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____________________________________________________________________________________ EMAIL:DEP_MATH@UT.EDU.SA Math1201 تـبوك جامـعة العلوـم. كلـية الرياضـيات. قـسم Lectures Note On the FUNDAMENTALS OF INTEGRAL CALCULUS (MATH1201)
3.1 3.2 3.3 3.4 3.5 ___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 2قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 4 2 1 INTEGRATION INFINITE SERIES CONTENTS APPLICATIONS OF THE DEFINITE INTEGRAL 1.1 1.2 1.3 1.4 1.5 2.1 2.2 2.3 2.4 2.5 4.1 4.2 4.3 Sequences Infinite Series Convergence Tests 4.4 Maclaurin and Taylor Polynomials Maclaurin and Taylor Series4.5 The Indefinite Integral Integration by Substitution The Definite Integral The Fundamental Theorem of Calculus Definite Integrals by Substitution Hyperbolic Functions and Hanging Cables Area Between Two Curves Volumes by Disks and Washers Length of A Plane Curve Area of A Surface of Revolution Integration by Parts Integrating Trigonometric Functions Trigonometric Substitutions Integrating Rational Functions by Partial Fractions. Improper Integrals 3 PRINCIPLES OF INTEGRAL EVALUATION
2 2 𝑥 √ 𝑥 − 1 ___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 3قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1.1 THE INDEFINITE INTEGRAL ________________________________ Table 1.1.1 No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Differentiation Formula ; a: constant Integration Formula ; a,c: constants𝑑 𝑑𝑥 (𝑎) = 0 𝑑 ( 𝑛 𝑛−1 𝑑𝑥𝑥)=𝑛𝑥 𝑑 ( 𝑑𝑥sin𝑥)=𝑐𝑜𝑠𝑥 𝑑 ( 𝑑𝑥−cos𝑥)=𝑠𝑖𝑛𝑥 𝑑 ( 2 𝑑𝑥tan𝑥)=𝑠𝑒𝑐𝑥 𝑑 ( 2 𝑑𝑥−cot𝑥)=𝑐𝑠𝑐𝑥 𝑑 ( 𝑑𝑥sec𝑥)=sec𝑥tan𝑥 𝑑 [ 𝑑𝑥−𝑐𝑠𝑐𝑥]=csc𝑥cot𝑥 𝑑 ( 𝑥 𝑥 𝑑𝑥𝑒)=𝑒 𝑑 𝑏𝑥 𝑥 𝑑𝑥( 𝑑𝑥𝑙𝑛|𝑥|)=𝑥 (ln𝑏)=𝑏 0<𝑏,𝑏≠1) 𝑑 1 ( 𝑑 (𝑑𝑥tan −1 1 𝑥)=1+𝑥2 𝑑 (𝑑𝑥sin −1 1√1−𝑥2 𝑥)= 𝑑 (𝑑𝑥sec −1 1 |𝑥|)= 𝑥√𝑥−1 ∫𝑎𝑑𝑥=𝑎𝑥+𝐶 ∫𝑥 𝑥𝑛+1 𝑛𝑑𝑥= +𝐶(𝑛≠−1) 𝑛+1∫cos𝑥𝑑𝑥=sin𝑥+𝐶 ∫sin𝑥𝑑𝑥=−cos𝑥+𝐶 ∫𝑠𝑒𝑐2𝑥𝑑𝑥=tan𝑥+𝐶 ∫𝑐𝑠𝑐2𝑥𝑑𝑥=−cot𝑥+𝐶 ∫sec𝑥tan𝑥𝑑𝑥=sec𝑥+𝐶 ∫csc𝑥cot𝑥 𝑑𝑥=−𝑐𝑠𝑐𝑥+𝐶 ∫𝑒𝑥𝑑𝑥=𝑒𝑥+𝐶 𝑏𝑥 ∫𝑏𝑥𝑑𝑥= ln𝑏+𝐶 (0<𝑏,𝑏≠1) 1 ∫ 𝑥𝑑𝑥=𝑙𝑛|𝑥|+𝐶 1 1+𝑥2 ∫ 𝑑𝑥 = tan−1 𝑥+𝐶 1 √1 − 𝑥2 ∫ 𝑑𝑥=sin−1𝑥+𝐶 1 ∫ 𝑑𝑥=sec−1|𝑥|+𝐶 1. 1.1 Definition A function given open interval if ′ is called an of a function for all on the interval. on a 1.1.2 Theorem. If𝐹(𝑥) is any antiderivative of 𝑓(𝑥) on an open interval, then for any constant C the function is also an antiderivative on that interval. Moreover, each antiderivative of on the interval can be expressed in the form by choosing the constant c appropriately. 𝐹(𝑥)+ 𝐶 𝐹 𝐹 (𝑥) = 𝑓(𝑥) 𝐹(𝑥) + 𝐶 𝑓(𝑥) 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑥 𝑓
