{"id":3475,"date":"2023-10-05T10:37:47","date_gmt":"2023-10-05T10:37:47","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3475"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-8-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-8-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 8 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

WEEK EIGHT
\n<\/strong>TOPIC : INTEGRATIO<\/em>N [INDEFINITE INTEGRALDEFINITE INTEGRAL AND AREA UNDER CURVE]<\/em>
\n\t\t\t<\/strong>The process of reversing differentiation is called Integration. If dy\/dx = 3x2<\/sup><\/strong>, then y could be x3<\/sup>,<\/strong> as the derivative of x3<\/sup><\/strong>is 3x2<\/sup><\/strong>.
\nWe say that x3 <\/sup>is an integral of 3x2<\/sup> with respect to x. The symbol for integration sign is given by \u222b . The expression to be integrated is put between the \u222b<\/strong> sign and dx.
\n\u222b 3×2<\/strong> dx could be x3<\/sup><\/strong>
\n\t\tSince differentiating any constant gives zero, the following also have derivation 3x2<\/sup><\/strong>.
\nX3<\/sup> + 2, x3<\/sup> + 4.5, x3<\/sup> \u2013 17<\/strong>etc
\nIn general, any function of the form x3 + c, where c is the constant has derivative of 3x2<\/sup><\/strong>
\n\t\tHence, \u222b 3x2<\/sup> dx = x3<\/sup> + C. C<\/strong> is called constant of integration. Because we do not know the actual or definite value of C, this is called INDEFINITE INTEGRAL.<\/em>
\n\t\t\t<\/strong>
\n\u00a0Let y = xn+1<\/sup>\/ n+1 , Differentiating dy\/dx = (n+1) xn+1<\/sup>\/n+1 = xn
\n<\/sup><\/em><\/strong>Reversing this,
\n\u222b xn<\/sup> dx = xn+1<\/sup>\/n+1 + C
\n<\/strong>\u222b kxn dx = K xn+1<\/sup>\/n+1 + C
\n<\/strong>To Integrate a sum, integrate each term as this is similar to differentiating a sum.
\nExamples:
\n<\/strong>Evaluate (i) \u222b(2x3<\/sup> + 3x2<\/sup> \u2013 4) dx
\n\u00a0\u00a0\u00a0\u00a02\/4x4<\/sup> + 3\/3x3<\/sup> \u2013 4x + C
\n\u00a0\u00a0\u00a0\u00a0\u00bd x2 + x3 \u2013 4x + C
\nEvaluate \u222b(4t3<\/sup> + 2t2 <\/sup>+ \u00bd t2<\/sup>) dt
\n4\/4 t4 + 2\/3 t3 + \u00bd \u00d7 \u00bd t2<\/sup> + C
\nt4 <\/sup>+ 2t3 + \u00bc t2 + C.<\/p>\n

\u00a0Integration of basic Trigonometric functions
\n<\/em><\/strong>Consider the table below for differentiating trigonometric functions.<\/p>\n

\n\n\n\n\n\n
Y<\/td>\nsinKx<\/td>\nCos Kx<\/td>\nTan Kx<\/td>\n<\/tr>\n
dy\/dx<\/td>\nkcoskx<\/td>\n-ksinkx<\/td>\nk\/(coskx)2<\/sup><\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0Consider a similar table as the one given above<\/p>\n

\n\n\n\n\n
Y<\/td>\nSin x<\/td>\nCos x<\/td>\nTan x<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
<\/td>\nCosx+C<\/td>\nSinx +C<\/td>\nTanx +C<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0<\/p>\n

\n\n\n\n\n
Y<\/td>\nsinkx<\/td>\ncoskx<\/td>\nTankx<\/td>\n<\/tr>\n
\u222bydx<\/td>\n-1\/k coskx + C<\/td>\n1\/k sinkx + C<\/td>\n1\/k tankx + C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0Example
\n<\/strong>\u222b(cos2x + sin3x) dx
\n\u222bcos2x dx + \u222b sin3x dx
\n\u00bd sin2x + -1\/3 cos3x
\n\u00bd sin2x \u2013 1\/3 cos3x + K.
\nSometimes,the value of the constant C can be found, if extra information is given.
\nIf dy\/dx = x2+2x-3, find y in terms of x given that x = 1, y =4
\n Y =\u222b(x2 +2x-3)dx
\nY =x3\/3 + 2x\/2 \u2013 3x + C
\nY = x3\/3 +x2 -3x + C
\nPutting x=1,and y = 4
\n4 = 1\/3 + 1 \u2013 3 + C. hence; C = 5 2\/
\n;. 1\/3×3 \u2013 x2 \u2013 3x + 5 2\/3.<\/p>\n

\u00a0If dy\/dx = 2Cos3x, find y given that y =2 and x = 1\/6\u220f
\n\u222b2Cos3x = 2\/3Sin3x + C
\n2 = 2\/3Sin3(\u220f\/6) + C = 2\/3 Sin\u220f\/2 + C
\nMultiplying by \u00bd \u220f (180\/\u220f) = 900<\/sup>
\n\t\tSin \u00bd \u220f – Sin 900<\/sup> = 1
\n2 = 2\/3 \u00d71 + C
\nC = 4\/3
\nY = 2\/3Sin3x + 4\/3 .<\/p>\n

