∫ e x Β  [f(x) + fβ€²(x)] dx

Β  = ∫ e x Β  f(x)dx + ∫ e x fβ€²(x) dx

Β  Using integration by parts

Β  = f(x) ∫ e x Β  dx - ∫ fβ€² (x) e x Β  dx + ∫ fβ€² (x) e x Β  dx

Β  = f(x) e x ∫ fβ€² (x) e x dx + ∫ fβ€² (x) e x dx

Β  = e x Β  f(x) + C

Based on the above information, answer any four of the following questions.

Β 

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Question 1

∫e x   (sin⁑ x + cos ⁑x) dx = ______________.

(a) e x   cos⁑x + c  

(b) e x   sin⁑x + c  

(c) e x + c Β 

(d) e x (-cos⁑x + sin⁑x ) + c  

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Question 2

∫e x   (x - 1)/x 2 )  dx =______________.

(a) e x + c Β 

(b) e x /x + c Β 

(c) e x /x 2 + c Β 

(d) -e x /x 2 + c

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Question 3

∫e x   (1 + x) dx =______________.

(a) xe x + cΒ 

(b) e x + c Β 

(c) e -x + c Β 

(d) none of these

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Question 4

∫ Ο€ 0 e x (tan⁑x + sec 2 ⁑x) 𝑑π‘₯ = _________.

(a) 0 Β 

(b) 1Β 

(c) -1Β 

(d) -e Ο€

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Question 5

∫xe x /(1 + x) 2 dx =______________.