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 4قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك ▪ PROPERTIES OF THE INDEFINITE INTEGRAL 1.1.3 Theorem. Solution. 6 Solution. Example 2 Example 1 Suppose that𝐹(𝑥)and𝐺(𝑥) areantiderivatives of respectively,andthat 𝑐 is a constant.Then: a) Aconstant factor can be moved through an integralsign; that is, ∫𝑐𝑓(𝑥)𝑑𝑥 = 𝑐𝐹(𝑥)+ 𝑐An antiderivative of a sum is the sum of the antiderivatives; that is, and ∫[𝑓(𝑥)+ 𝑔(𝑥)]𝑑𝑥 = 𝐹(𝑥) + 𝐺(𝑥) + 𝑐An antiderivative of a difference is the difference of the antiderivatives; that is, Evaluate Evaluate ∫(3𝑥 − 2𝑥2 + 7𝑥 + 1) 𝑑𝑥 (𝑎) ∫4cos𝑥𝑑𝑥. ∫(3𝑥6−2𝑥2+7𝑥+1)𝑑𝑥=3∫𝑥6 3 𝑥7 7 −2 𝑥3 = +7 𝑥 +𝑥+𝐶 2 3 2 (𝑎) ∫4cos𝑥𝑑𝑥=4∫cos𝑥𝑑𝑥=4sin𝑥+𝐶 𝑏) 𝑑𝑥=∫𝑥𝑑𝑥+∫𝑥𝑑𝑥=𝑥2 ( ∫(𝑥+𝑥2) 2 +2 𝑑𝑥−2∫𝑥2 + 𝐶 ∫[𝑓(𝑥)− 𝑔(𝑥)]𝑑𝑥 = 𝐹(𝑥) − 𝐺(𝑥) + 𝑐 𝑑𝑥+7∫𝑥𝑑𝑥+∫1𝑑𝑥 (𝑏) 𝑓(𝑥) 𝑔(𝑥), ∫(𝑥 + 𝑥2) 𝑑𝑥. 𝑥 3 3 c) b)
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 5قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك Solution. Evaluate the integral Evaluate the integral PRACTICAL SET 1.1 HOMEWORK SET 1.1 Example 3 Evaluate 𝑠𝑒𝑐𝜃 𝑐𝑜𝑠𝜃 1.∫ 𝑑𝜃 cos𝑥 (𝑎) ∫ 𝑑𝑥sin 𝑥2 2 . ∫ 𝑥 𝑑𝑥 𝑡2 − 2𝑡4 𝑡4(𝑏) ∫ 𝑑𝑡 c o s 𝑥 1 c o s 𝑥 sin𝑥sin𝑥 (𝑎) ∫ 𝑑𝑥=∫ 𝑑𝑥=∫csc𝑥cot𝑥𝑑𝑥=−csc𝑥+𝐶sin 𝑥2 𝑡2 − 2𝑡4 𝑡 2𝑡 1 𝑡2 (𝑏) ∫ 𝑑𝑡=∫( − )𝑑𝑡=∫( −2)𝑑𝑡=∫(𝑡−2−2)𝑑𝑡𝑡4 𝑡 𝑡 =∫𝑡−2𝑑𝑡−2∫𝑑𝑡= 𝑡−1 −1 1 𝑡 −2𝑡+𝐶=− −2𝑡+𝐶 𝑥2 2 𝑥2+1−1 𝑥 +12 𝑥2+1 𝑥 +12 1 𝑥 +12(𝑐) ∫ 𝑑𝑥=∫ 𝑑𝑥=∫ 𝑑𝑥−∫ 𝑑𝑥 𝑥+1 =∫𝑑𝑥−∫1𝑑𝑥=𝑥−tan−1𝑥+𝐶 𝑥 +12 𝑥 (𝑐) ∫ 𝑑𝑥 𝑥+1 3.∫𝑥3√𝑥 𝑑𝑥 2 4 4 4 5 7 2 2 ---------------------------------------------------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------------------------------------------------- 1. ∫x(2−𝑥)2𝑑𝑥 𝑠𝑖𝑛𝑥 2. ∫ 𝑐𝑜𝑠 𝑥 𝑑𝑥2 1 3
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 6قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1.2 INTEGRATION BY SUBSTITUTION_____________________________ Example 2 Evaluate Solution. Let Or since F is an antiderivative of Which can write in integralform as Suppose that𝐹 an antiderivative of implies that the derivative of Example 1 Evaluate Solution. ∫(𝑥2+1)50 Let 𝑢=𝑥2+1 ⟹𝑑𝑢=2𝑥 𝑑𝑥. Thus, the given integral can be written as Let 𝑢=𝑔(𝑥)⟹ 𝑑𝑢 ′ ′ 𝑑𝑥With this notation (2) can be expressed as and thatg is a differentiable function. The chain rule can be expressed ∫ 𝑢 ∫ ( + 𝑠𝑒𝑐2 𝑢=𝜋𝑥 ⟹ 𝑑𝑢=𝜋𝑑𝑥⟹𝑑𝑥= 𝑓 𝐹(g(𝑥)) 𝑑 [𝐹(g(𝑥))] = 𝐹′(g(𝑥))g′(𝑥) 𝑑𝑥 𝜋𝑥)𝑑𝑥 ⋅𝑑𝑢= ⋅2𝑥 𝑑𝑥. = 𝑔 (𝑥) ⟹ 𝑑𝑢 = 𝑔 (𝑥)𝑑𝑥. + 𝐶 = ∫𝑓(u)𝑑𝑢 = 𝐹(u)+ 𝐶 ∫𝐹′(g(𝑥))g′(𝑥)𝑑𝑥 = 𝐹(g(𝑥))+ 𝐶𝑓, ∫𝑓(g(𝑥))g′(𝑥)𝑑𝑥 = 𝐹(g(𝑥))+ 𝐶 + 𝐶 (3) (2) (1) 1 𝑥 50 𝑑𝑢 𝜋 𝑢 51 (𝑥 +1)2 51 51 51
