Chapter 7 Class 12 Integrals
Serial order wise

∫ e x   [f(x) + f′(x)] dx

  = e x   f(x)dx + ∫ e x f′(x) dx

  Using integration by parts

  = f(x) ∫ e x   dx - ∫ f′ (x) e x   dx + ∫ f′ (x) e x   dx

  = f(x) e x ∫ f′ (x) e x dx + ∫ f′ (x) e x dx

  = e x   f(x) + C

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

 

Slide11.JPG

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  

(d) e x (-cos⁡x + sin⁡x ) + c  

Slide12.JPG

Question 2

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

(a) e x + c  

(b) e x /x + c  

(c) e x /x 2 + c  

(d) -e x /x 2 + c

Slide13.JPG Slide14.JPG

Question 3

∫e x   (1 + x) dx =______________.

(a) xe x + c 

(b) e x + c  

(c) e -x + c  

(d) none of these

Slide15.JPG

Question 4

π 0 e x (tan⁡x + sec 2 ⁡x) 𝑑π‘₯ = _________.

(a) 0  

(b) 1 

(c) -1 

(d) -e π

Slide16.JPG Slide17.JPG

Question 5

∫xe x /(1 + x) 2 dx =______________.

(a) xe x + c 

(b) e x /(x + 1) 2 + c  

(c) x e x /x + 1 + c 

(d) e x /x + 1 + c 

Slide18.JPG Slide19.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.