. Evaluate \u222b cos 7x dx A. 7sin 7x B. 1\/7 sin 7x + c C. 7sin 7x + c
    \n3. Integrate the function; (3x + 5)5<\/sup>wrt x .A 12(2x+3)6<\/sup> + c B. (2x +3)6<\/sup>+ c C. (3x + 5)6<\/sup> + c
    \n 12 18
    \n4. Find \u222b(x+1)(x2<\/sup>– 2)dx A. x4<\/sup> + x3<\/sup> \u2013 x2<\/sup> \u2013 2x + c B. x4<\/sup> \u2013 x3<\/sup> + x2<\/sup> + c C. x + x4<\/sup> \u2013 x3<\/sup> \u2013 x2<\/sup> + c
    \n 4 3 3 4<\/p>\n

    \u00a05. Integrate 1\/x5 <\/sup>wrt x. A. x 6<\/sup> + c B .x-4 <\/sup>+ c C. x-4<\/sup>+ c
    \n 6 – 4 5
    \nTheory:
    \n<\/strong>\"\"\/1. Find \u222bx sin2xdx 2. Evaluate \u222b dx
    \nx ( x + 2)<\/p>\n

    \u00a0
    \n\t\t\t\t\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

    WEEK SEVEN TOPIC: INTEGRATION Integration: This is defined as anti- differentiation. Suppose, y = x3…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,275],"tags":[],"class_list":["post-3477","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3477","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3477"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3477\/revisions"}],"predecessor-version":[{"id":3478,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3477\/revisions\/3478"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3477"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3477"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3477"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}