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 7قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1. 2. 3. 𝑢 = 𝑥 − 1 So, that ∫ = ∫ 𝑒 ∫(4𝑥−3)9 ∫ 𝑒2𝑥 𝑑𝑥. ∫2𝑥(𝑥2+1) ∫sin7𝑥𝑑𝑥. sin√𝑥𝑑𝑥 ; 𝑐𝑜𝑠𝑥 𝑑𝑥. 𝑑𝑢 = 𝑑𝑥 1 𝑥 ∫ ( + 𝑠𝑒𝑐2 𝑑𝑥; 𝑢 = √𝑥. 𝑢=𝑥2+1. 𝑥2=(𝑢+1) =𝑢2+2𝑢+12 1 1 𝜋 𝜋𝑥) 𝑑𝑥 = ln|𝑥| + ∫ 𝑠𝑒𝑐2𝑢 𝑑𝑢 = ln|𝑥| + tan 𝑢 + 𝐶 𝜋=ln|𝑥|+1𝑡𝑎𝑛𝜋𝑥+𝑐 𝜋∫𝑥2√𝑥−1𝑑𝑥 ∫ 𝑥2 √𝑥 − 1𝑑𝑥 = ∫(𝑢2 + 2𝑢 + 1)√𝑢 𝑑𝑥 = ∫ (𝑢 + 2𝑢 + 𝑢 ) 𝑑𝑥 𝑢7⁄2+4𝑢5⁄2+2𝑢3⁄2+𝑐=2(𝑥−1) ⁄2+4(𝑥−1) ⁄2+2(𝑥−1) ⁄2+𝑐7 5 3 5 3 7 5 3 2 7 1 √𝑥 23 5⁄2 3⁄2 1⁄2 PRACTICAL SET 1.2 Example 3 Evaluate Solution. so that 3. 𝑠𝑖𝑛𝑥 HOMEWORK SET1.2 Evaluate the integrals using the substitutions. Evaluate the integrals using appropriatesubstitutions. ---------------------------------------------------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------------------------------------------------- 1. 2. 𝑑𝑥.
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 8قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1.3 THE DEFINITE INTEGRAL __________________________________ Example 1 integrable on interval If a function 𝑓 is continuous on an interval, then 𝑓 is and the net signed area A between the graph ofand the [𝑎,𝑏], [𝑎,𝑏] 𝑖𝑠 ∫𝑥2 𝑑𝑥=0 𝐴=∫𝑓(𝑥)𝑑𝑥𝑎 [𝑎,𝑏], 𝑓 1 1 𝑏 (b) If 1.3.1 THEOREM. is integrable on 1.3.2 DEFINTION. (a) If is in the domain of , we define 𝑎then we define 𝑎 𝑓 𝑓 [𝑎,𝑏], ∫ 𝑓(𝑥)𝑑𝑥 = 0 ∫𝑓(𝑥)𝑑𝑥=−∫𝑓(𝑥)𝒅𝒙𝑏 𝑎 1.3.3 𝑇𝐻𝐸𝑂𝑅𝐸𝑀.𝐼𝑓 𝑓 𝑎𝑛𝑑 𝑔 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒 𝑜𝑛 [𝑎,𝑏],𝑎𝑛𝑑 𝑖𝑓 𝑐 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,𝑡ℎ𝑒𝑛 𝑐𝑓,𝑓+𝑔,𝑎𝑛𝑑 𝑓−𝑔 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒 𝑜𝑛 [𝑎,𝑏],𝑎𝑛𝑑 (a) ∫𝑏𝑐𝑓(𝑥)𝑑𝑥=𝑐∫𝑏 𝑎 𝑎𝑓(𝑥)𝑑𝑥 (b) ∫𝑏[𝑓(𝑥)+𝑔(𝑥)]𝑑𝑥=∫𝑏𝑓(𝑥)𝑑𝑥+∫𝑏 𝑎 𝑎 𝑎𝑔(𝑥)𝑑𝑥 (c) ∫𝑏[𝑓(𝑥)−𝑔(𝑥)]𝑑𝑥=∫𝑏𝑓(𝑥)𝑑𝑥−∫𝑏 𝑎 𝑎 𝑎𝑔(𝑥)𝑑𝑥 𝑎 𝑎 𝑏
∫ ] ___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 9قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1.4 THE FUNDAMENTAL THEOREM OF CALCULUS _________________ (b) Example 1 (a) 2 5 𝑒 ∫𝑥√𝑥𝑑𝑥=∫𝑥2 𝑥 𝑑𝑥=5[𝑒 𝑥 =5(𝑒 𝑑𝑥=∫𝑥 𝑛𝑜 𝑚𝑎𝑡𝑡𝑒𝑡𝑟 ℎ𝑜𝑤 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒 𝑜𝑟𝑑𝑒𝑟𝑒𝑑. 𝑑𝑥= ∫𝑓(𝑥)𝑑𝑥=∫𝑓(𝑥)𝑑𝑥+∫𝑓(𝑥)𝑑𝑥 𝑎 𝑎 𝑐 −𝑒0)=5(3−1)=10 [𝑥]= (2187−128)= 1.3.4 𝑇𝐻𝐸𝑂𝑅𝐸𝑀..𝐼𝑓𝑓 𝑖𝑠 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒 𝑜𝑛 𝑎 𝑐𝑙𝑜𝑠𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑒𝑝𝑜𝑖𝑛𝑡𝑠 𝑎,𝑏,𝑎𝑛𝑑 𝑐,𝑡ℎ𝑒𝑛 𝑏 𝑐 𝑏 𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 [𝑎,𝑏] 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑛𝑦 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓 𝑜𝑛 [𝑎,𝑏],𝑡ℎ𝑒𝑛 𝑏 ∫𝑓(𝑥)𝑑𝑥=𝐹(𝑏)−𝐹(𝑎)𝑎 =588 9 4 ln3 0 9 4 𝑥ln3 0 9 4 ln3 2 7 9 4 2 7 4118 7 2 7 1 2 5 2 7 2 𝟏.𝟒.𝟏 𝑻𝑯𝑬𝑶𝑹𝑬𝑴 (𝑻𝒉𝒆 𝑭𝒖𝒏𝒅𝒂𝒎𝒆𝒏𝒕𝒂𝒍 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝒐𝒇 𝑪𝒂𝒍𝒄𝒖𝒍𝒖𝒔,𝑷𝒂𝒓𝒕 𝟏).
