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16.Integration 16. INTEGRATION Defining Integration � (�$ + 6) = 2� Integration is defined as the reverse process of �� differentiation. When we differentiate a function, the � result is called the differential. When we integrate a (�$ − 4) = 2� function, the result is called the integral or the �� antiderivative. The integral of 2� can be any of the functions: �$+ 0 Notation �$+ 6, The symbol for integration is ò and the term or �$ − 4 or in general $ expression that is to be integrated is called the � + some constant. integrand. If the integrand is, say, a function of x, In fact, the integral of 2� can be represented by a then we must integrate with respect to x and this is family of curves and not a definite curve. Let us written as, examine the graphs of these curves. ∫ f (x)dx The integrand can be a function in any variable and so we can have integrands such as ∫ f (r)dr g(t)dt ∫∫ h(θ)dθ Law of integration To derive the law of integration, we are interested in deriving a rule for the reverse process of differentiation. Recall the rule for differentiation: d ()xnxnn= -1 dx To differentiate xn , multiply by n and decrease the power n by 1. To integrate xn , increase the power n by 1 and divide by the new power, n+1. The law of integration can now be formally stated. We can deduce that $ n + 1 ∫ 2� �� = � + �, where C is a constant to be n x xdx=+ C, where C is the constant of determined. ò n +1 Since the result of this integral is dependent on the integration andwww.faspassmaths.com n ¹-1 . value of C, we refer to integrals such as these as indefinite integrals. At a later stage, we explain how The constant of Integration to evaluate this constant in a given situation. When we differentiate any constant, the result is zero. Consider the following family of functions and their We noted earlier that the differential of a constant is derivatives. zero, we can also conclude that the integral of zero is � (�$) = 2� a constant. �� Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 7 The integral �0� The integral of axn The fundamental law of integration is not applicable When we have a coefficient of xn other than 1, we to the integrand of �0�. This is the one case when the can move it to outside the integral sign and integrate rule, shown above, cannot be applied, that is, the as usual. However, the coefficient need not exception to the law. necessarily be moved to attain the desired result. For If we tried to apply the law to, ∫ �02, we would example, we may choose any one of the following ways to handle coefficients. xx-11+ 0 1 obtain +=CCC +=+. -11+ 0 0 3 3 3 + 1 6x dx 6 x dx 3 6x ∫ = ∫ 6x dx = + C This result is undefined or tends to infinity. So, we ∫ 3+1 x4 can see that the law is inapplicable for � = −1. = 6. + C 6x4 4 = + C 4 1 4 When we have to integrate with respect to x, we x 4 x = 3 + C 3x 2 = + C obtain the following result. 2 1 dx=+ln x C , where C is a constant. 9 ò x The integral of �� is defined as follows: n n Since we can only find the log of a positive value, x ax dx = a x dx , where a is a constant ò ò has a modulus sign. Note also that ln is really loge. Example 3 Integration of �� and ��� 3 Just as in differentiation, if an expression is not in the ∫ dx form that is conformable to the use of the law of 4 x integration, we use algebraic techniques to express it in such a form. Solution 33 dx= dx Example 1 òò4 x 1 4x 2 1 11 3 dx 33-- ∫ x ==x22 dx x dx òò44 1 -+1 Solution 33x 2 æö21 =+=. CxCç÷´2 + 1 −3 441 èø1 dx = x dx The law of integration can now -+1 3 ∫ ∫ x 2 3 be applied. =+xC(where C is a constant) x−3 + 1 x−2 2 x−3 dx = + C = + C ∫ −3+1 −2 1 Integrating a sum C (where C is a constant) = − 2 + When the integrand is composed of more than one 2x term, it must be written within brackets. Each term is to be integrated individually and then the constant of Example 2 www.faspassmaths.comintegration is added on in the final result. x dx ∫ òòò()58xxdxxdxxdx22+º 5+ 8 Solution 58xx32 5 x 3 1 3 2 + 1 =++=++CxC4 1 x 2 x 2 32 3 x dx = x 2 dx = + C = + C ∫ ∫ 1 3 (whereC is constant) 2 +1 2 2 x3 C (where C is a constant) = + 3 Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 2 of 7 The integral of a constant, k Solution To evaluate k dx . We write kx0 = k because We expand the integrand to get: ∫ 2 ()()3x- 1 x+ 4 dx =( 3 x + 11 x- 4) dx x0 = 1 . So, òò We proceed to integrate each term separately. 