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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 27 Evaluate the following integrals: (i) ∫_2^3β–’π‘₯^2 𝑑π‘₯ Step 1 ∫1β–’γ€–π‘₯^2 𝑑π‘₯γ€—=(π‘₯^(2 + 1) )/(2 + 1)=(π‘₯^3 )/3 Hence F(π‘₯)=π‘₯^3/3 Step 2 ∫_2^3β–’γ€–π‘₯^2 𝑑π‘₯γ€—=𝐹(3)βˆ’πΉ(2) =3^3/3βˆ’2^3/3=27/3βˆ’8/3=πŸπŸ—/πŸ‘ Example 27 Evaluate the following integrals: (ii) ∫_4^9β–’βˆšπ‘₯/((30 βˆ’ π‘₯^(3/2) )^2 ) 𝑑π‘₯ Step 1 :- ∫1β–’βˆšπ‘₯/(30 βˆ’ π‘₯^(3/2) )^2 𝑑π‘₯ Let 30βˆ’π‘₯^(3/2)=𝑑 Differentiating w.r.t. π‘₯ both sides 𝑑(30 βˆ’ π‘₯^(3/2) )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ βˆ’3/2 π‘₯^(3/2 βˆ’1)=𝑑𝑑/𝑑π‘₯ (βˆ’3)/2 π‘₯^(1/2 )=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’ 3/2 γ€– π‘₯γ€—^( 1/2 ) ) 𝑑π‘₯=(βˆ’2𝑑𝑑)/(3√π‘₯) Therefore, our equation becomes ∫1β–’γ€–(√π‘₯ 𝑑π‘₯ )/(30βˆ’π‘₯^( 3/2) )^2 =∫1β–’γ€–βˆšπ‘₯/𝑑^2 (βˆ’2 𝑑𝑑)/(3 √π‘₯)γ€—γ€— =(βˆ’2)/( 3) ∫1β–’( 𝑑𝑑)/𝑑^2 =(βˆ’2)/( 3) ∫1▒〖𝑑^(βˆ’2) 𝑑𝑑〗 =(βˆ’2)/( 3) 𝑑^(βˆ’ 2 + 1)/(βˆ’ 2 + 1) =(βˆ’2)/( 3) 𝑑^(βˆ’ 1)/(βˆ’1) =2/3 𝑑^(βˆ’1) =2/3𝑑 Putting 𝑑=(30βˆ’π‘₯^(3/2) ) =2/3(30 βˆ’ π‘₯^(3/2) ) Hence F(π‘₯)=2/3(30 βˆ’ π‘₯^(3/2) ) Step 2 :- ∫_4^9β–’βˆšπ‘₯/((30 βˆ’ π‘₯^( 3/2) ) ) 𝑑π‘₯=𝐹(9)βˆ’πΉ(4) =2/3(30 βˆ’ (9)^(3/2) ) βˆ’2/3(30 βˆ’ (4)^(3/2) ) = 2/(3 (30 βˆ’ (3^2 )^(2/3) ) )βˆ’2/(3 (30 βˆ’ (2^2 )^(2/3) ) ) =2/3 [1/(30 βˆ’ 3^3 )βˆ’1/(30 βˆ’ 2^3 )] =2/3 [1/(30 βˆ’ 27)βˆ’1/(30 βˆ’ 8)] =2/3 [1/3βˆ’1/22] =2/3 [(22 βˆ’ 3)/(3 Γ— 22)] =2/3 (19/66) =19/(3 (33)) =πŸπŸ—/πŸ—πŸ— Example 27 Evaluate the following integrals: (iii) ∫_1^2β–’(π‘₯ 𝑑π‘₯)/((π‘₯ + 1) (π‘₯ + 2) ) 𝑑π‘₯ 𝐹(π‘₯)=∫1β–’(π‘₯ 𝑑π‘₯)/(π‘₯ + 1)(π‘₯ + 2) We can write the integrate as : π‘₯/(π‘₯ + 1)(π‘₯ + 2) =A/(π‘₯ + 1)+B/(π‘₯ + 2) π‘₯/(π‘₯ + 1)(π‘₯ + 2) =(A(π‘₯ + 2) + B(π‘₯ + 1))/(π‘₯ + 1)(π‘₯ + 2) Canceling denominators π‘₯=A(π‘₯+2)+B(π‘₯+1) Putting 𝒙=βˆ’πŸ βˆ’2=A(2+2)+B(2+1) βˆ’2=A Γ—0+B(βˆ’1) βˆ’2=βˆ’B B=2 Putting 𝒙=βˆ’πŸ βˆ’1=A(βˆ’1+2)+B(βˆ’1+1) βˆ’1=A(1)+B(0) βˆ’1=A A=βˆ’1 Therefore π‘₯/(π‘₯ + 1)(π‘₯ + 2) =(βˆ’1)/(π‘₯ + 1)+2/(π‘₯ + 2) Integrating w.r.t.π‘₯ ∫1β–’γ€–π‘₯/(π‘₯+1)(π‘₯+2) =∫1β–’γ€–(βˆ’1)/((π‘₯+1) ) 𝑑π‘₯γ€—+∫1β–’γ€–2/(π‘₯+2) 𝑑π‘₯γ€—γ€— =βˆ’π‘™π‘œπ‘”|π‘₯+1|+2π‘™π‘œπ‘”|π‘₯+2| =βˆ’π‘™π‘œπ‘”|π‘₯+1|+π‘™π‘œπ‘”|π‘₯+2|^2 =π‘™π‘œπ‘”|π‘₯+2|^2βˆ’π‘™π‘œπ‘”|π‘₯+1| =π‘™π‘œπ‘”|(π‘₯ + 2)^2/(π‘₯ + 1)| Hence 𝐹(π‘₯)=π‘™π‘œπ‘”|(π‘₯ + 2)^2/(π‘₯ + 1)| Now, ∫_1^2β–’γ€–π‘₯/(π‘₯ + 1)(π‘₯ + 2) 𝑑π‘₯γ€—=𝐹(2)βˆ’πΉ(1) ∫_1^2β–’γ€–π‘₯/(π‘₯ + 1)(π‘₯ + 2) 𝑑π‘₯γ€—=π‘™π‘œπ‘”|(2 + 2)^2/(2 + 1)|βˆ’π‘™π‘œπ‘”|(1 + 2)^2/(1 + 1)| (𝑏 logβ‘γ€–π‘Ž=logβ‘γ€–π‘Ž^𝑏 γ€— γ€— ) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡) =π‘™π‘œπ‘”|(4)^2/3|βˆ’π‘™π‘œπ‘”|(3)^2/2| =π‘™π‘œπ‘”|(4^2/3)/(3^2/2)| =π‘™π‘œπ‘”|4^2/3 Γ— 2/3^2 | =π‘™π‘œπ‘”|16/3 Γ— 2/9| =π‘™π‘œπ‘”|32/27 | =π₯𝐨𝐠⁑(πŸ‘πŸ/πŸπŸ•) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡) Example 27 Evaluate the following integrals: (iv) ∫_0^(πœ‹/4)β–’γ€–sin^3⁑2𝑑 cos⁑2 𝑑〗 𝑑𝑑 Let F(π‘₯)=∫1▒〖𝑠𝑖𝑛^3 2𝑑 π‘π‘œπ‘  2𝑑 𝑑𝑑〗 Let s𝑖𝑛 2𝑑=𝑒 Differentiating w.r.t.π‘₯ (𝑑(sin⁑2𝑑))/𝑑𝑑=𝑑𝑒/𝑑𝑑 2cπ‘œπ‘  2𝑑 =𝑑𝑒/𝑑𝑑 𝑑𝑑=𝑑𝑒/(2 π‘π‘œπ‘  2𝑑) Putting value of u and du in our integral ∫1▒〖𝑠𝑖𝑛^3 2𝑑 π‘π‘œπ‘  2𝑑 𝑑𝑑〗=∫1▒〖𝑒^3 π‘π‘œπ‘  2𝑑 Γ— 𝑑𝑒/(2 π‘π‘œπ‘  2𝑑)γ€— =1/2 ∫1▒〖𝑒^3 𝑑𝑒〗 =1/2 𝑒^(3+1)/(3+1)=1/2 𝑒^4/4= 𝑒^4/8 Putting back 𝑒=𝑠𝑖𝑛 2𝑑 =1/8 𝑠𝑖𝑛^4 2𝑑 Hence, F(𝑑)=1/8 𝑠𝑖𝑛^4 2𝑑 Now, ∫_0^(πœ‹/4)▒〖𝑠𝑖𝑛^3 2𝑑 π‘π‘œπ‘  2𝑑=𝐹(πœ‹/4)βˆ’πΉ(0) γ€— =1/8 𝑠𝑖𝑛^4 2(πœ‹/4)βˆ’1/8 𝑠𝑖𝑛^4 2(0) =1/8 𝑠𝑖𝑛^4 πœ‹/2βˆ’1/8 𝑠𝑖𝑛^4 (0) =1/8 Γ—1^4βˆ’1/8 Γ—0^4 =1/8 Γ—1βˆ’0 =𝟏/πŸ–

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.