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 10قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك Example 2 Solution. From theorem1.3.4 Example 3 Solution. 𝐹𝑖𝑛𝑑 Evaluate = [ [∫ 𝑡 ∫ 𝑡 𝑑𝑡] ] + [ ∫ 𝑓(𝑥)𝑑𝑥 𝑑𝑡=[ 𝑖𝑓 𝑑 𝑑𝑥 [∫ 𝑡 𝑑𝑡] = 𝑥3 1 𝑥4 4 4 ] = −1, 𝑑[ 4 𝑑𝑥 2 𝑓(𝑥) = {𝑥 , 3𝑥 − 2, ∫𝑓(𝑥)𝑑𝑥=∫𝑓(𝑥)𝑑𝑥+∫𝑓(𝑥)𝑑𝑥=∫𝑥0 0 ∫𝑓(𝑥)𝑑𝑥=∫𝑓(𝑥)𝑑𝑥+∫𝑓(𝑥)𝑑𝑥𝑎 𝑎 𝑐 𝑥 < 2 𝑥 ≥ 2 − ] = 𝑥 3 −2𝑥]=(−0)+(42−2)= 𝑑𝑥+∫(3𝑥−2)𝑑𝑥 𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 𝑎𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝐼,𝑡ℎ𝑒𝑛 𝑓 ℎ𝑎𝑠 𝑎𝑛 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑛 𝑡ℎ𝑎𝑡 𝐼.𝐼𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟, 𝑖𝑓 𝑎 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛 𝐼,𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐹 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑥 𝐹(𝑥)=∫𝑓(𝑡)𝑑𝑡𝑎 𝑖𝑠 𝑎𝑛 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓 𝑜𝑛 𝐼: 𝑡ℎ𝑎𝑡 𝑖𝑠,𝐹′(𝑥)=𝑓(𝑥) 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑥 𝑖𝑛 𝐼,𝑜𝑟 𝑖𝑛 𝑎𝑛 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 𝑛𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑑 𝑑𝑥 𝑥 [∫𝑓(𝑡)𝑑𝑡]=𝑓(𝑥)𝑎 6 𝑑 𝑑𝑥 𝑥 1 𝑥 1 6 0 3 3 0 𝑥32 3 0 2 𝑏 3𝑥 2 2 𝑡 𝑥 4 1 2 6 2 6 𝑐 𝑥 3 8 3 𝑥 4 4 𝑏 2 2 1 4 6 2 128 3 𝟏.𝟒.𝟐 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 (𝑻𝒉𝒆 𝑭𝒖𝒏𝒅𝒂𝒎𝒆𝒏𝒕𝒂𝒍 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝒐𝒇 𝑪𝒂𝒍𝒄𝒖𝒍𝒖𝒔,𝑷𝒂𝒓𝒕 𝟐).
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 11قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 2. (c) (a) (b) Let PRACTICAL SET 1.4 HOMEWORK SET 1.4 Evaluate the following integral 1. Evaluate the following integral Find ∫ 𝐹(4) 𝐹′(4) ---------------------------------------------------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------------------------------------------------- 1. 2. 𝑑𝑥 𝑑𝑥 𝐹′′(4) ∫ 𝑡sec𝑡𝑑𝑡 𝐹(𝑥) = ∫ √𝑡 + 9 𝑑𝑡.2 𝑑 𝑑𝑥 ∫441𝑥2 4 1 1𝑥√𝑥 0 𝑥 𝑥 4
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 12قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 1.5 DEFINITE INTEGRALS BY SUBSTITUTION _________________ TWO METHODS FOR MAKING SUBSTITUTIONS IN DEFINITE INTEGRALS Method 1. First evaluate the indefinite integral If we let 2 Then we obtain Solution by Method 1 𝑎to evaluate the definite integral. by substitution and then use the relationship Example 1 Use the two methods above to evaluate. Method 2. Make the substitution relationship ′ directly in the definite integral and then use the to replace the x-limits, by corresponding u-limits, This produces a new definite integral 𝑢 = 𝑔(𝑥),𝑑𝑢 = 𝑔 (𝑥)𝑑𝑥 𝑢 = 𝑔(𝑥) 𝑢=𝑔(𝑎) & 𝑢=𝑔(𝑏). 𝑢=𝑥+1 ⟹ 𝑑𝑢= 2𝑥𝑑𝑥 ⟹𝑥𝑑𝑥= 𝑑𝑢 𝑓(𝑢)𝑑𝑢 ∫𝑓(g(𝑥))g′(𝑥)𝑑𝑥 ∫𝑥(𝑥2+1) 𝑑𝑥.3 0 𝑥=𝑎 & 𝑥=𝑏, ∫ 𝑓(g(𝑥))g′(𝑥)𝑑𝑥 = [∫ 𝑓(g(𝑥))g′(𝑥)𝑑𝑥]𝑎 𝑏 2 ∫𝑔(𝑎) 𝑔(𝑏) 1 2 𝑏