\u00a0Evaluation:
\n<\/strong>Evaluate these indefinite integrals<\/p>\n

    \n
  1. \u222b(y2 \u2013 7y)dy\n<\/li>\n
  2. \u222b(3×2 \u2013 2x \u2013 1)dx\n<\/li>\n
  3. \u222b(Cos4x)dx\n<\/li>\n
  4. \u222b(3cos2x + 4sin3x)dx\n<\/li>\n
  5. If dy\/dx = 4×2 + 1 and y = 2 when x =3, find y in terms of x\n<\/li>\n
  6. \n
    If dy\/dx = 2Sin1\/3x and y = 4 when x =\u220fm, find y\n<\/div>\n

    \u00a0DEFINITE INTEGRALS
    \n<\/strong>In this part, the constant is removed. If a definite integration is performed, the function is evaluated between the values called limits. Upper and lower ie
    \n<\/strong>Example: Evaluate
    \n =x3<\/sup> + C , (33<\/sup> + c) \u2013 (23<\/sup> + c)
    \n(27 + C) \u2013 (8 + C)
    \n27 + C -8 \u2013 C
    \n19.
    \n = 19.
    \n = { \u00bd Sin2}
    \n(\u00bd sin\u00d7 0)
    \n\u00bd sin
    \n = \u00bd <\/p>\n

    \u00a0
    \n\u00a0Area under curve using Definite Integral
    \n<\/strong>\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/Given in the diagram, the area between the curve and the x axis from x = a and to x = b. The area is given by <\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0 Area =
    \n<\/strong>
    \n\u00a0The area can be explained as: Area = \u222b y \u00d7 dx = Sum of height of rectangle \u00d7 width of rectangle
    \n\"\"\/y
    \n<\/strong>
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0a b
    \n<\/strong>
    \n\u00a0Ex 1: Find the Area between the curve y = x3 \u2013 x and the x axis when x =2 and x = 4.
    \n\"\"\/y
    \n<\/strong>
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a00 2 4 x<\/p>\n

    \u00a0
    \n\t\t\t\t{ \u00bc (4)4<\/sup> \u2013 \u00bd(4)2} \u2013 { \u00bc X 24 <\/sup>\u2013 \u00bd (2)2<\/sup>} <\/p>\n

    \u00a064 \u2013 8 -4 + 2 = 54 UNITS.<\/p>\n

    \u00a0Ex 2 : Find the area in the diagram shown below
    \n\"\"\/<\/p>\n

    \u00a0
    \n\u00a0\"\"\/y = 4 \u2013 x2<\/sup>
    \n\t\t\t\t 4<\/p>\n

    \u00a0\"\"\/
    \n\t\t\t\t -2 0 2
    \n\u222b(4 \u2013 x2 )dx
    \n{4x \u2013 x3\/3}2<\/sub>2<\/sup>.
    \n4(2) \u2013 23<\/sup>\/3 – 4(-2) \u2013 (-2)3<\/sup>\/3
    \n(8- 8\/3) \u2013 (-8 + 8\/3)
    \n8 \u2013 8\/3 + 8 \u2013 8\/3
    \n32\/3.square unitrs<\/p>\n

    \u00a0Evaluation:
    \n<\/strong>Find the area between the values as shown below
    \n\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/y
    \n<\/strong>
    \n\u00a0 y = 3x2<\/sup>
    \n\t\t\t\t\t<\/strong>
    \n\u00a0 0 2 4 x
    \n<\/strong>
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0Find the area between the curve y = x2<\/sup> + 3 At the xs axis and when x = -1 and x = 3.<\/p>\n

    \u00a0Evaluation
    \n<\/strong>1. x2<\/sup> and the line y = 2.<\/p>\n

    \u00a0General Evaluation:
    \n<\/strong><\/li>\n

  7. \u222b(3x \u2013 1)(x + 2) dx\n<\/li>\n
  8. \u222b5cos4x (dx)\n<\/li>\n
  9. \n
    \n\t\t\t\t<\/div>\n

    \u00a0Reading Assignment :<\/strong>Solve the evaluation questions given above<\/p>\n

    \u00a0Weekend Assignment:
    \n<\/strong><\/li>\n

  10. Evaluate A. 2\/3 B. -2\/3 C. -6 2\/3 D. 6 2\/3\n<\/li>\n
  11. Evaluate A. 4 B. 2 C. 4\/3 D. 1\/3\n<\/li>\n
  12. Evaluate A. – \u00bd B. 1 C. -1 D. 0\n<\/li>\n
  13. Find the area enclosed by by the curve y = x2 <\/sup>, X = 0 and X = 3 A. 9 B. 7 C. 5\/2 D. 5\n<\/li>\n
  14. \n
    Given y = 3x -2, x=3, x=4. Find the area under the curve A. 4\/3 B. 17\/2 C. 6 D. 3\n<\/div>\n

    Theory
    \n<\/strong><\/li>\n

  15. Find the area enclosed between the curve y =x2 + x -2 and the x axis\n<\/li>\n
  16. \n
    Find the area enclosed by the curve y = x2 \u2013 3x + 3 and the y = 1.\n<\/div>\n

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

    WEEK EIGHT TOPIC : INTEGRATION [INDEFINITE INTEGRALDEFINITE INTEGRAL AND AREA UNDER CURVE] The process of…<\/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-3475","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\/3475","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=3475"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3475\/revisions"}],"predecessor-version":[{"id":3476,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3475\/revisions\/3476"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3475"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3475"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3475"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}