(a) xe x + cΒ 

(b) e x /(x + 1) 2 + c Β 

(c) x e x /x + 1 + cΒ 

(d) e x /x + 1 + cΒ 

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  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Question 2 ∫1▒〖𝑒^π‘₯ [𝑓(π‘₯)+𝑓^β€² (π‘₯)] γ€— 𝑑π‘₯ = ∫1▒〖𝒆^𝒙 𝒇(𝒙)𝒅𝒙〗+∫1▒〖𝑒^π‘₯ 𝑓′(π‘₯)𝑑π‘₯γ€— Using integration by parts = 𝒇(𝒙) ∫1▒𝒆^𝒙 π’…π’™βˆ’βˆ«1▒〖𝒇^β€² (𝒙) 𝒆^𝒙 γ€— 𝒅𝒙 + ∫1▒〖𝑓^β€² (π‘₯) 𝑒^π‘₯ γ€— 𝑑π‘₯ = 𝑓(π‘₯) 𝑒^π‘₯βˆ’βˆ«1▒〖𝑓^β€² (π‘₯) 𝑒^π‘₯ γ€— 𝑑π‘₯ + ∫1▒〖𝑓^β€² (π‘₯) 𝑒^π‘₯ γ€— 𝑑π‘₯ = 𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Based on the above information, answer any four of the following questions. Question 1 ∫1▒〖𝑒^π‘₯ (sin⁑〖π‘₯ γ€—+cos⁑π‘₯)γ€— 𝑑π‘₯=______________. (a) 𝑒^π‘₯ cos⁑π‘₯+𝑐 (b) 𝑒^π‘₯ sin⁑π‘₯+𝑐 (c) 𝑒^π‘₯+𝑐 (d) 𝑒^π‘₯ (βˆ’cos⁑π‘₯+sin⁑π‘₯ )+c ∫1▒〖𝑒^π‘₯ (sin⁑〖π‘₯ γ€—+cos⁑π‘₯)γ€— 𝑑π‘₯ Putting 𝒇(𝒙)=π’”π’Šπ’ 𝒙" " ∴ 𝒇^β€² (𝒙)=cos⁑π‘₯ Thus, ∫1▒〖𝑒^π‘₯ (sin⁑〖π‘₯ γ€—+cos⁑π‘₯)γ€— 𝑑π‘₯=𝒆^𝒙 π’”π’Šπ’ 𝒙+π‘ͺ So, the correct answer is (b) Question 2 ∫1▒〖𝑒^π‘₯ ((π‘₯ βˆ’ 1)/π‘₯^2 ) γ€— 𝑑π‘₯=______________. (a) 𝑒^π‘₯+𝑐 (b) 𝑒^π‘₯/π‘₯+𝑐 (c) 𝑒^π‘₯/π‘₯^2 +𝑐 (d) γ€–βˆ’π‘’γ€—^π‘₯/π‘₯^2 +𝑐 ∫1▒〖𝑒^π‘₯ ((π‘₯ βˆ’ 1)/π‘₯^2 ) γ€— 𝑑π‘₯= ∫1▒〖𝑒π‘₯" " (π‘₯/π‘₯^2 βˆ’ 1/π‘₯2) 𝑑π‘₯γ€— = ∫1▒〖𝒆𝒙" " (𝟏/𝒙 βˆ’ 𝟏/π’™πŸ) 𝒅𝒙〗 Putting 𝒇(𝒙)=𝟏/𝒙 ∴ 𝒇^β€² (𝒙)=(βˆ’1)/π‘₯^2 Thus, ∫1▒〖𝑒π‘₯" " (1/π‘₯ βˆ’ 1/π‘₯2) 𝑑π‘₯γ€— = 𝒆^𝒙/𝒙+π‘ͺ So, the correct answer is (b) Question 3 ∫1▒〖𝑒^π‘₯ (1+π‘₯) γ€— 𝑑π‘₯=______________. (a) π‘₯𝑒^π‘₯+𝑐 (b) 𝑒^π‘₯+𝑐 (c) 𝑒^(βˆ’π‘₯)+𝑐 (d) none of these ∫1▒〖𝑒^π‘₯ (1+π‘₯)γ€— 𝑑π‘₯=∫1▒〖𝒆^𝒙 (𝒙+𝟏)γ€— 𝒅𝒙 Putting 𝒇(𝒙)=𝒙 ∴ 𝒇^β€² (𝒙)=1 Thus, ∫1▒〖𝑒^π‘₯ (π‘₯+1)γ€— 𝑑π‘₯=𝒆^𝒙 Γ— 𝒙+π‘ͺ So, the correct answer is (a) Question 4 ∫1_0^πœ‹β–’π‘’^π‘₯ (tan⁑π‘₯+sec^2⁑π‘₯) 𝑑π‘₯ = _________. (a) 0 (b) 1 (c) βˆ’1 (d) βˆ’π‘’^πœ‹ ∫1_0^πœ‹β–’π‘’^π‘₯ (tan⁑π‘₯+sec^2⁑π‘₯) 𝑑π‘₯ Putting 𝒇(𝒙)=𝒕𝒂𝒏 𝒙" " ∴ 𝒇^β€² (𝒙)=〖𝑠𝑒𝑐〗^2⁑π‘₯ Thus, ∫1_0^πœ‹β–’π‘’^π‘₯ (tan⁑π‘₯+sec^2⁑π‘₯) 𝑑π‘₯ = γ€–[𝒆^𝒙 𝒕𝒂𝒏 𝒙]γ€—_0^πœ‹ =[𝒆^𝒙 𝒕𝒂𝒏 πœ‹βˆ’π’†^𝒙 𝒕𝒂𝒏 𝟎] =[𝒆^𝒙 𝒕𝒂𝒏 πœ‹βˆ’π’†^𝒙 𝒕𝒂𝒏 𝟎] =[𝒆^𝒙 Γ— πŸŽβˆ’π’†^𝒙 Γ— 𝟎] =0βˆ’0 =0 So, the correct answer is (a) Question 5 ∫1β–’γ€–γ€–π‘₯𝑒〗^π‘₯/(1 + π‘₯)^2 γ€— 𝑑π‘₯=______________. (a) π‘₯𝑒^π‘₯+𝑐 (b) 𝑒^π‘₯/(π‘₯ + 1)^2 +𝑐 (c) (π‘₯ 𝑒^π‘₯)/(π‘₯ + 1)+𝑐 (d) 𝑒^π‘₯/(π‘₯ + 1)+𝑐 ∫1▒〖𝑒^π‘₯ ((π‘₯ )/(1 + π‘₯)^2 ) γ€— 𝑑π‘₯= ∫1▒〖𝑒π‘₯" " ((1 + π‘₯ βˆ’ 1 )/(1 + π‘₯)^2 ) 𝑑π‘₯γ€— =" " ∫1▒〖𝑒π‘₯" " ((1 + π‘₯ )/(1 + π‘₯)^2 βˆ’1/(1 + π‘₯)^2 ) 𝑑π‘₯γ€— =" " ∫1▒〖𝒆𝒙" " (𝟏/((𝟏 + 𝒙))βˆ’πŸ/(𝟏 + 𝒙)^𝟐 ) 𝒅𝒙〗 It is of form ∫1▒〖𝒆^𝒙 (𝒇(𝒙)+𝒇^β€² (𝒙)) γ€— 𝒅𝒙=𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Putting 𝒇(𝒙)=𝟏/((𝟏 + 𝒙)) ∴ 𝒇^β€² (𝒙)=(βˆ’1)/(𝟏 + 𝒙)^2 Thus, ∫1▒〖𝑒π‘₯" " (1/((1 + π‘₯))βˆ’1/(1 + π‘₯)^2 ) 𝑑π‘₯γ€— = 𝒆^𝒙/((𝟏 + 𝒙))+π‘ͺ So, the correct answer is (d)

Chapter 7 Class 12 Integrals (Term 2)
Serial order wise

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.