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 13قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 𝐿𝑒𝑡 𝑔(𝑥) 𝑖𝑓 ∫si2 0 n 2𝑥cos2𝑥𝑑𝑥 ∫ (2𝑥 − 5) (𝑥 − 3) 𝑑𝑥9 𝑢=𝑥2+1 ⟹ 𝑑𝑢= 2𝑥𝑑𝑥 ⟹𝑥𝑑𝑥= 𝑥=0 ⟹ 𝑢=1 ∫ 𝑥 (𝑥2 + 1) 𝑑𝑥 = ∫ 𝑢33 𝑑𝑢 1 2 ∫ 𝑥(𝑥2 + 1) 𝑑𝑥 = ∫ 𝑢33 & 𝑔′ 𝑓 𝑎≤𝑥≤𝑏 , 𝒃 ∫𝒇(𝒈(𝒙))𝒈′(𝒙)𝒅𝒙=∫𝒂 (3) 𝑑𝑢=[ 𝑖𝑓 𝑢=sin2𝑥 ⟹𝑑𝑢=2cos2𝑥𝑑𝑥 ⟹𝑑𝑥= ] = 𝒇(𝒖)𝒅𝒖 𝑢4 (𝑥2 + 1)4 𝑑𝑢= = 8 8 𝑥=2 ⟹ 𝑢=5 − =78 2 (𝑥2 + 1) (4 + 1) 14 625 14 ∫𝑥(𝑥2+1) 𝑑𝑥=[3 ]= − = − =78 0 8 0 8 8 8 8 5 2 2 0 1 2 1 5 2 1 4 2 𝒈(𝒂) 𝒈(𝒃) 𝑢 54 8 1 𝑑𝑢 2cos2𝑥 . 625 8 1 8 Thus, Thus, Solution. (a) the values of 𝟏.𝟓.𝟏𝑻𝒉𝒆𝒐𝒓𝒆𝒎 . Solution by Method2 Let Example 2 Evaluate If is continuous on [a, b] and for then is continuous on interval containing 𝜋 8 (a) (b)
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 14قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 𝐿𝑒𝑡 𝑖𝑓 𝑖𝑓 𝑑𝑥 1 − 𝑥 𝑎)∫ 0 𝑥=0 ⟹ 𝑢=sin0=0 𝑥=2 ⟹ 𝑢=2−3=−1 & & 𝑖𝑓 𝑖𝑓 𝑑𝑢 1 1 6 ∫ sin2 2𝑥 cos 2𝑥 𝑑𝑥 = ∫ 𝑢2 cos 2𝑥 = ∫ 𝑢2 𝑑𝑢 = [𝑢 ] 2cos2𝑥 2 1 1 1 1 1 12√2 = [ −0]=3 ( )= 6 (√2) 62√2 𝑏) ∫0 ∫(2𝑥−5)(𝑥−3) 𝑑𝑥=∫(2𝑢+1)𝑢9𝑑𝑢=∫(2𝑢9 +𝑢9)𝑑𝑢 2 −1 2 (2)11 −1 2𝑢11 𝑢102 210 2(−1)11 (−1)10 10=[ + ] = ( + )−( + ) 11 10 −1 11 10 11=(4096+1024)−(−2+1)=4094+1025=40940+11275=52215=474.68 11 10 11 10 11 10 110 110 𝑒 𝑥 (1+𝑒 𝒂) 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛: 𝑢=1−𝑥 ⟹𝑑𝑢=−𝑑𝑥 ⟹𝑑𝑥=−𝑑𝑢 3 4 3 4 1 4 𝑖𝑓 𝑥=0 ⟹ 𝑢=1−0=1 & 𝑖𝑓 𝑥= ⟹ 𝑢=1−= So, 𝜋 𝜋 1 √2 𝑥= ⟹ 𝑢=sin= 8 4 𝑥=5 ⟹ 𝑢=5−3=2 ) 𝑑𝑥 𝑢=𝑥−3 ⟹𝑑𝑢=𝑑𝑥 & 𝑥=𝑢+3 ⟹ 2𝑥−5=2(𝑢+3)−5=2𝑢+1 So, So, (b) Solution. Example 3 Evaluate 5 𝜋 8 0 3 4 1 √2 0 2 2 1 √2 0 10 1 3√2 0 ln3 1 2𝑥
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 15قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك ∫0 𝑒 𝒃) 𝐿𝑒𝑡 So 𝑥 (1+𝑒 𝑑𝑥 ∫𝑥√1+𝑥𝑑𝑥 sin√𝑥𝑑𝑥 = )𝑑𝑥=∫𝑢 2 2 3 23 𝑑𝑢=[𝑢]= [4 −2 ]=2 2 [8− 8] 3 3 3 √ 𝑑𝑥 1 1 ∫ = −∫ 𝑑𝑢 = −[𝑙𝑛|𝑢|] = −[𝑙𝑛( ) − 𝑙𝑛(1)] = 𝑙𝑛(4) 1 − 𝑥 𝑢 4 𝑢=1+𝑒 ⟹𝑑𝑢=𝑒𝑑𝑥 𝑖𝑓 𝑥=0 ⟹ 𝑢=1+𝑒0=2 & 𝑖𝑓 𝑥=ln3 ⟹ 𝑢=1+𝑒ln3=4 ln3 8 0 2 ∫4𝜋 1 𝜋2 𝑥 ∫ln3 𝑒 −ln3𝑒 +4𝑥 1 2𝑥 3 4 0 𝑥 4 1 2 2 16−4√2 .3 1 4 1 𝑥 3 4 2 2 1 4 1 2. PRACTICAL SET 1.5 HOMEWORK SET 1.4 Evaluate the definite integral. 1. ---------------------------------------------------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------------------------------------------------