0 ∫ 4 dx = ∫ 4x dx 0 + 1 x 2 = 4 + C = 4x + C (where C is a constant) ∫()3x −1 ()x + 4 dx = ∫(3x +11x − 4) dx 0 +1 3x3 11x2 = + − 4x + C Example 4 3 2 1 3 11 2 − dx = x + x − 4x + C (where C is a constant) ∫ 2 2 Solution Example 7 3 1 1 0 æö − dx = − x dx 43xx- ∫ 2 ∫ 2 ò ç÷dx èøx 1 x0 + 1 1 = − + C = − x + C (where C is a constant) 2 0 +1 2 Solution We can now deduce, that when A is a constant, then We divide each term by x, to get: æöæö43xx33- 4 x 3 x Adx=+ Ax C(isaconstant) C òòç÷ç÷dx º-dx ò èøèøxxx =43xdx2 - Integral of products and quotients ò () We do not have a product rule or a quotient for 4x3 =-3xC+ (where C is a constant) integration. In simple cases involving these 3 n situations, we apply the distributive law to remove The integral of ()ax+ b the brackets and convert the expression to a sum of n terms. To find 2x + 3 dx , when n = 2, we may choose ∫() to expand as was done in the examples above. Example 5 As n becomes larger, say 3 or greater, the expansion ò xx2 ()34- dx will result in many terms and integration can become tedious. In such cases, the method of substitution is very helpful. Solution n First, we expand the integrand to get ò ()ax+ b dx 232 òòxx()34- dxx=() 3- 4 xdx dt dt Let t=+ ax b = a dx = Now we integrate each term separately: dx a n + 1 n dt t www.faspassmaths.com∴ ()ax + b dx ≡ t n . = + C x2 3x 4 dx 3x3 4x2 dx ∫ ∫ a n +1 a ∫ ()− = ∫()− () 3x4 4x3 Re-substituting for t, we get the following rule: = − + C (where C is a constant) 4 3 n + 1 n ()ax+ b ()ax+= b dx +CC(where is a constant) Example 6 ò ()na+1 ò ()()31xxdx-+ 4 Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 3 of 7 We can apply the above rule in evaluating the Using integration to find the equation of a integral of functions of the form (�� + �)9 as curve illustrated in the following examples. We can use integration to find the equation of a curve Example 8 when we know its gradient function and one point on 10 ()2x −1 dx the curve. The point will help us to find the numerical ∫ value of the constant of integration. Solution Example 16 dy The gradient function, , of a curve is given by dx 2 665xx+- Find the equation of the curve if the point()2, 12 lies on the curve. Example 9 4x −1 dx Solution ∫ The equation of the curve will be Solution yxxdx=+6652 - 1 ò () 4x −1 dx ≡ ()4x −1 2 dx 66xx32 ∫ ∫ yx=+-5+CC (where is a constant) 1 32 In this example, ab==4,- 1 and n = . 2 yx=+23532 x- xC+ 1 3 + 1 ()4x −1 2 ()4x −1 2 Since ()2, 12 lies on the curve, then the value of C, ∴ 4x −1 dx = + C = + C ∫ ⎛ 1 ⎞ 6 +1 × 4 can be found by substituting x = 2 and y = 12 ⎝⎜ 2 ⎠⎟ 3 2 1 3 12 = 2 2 + 3 2 − 5 2 + C = 16 +12 −10 + C = 4x −1 + C (where C is a constant) () () () 6 () 12 = 18 + C Example 10 C = −6 1 dx ∫ 3 2x 1 \The equation of the curve is, − 32 yx=+2356 x-- x. Solution 1 1 − Indefinite and definite integrals dx ≡ 2x −1 3 dx ∫ 3 ∫() 2x −1 All the integration examples done so far are examples 1 of what is referred to as indefinite integrals. The Using the formula, a = 2, b = −1 and n = − , we 3 results were written in terms of the variable together have www.faspassmaths.comwith the constant of integration added on. n 1 2 − + 1 For example, an integral such as x dx = 3 3 ò 1 ()2x +1 ()2x +1 dx = + C = + C n + 1 ∫ 3 2x 1 ⎛ 1 ⎞ 4 x − − +1 × 2 + C is said to be of indefinite value. ⎝⎜ 3 ⎠⎟ 3 n +1 ⎛ 2 ⎞ As we shall explain in a later chapter on the meaning 33 2x +1 ⎝⎜ () ⎠⎟ of integration, indefinite integrals cannot be C (where C is a constant) = + evaluated and are non-numerical in nature.
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