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 16قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 2.1 HYPERBOLIC FUNCTIONS AND HANGING CABLES _________ =𝑥 −𝑥 Hyperboliccosine =𝑥 −𝑥 Hyperbolic tangent =GRAPHS OF THE HYPERBOLIC FUNCTIONS Hyperbolic secant ▪ DEFINITIONS OF HYPERBOLIC FUNCTIONS Definition. Hyperbolicsine 𝒙 −𝒙 Hyperbolic cosecant Hyperbolic cotangent 𝐬𝐢𝐧𝐡 𝒙 = tanh 𝑥 = cosh 𝑥 = = sech 𝑥 = csch 𝑥 = coth 𝑥 = 1 sinh 𝑥 1 cosh 𝑥 cosh 𝑥 sinh 𝑥 ▪ 2 𝑒−𝑒 2 𝑒+𝑒 𝑒+𝑒 𝑒−𝑒 𝒆−𝒆 𝟐 𝑒 +𝑒𝑥 2 sinh 𝑥 cosh 𝑥 𝑒−𝑒 𝑒+𝑒 𝑥 𝑥 −𝑥 −𝑥 −𝑥 𝑥 𝑥 −𝑥 −𝑥
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 17قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك ▪ ▪ HYPERBOLIC IDENTITIES cosh2 𝑥 − sinh2 𝑥 = 1 1−tanh2𝑥=sech2𝑥 coth2 𝑥 − 1 = csch2 𝑥 DERIVATIVE AND INTEGRALFORMULAS 𝟐.𝟏.𝟏𝑻𝒉𝒆𝒐𝒓𝒆𝒎. 𝑑 𝑑𝑥 𝑑𝑢 𝑑𝑥 [sinh𝑢]=cosh𝑢 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑𝑢 𝑑𝑥 [cosh𝑢] = sinh𝑢 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 [tanh𝑢]= 𝑠𝑒𝑐ℎ2 𝑢 [coth𝑢]=−𝑐𝑠𝑐ℎ2 𝑢 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑𝑢 𝑑𝑥 [sech𝑢] = −sech𝑢tanh𝑢 𝑑𝑢 𝑑𝑥 [𝑐𝑠𝑐ℎ 𝑢]=−𝑐𝑠𝑐ℎ 𝑢 𝑐𝑜𝑡ℎ 𝑢 ∫ 𝑐𝑠𝑐ℎ2 ∫ 𝑠𝑒𝑐ℎ2 ∫cosh𝑢𝑑𝑢=sinh𝑢+ 𝐶 ∫sinh𝑢𝑑𝑢=cosh𝑢+ 𝐶 𝑢𝑑𝑢 =tanh𝑢+ 𝐶 𝑢𝑑𝑢 = −coth𝑢+ 𝐶 ∫sech𝑢tanh𝑢𝑑𝑢= −sech𝑢+ 𝐶 ∫ 𝑐𝑠𝑐ℎ 𝑢 𝑐𝑜𝑡ℎ 𝑢𝑑𝑢 = −𝑐𝑠𝑐ℎ 𝑢 + 𝐶
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 18قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك ) Substituting Example 1 Example 2 Evaluate 𝑑 𝑑𝑥 ∫tanh𝑥𝑑𝑥=∫ ∫𝑠𝑖𝑛ℎ5 𝑥cosh𝑥𝑑𝑥 𝑢=sinh𝑥 ⟹ 𝑑𝑢=cosh𝑥𝑑𝑥 ∫𝑠𝑖𝑛ℎ5 𝑥𝑐𝑜𝑠ℎ 𝑥𝑑𝑥=∫𝑢5 [cosh(𝑥3 )]=sinh(𝑥3 [𝑙𝑛(𝑡𝑎𝑛ℎ 𝑥)] = 1𝑡𝑎𝑛ℎ 𝑥 𝑑 𝑑𝑥 )· [𝑥3]= 3𝑥2 ·𝑑 [𝑡𝑎𝑛ℎ 𝑥] = 𝑑𝑥 sinh(𝑥3 2 𝑢6 𝑠𝑖𝑛ℎ6𝑥 𝑑𝑢 = + 𝐶 = + 𝐶. 6 6𝑑𝑥=ln|cosh𝑥|+𝑐=ln(cosh𝑥)+𝑐 𝑢=cosh𝑥 𝑑𝑢=sinh𝑥𝑑𝑥 𝑑 𝑑𝑥 sinh𝑥 cosh𝑥 𝑠𝑒𝑐ℎ 𝑥 tanh𝑥 ▪ ▪
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 19قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك ▪ ▪ INVERSES OF HYPERBOLIC FUNCTIONS LOGARITHMIC FORMS OF INVERSE HYPERBOLIC FUNCTIONS 𝟐.𝟏.𝟐 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 .𝑇ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝𝑠 ℎ𝑜𝑙𝑑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑒𝑑 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠: 𝑠𝑖𝑛ℎ−1𝑥=ln(𝑥+√𝑥 +1)2 𝑐𝑜𝑠ℎ−1𝑥=ln(𝑥+√𝑥 −1)2 −1 1 1 + 𝑥 1 𝑥 + 1 𝑡𝑎𝑛ℎ 𝑥= ln( ) 𝑐𝑜𝑡ℎ−1𝑥= ln( ) 2 1−𝑥 2 𝑥−1 1 + √1 − 𝑥2 1 √1 + 𝑥2 𝑠𝑒𝑐ℎ−1𝑥=ln( ) 𝑐𝑠𝑐ℎ−1𝑥=ln( + 𝑥 𝑥 | ) 𝑥|
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 20قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك Example 3 𝑑 𝑑𝑥 (𝑠𝑖𝑛ℎ 𝑑 (𝑐𝑜𝑠ℎ 𝑑𝑥 𝑑 (𝑡𝑎𝑛ℎ 𝑑𝑥 𝑠𝑖𝑛ℎ 𝑡𝑎𝑛ℎ 1 𝑑𝑢 𝑢)= √1 + 𝑢 𝑑𝑥2 1 𝑑𝑢 √𝑢 − 1 𝑑𝑥2 𝑢)= ,𝑢>1 1 𝑑𝑢 𝑢)= 1− 2 ,|𝑢|<1 𝑢𝑑𝑥 ( ) = ln( ) = ln3≈0.5493 1=ln(1+√1 +1)=ln(1+√2) ≈0.88142 𝐼𝑓 𝑎≥0,𝑡ℎ𝑒𝑛 𝑑𝑢 =sinh−1𝑢 ∫ ()+𝑐 𝑜𝑟 ln(𝑢+√𝑎 +𝑢 )+𝑐2 2 √𝑎 + 𝑢2 2 𝑎 𝑑𝑢 ∫ =cosh−1𝑢()+𝑐 𝑜𝑟 ln(𝑢+√𝑢 −𝑎 )+𝑐,𝑢>𝑎2 2 √𝑢 − 𝑎2 2 𝑎 1 𝑢h−1 𝑑𝑢 tan ( )+ 𝑐,|𝑢| < 𝑎 ∫ 𝑎 1 𝑎+𝑢 2 ={𝑎2 1 𝑎 𝑢 𝑎 or ln( )+𝑐,|𝑢|≠𝑎 𝑎−𝑢 coth−1 ( ) + 𝑐, |𝑢| > 𝑎 2𝑎 𝑎 − 𝑢 𝑑𝑢 1 𝑎 𝑢 𝑎 1 𝑎 + √𝑎 − 𝑢2 2 ∫ =− sech−1||+𝑐 𝑜𝑟 − ln( )+𝑐,0< |𝑢| +𝑐,𝑢≠0 | <𝑎 𝑢√𝑎 −𝑢2 2 𝑎 1 𝑎 + √𝑎 + 𝑢2 2 𝑢| 𝑑𝑢 1 𝑎 𝑢 𝑎 ∫ =− csch−1||+𝑐 𝑜𝑟 − ln(𝑢√𝑎 + 𝑢2 2 𝑎 |𝑢| ) 𝑑 1 𝑑𝑢 1 − 𝑢 𝑑𝑥2 (𝑐𝑜𝑡ℎ−1𝑢) = , |𝑢| > 1 𝑑𝑥 𝑑 1 𝑑𝑢 𝑢√1 − 𝑢 𝑑𝑥2 (𝑠𝑒𝑐ℎ−1𝑢)=− ,0<𝑢<1 𝑑𝑥 𝑑 1 𝑑𝑢 |𝑢|√1 + 𝑢 𝑑𝑥2 (𝑐𝑠𝑐ℎ−1𝑢)=− .𝑢≠0 𝑑𝑥 −1 −1 −1 −1 −1 1 2 1 2 1+ 1− 1 2 1 2 1 2 𝟐. 𝟏. 𝟑 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 . 𝟐. 𝟏. 𝟒 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 .
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 21قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك a. b. PRACTICAL SET 2.1 Example 4 Evaluate Solution. Let HOMEWORK SET 2.1 Evaluate the integrals Evaluate the integrals and ∫ = ∫ ∫ 1. ∫coth2𝑥csch2𝑥𝑑𝑥 2. ∫cosh(2𝑥−3)𝑑𝑥 Find 𝑑𝑦. 𝑑𝑥 𝑦=sinh−1(1) 𝑥 1. ∫sinh6𝑥cosh𝑥𝑑𝑥 1⁄2 𝑑𝑥 2. ∫ 1−𝑥2 0 Find 𝑑𝑦. 𝑑𝑥 𝑦=sinh(4𝑥−8) , 𝑢=2𝑥 ⟹ 𝑑𝑢=2𝑑𝑥 ⟹𝑑𝑥= 2 = ∫ 𝑥 > 2 = cosh ( )+ 𝐶 = cosh ) + 𝐶 𝑑𝑥 √4𝑥 −92 √𝑢−3 𝑑𝑥 2 √4𝑥−9 1 2 3 2 𝑑𝑢 2 𝑑𝑢 √𝑢 −32 1 2 −1 𝑢 3 1 2 −1 (2𝑥3 𝑑𝑢 2 2 a. b. ---------------------------------------------------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------------------------------------------------
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 22قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك 2.2 Area Between Two Curves ______________________________________ Example 1. Find the area of region bounded above by on the sides by the lines Solution. 𝑥=0 & 𝑥=2. bounded below by and bounded 𝐴=∫[(𝑥+6)−𝑥2 0 4 8 12 + 72 − 16 68 34 = + 12 − = = = 2 3 6 6 3 𝑥2 𝑥 ] 𝑑 𝑥 = [ + 6 𝑥 − ] 2 3 0 𝑦=𝑥+6, 𝑦=𝑥2 , 𝐼𝑓 𝑤 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑛𝑑 𝑖𝑓 𝑤(𝑦)≥𝑣(𝑦).𝐹𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 [𝑐,𝑑],𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑏𝑦 𝑥=𝑣(𝑦),𝑜𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑏𝑦 𝑥=𝑤(𝑦), 𝑏𝑒𝑙𝑜𝑤 𝑏𝑦 𝑙𝑖𝑛𝑒 𝑦=𝑐,𝑎𝑛𝑑 𝑎𝑏𝑜𝑣𝑒 𝑏𝑦 𝑦=𝑑 𝑖𝑠. 𝑑 𝐴=∫[𝑤(𝑦)−𝑣(𝑦)]𝑑𝑦.𝑐 𝑎𝑛𝑑 𝑖𝑓 𝑓(𝑥)≥𝑔(𝑥) 𝐹𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 [𝑎,𝑏],𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑏𝑦 𝑦=𝑓(𝑥),𝑏𝑒𝑙𝑜𝑤 𝑏𝑦 𝑦=𝑔(𝑥), 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑏𝑦 𝑙𝑖𝑛𝑒 𝑥=𝑎,𝑎𝑛𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑏𝑦 𝑡ℎ𝑒 𝑥=𝑏 𝑖𝑠 𝑏 𝐴=∫[𝑓(𝑥)−𝑔(𝑥)]𝑑𝑥.𝑎 2 3 2 First Area Formula. SECOND AREA FORMULA. 𝐼𝑓 𝑓 𝑎𝑛𝑑 𝑔 𝑎𝑟𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑛 𝑎𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [𝑎,𝑏]
, ___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 23قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك HOMEWORK SET 2.2 Example 2 Find the area of the region enclosed by Solution. We rewrite the equation as So, 2 2 Sketch the region enclosed by the curves andfind its area. integrating with recpect to. 1. 2. 𝑦=𝑥2 , 𝑥=𝑠𝑖𝑛𝑦,3. 𝑦=𝑒𝑥 𝑦 = √𝑥, 𝑥=0, 𝑦 = 𝑒2𝑥, 𝑥=𝑦2 1 4 𝑥= , 𝑥=1 𝜋 3𝜋 4 𝑦= , 𝑦= 4 𝑥=0,𝑥=𝑙𝑛2 2 𝑦2 𝑦𝐴=∫[(𝑦+2)−𝑦2]𝑑𝑦=[ +2𝑦− ]−1 2 3 −1 4 8 3 1 2 1 8 3 1 2 1 3 =[( +4− )−( −2+ )]=8− − − 2 3 48 − 16 − 3− 2 27 9 = = = 6 6 2 & 𝑦=𝑥−2, 𝑦=𝑥−2 𝑥=𝑦+2 & 𝑥=𝑦2𝑦=𝑦+2 ⟹𝑦−𝑦−2=0⟹(𝑦−2)(𝑦+1)=0 ⟹𝑦=2 & 𝑦=−1 𝑦 3 2 ----------------------------------------------------------------------------------------------------------------------------------------
___________________________________________________________________________________________ EMAIL: DEP_MATH@UT.EDU.SA Math1201 24قـسم الرياضـيات. كلـية العلوـم. جامـعة تـبوك QUIZ 1 Fundamental ofIntegral Calculus Answer the following questions Math1201 1. 2. 3. 4. 5. 6. 7. 8. (b) (b) (b) (a)Find the area of the region enclosed by to 𝑥=𝑦2𝑦. (d) Find the area of the region enclosed by the curves. (c) (c) Serial Number……….. ……………………………………….………………. Student Name: …………………………………………………………………… Student Academic Number: ………………..……………………..…………. integrating with respect (𝑎) Let 𝜋 8 (a) 0 (a) 𝑥 2 4 Find الرقم المتسلسل اسم الطالب: الرقم االالكاديم:ي (𝑎) 5 𝑒 ∫tanh𝑥𝑑𝑥 (𝑎) ∫ 2𝑥(𝑥2 + 1) (𝑎) ∫2cos𝑥𝑑𝑥. 𝑑𝑥 [cosh(𝑥3 𝑑𝑥; ∫ sin2 2𝑥 cos 2𝑥 𝑑𝑥 )] 𝐹(𝑥) = ∫ √𝑡 + 9𝑑𝑡. (𝑎) 𝐹(4) (𝑏) 𝑥 = 𝑠𝑖𝑛𝑦, 𝑢=𝑥2+1. (𝑏) 𝑥=0, 𝐹′(4) ∫(3𝑥 + 𝑥2) 𝑑𝑥 ∫ cosh(2𝑥 − 3) 𝑑𝑥 [∫ 𝑡 (𝑏) ∫sin7𝑥𝑑𝑥. 𝑑𝑡] 𝜋 𝑦 = , 4 & 𝑦 = 𝑥 −2, (𝑐) [ln(tanh𝑥)] 3𝜋 𝑦 = 4 ∫ 𝑑𝑥 (𝑐)∫ 𝑥√1 + 𝑥𝑑𝑥 𝐹′′(4) (𝑏) ∫(2𝑥−5)(𝑥−3) 𝑑𝑥9 (𝑐) ∫ 𝑥2 √𝑥 − 1𝑑𝑥. ∫ln30 𝑑 𝑑𝑥 𝑥 23 𝑑 𝑑𝑥 𝑥 1 3 𝑑 𝑑𝑥 5 2 8 0 1 ∫⁄2 𝑑𝑥 0 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥2