Example 27 - Chapter 7 Class 12 Integrals

Transcript

Example 27 Evaluate the following integrals: (i) ∫_2^3▒𝑥^2 𝑑𝑥 Step 1 ∫1▒〖𝑥^2 𝑑𝑥〗=(𝑥^(2 + 1) )/(2 + 1)=(𝑥^3 )/3 Hence F(𝑥)=𝑥^3/3 Step 2 ∫_2^3▒〖𝑥^2 𝑑𝑥〗=𝐹(3)−𝐹(2) =3^3/3−2^3/3=27/3−8/3=𝟏𝟗/𝟑 Example 27 Evaluate the following integrals: (ii) ∫_4^9▒√𝑥/((30 − 𝑥^(3/2) )^2 ) 𝑑𝑥 Step 1 :- ∫1▒√𝑥/(30 − 𝑥^(3/2) )^2 𝑑𝑥 Let 30−𝑥^(3/2)=𝑡 Differentiating w.r.t. 𝑥 both sides 𝑑(30 − 𝑥^(3/2) )/𝑑𝑥=𝑑𝑡/𝑑𝑥 −3/2 𝑥^(3/2 −1)=𝑑𝑡/𝑑𝑥 (−3)/2 𝑥^(1/2 )=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(− 3/2 〖 𝑥〗^( 1/2 ) ) 𝑑𝑥=(−2𝑑𝑡)/(3√𝑥) Therefore, our equation becomes ∫1▒〖(√𝑥 𝑑𝑥 )/(30−𝑥^( 3/2) )^2 =∫1▒〖√𝑥/𝑡^2 (−2 𝑑𝑡)/(3 √𝑥)〗〗 =(−2)/( 3) ∫1▒( 𝑑𝑡)/𝑡^2 =(−2)/( 3) ∫1▒〖𝑡^(−2) 𝑑𝑡〗 =(−2)/( 3) 𝑡^(− 2 + 1)/(− 2 + 1) =(−2)/( 3) 𝑡^(− 1)/(−1) =2/3 𝑡^(−1) =2/3𝑡 Putting 𝑡=(30−𝑥^(3/2) ) =2/3(30 − 𝑥^(3/2) ) Hence F(𝑥)=2/3(30 − 𝑥^(3/2) ) Step 2 :- ∫_4^9▒√𝑥/((30 − 𝑥^( 3/2) ) ) 𝑑𝑥=𝐹(9)−𝐹(4) =2/3(30 − (9)^(3/2) ) −2/3(30 − (4)^(3/2) ) = 2/(3 (30 − (3^2 )^(2/3) ) )−2/(3 (30 − (2^2 )^(2/3) ) ) =2/3 [1/(30 − 3^3 )−1/(30 − 2^3 )] =2/3 [1/(30 − 27)−1/(30 − 8)] =2/3 [1/3−1/22] =2/3 [(22 − 3)/(3 × 22)] =2/3 (19/66) =19/(3 (33)) =𝟏𝟗/𝟗𝟗 Example 27 Evaluate the following integrals: (iii) ∫_1^2▒(𝑥 𝑑𝑥)/((𝑥 + 1) (𝑥 + 2) ) 𝑑𝑥 𝐹(𝑥)=∫1▒(𝑥 𝑑𝑥)/(𝑥 + 1)(𝑥 + 2) We can write the integrate as : 𝑥/(𝑥 + 1)(𝑥 + 2) =A/(𝑥 + 1)+B/(𝑥 + 2) 𝑥/(𝑥 + 1)(𝑥 + 2) =(A(𝑥 + 2) + B(𝑥 + 1))/(𝑥 + 1)(𝑥 + 2) Canceling denominators 𝑥=A(𝑥+2)+B(𝑥+1) Putting 𝒙=−𝟐 −2=A(2+2)+B(2+1) −2=A ×0+B(−1) −2=−B B=2 Putting 𝒙=−𝟏 −1=A(−1+2)+B(−1+1) −1=A(1)+B(0) −1=A A=−1 Therefore 𝑥/(𝑥 + 1)(𝑥 + 2) =(−1)/(𝑥 + 1)+2/(𝑥 + 2) Integrating w.r.t.𝑥 ∫1▒〖𝑥/(𝑥+1)(𝑥+2) =∫1▒〖(−1)/((𝑥+1) ) 𝑑𝑥〗+∫1▒〖2/(𝑥+2) 𝑑𝑥〗〗 =−𝑙𝑜𝑔|𝑥+1|+2𝑙𝑜𝑔|𝑥+2| =−𝑙𝑜𝑔|𝑥+1|+𝑙𝑜𝑔|𝑥+2|^2 =𝑙𝑜𝑔|𝑥+2|^2−𝑙𝑜𝑔|𝑥+1| =𝑙𝑜𝑔|(𝑥 + 2)^2/(𝑥 + 1)| Hence 𝐹(𝑥)=𝑙𝑜𝑔|(𝑥 + 2)^2/(𝑥 + 1)| Now, ∫_1^2▒〖𝑥/(𝑥 + 1)(𝑥 + 2) 𝑑𝑥〗=𝐹(2)−𝐹(1) ∫_1^2▒〖𝑥/(𝑥 + 1)(𝑥 + 2) 𝑑𝑥〗=𝑙𝑜𝑔|(2 + 2)^2/(2 + 1)|−𝑙𝑜𝑔|(1 + 2)^2/(1 + 1)| (𝑏 log⁡〖𝑎=log⁡〖𝑎^𝑏 〗 〗 ) (log⁡A−log⁡B=𝑙𝑜𝑔 A/𝐵) =𝑙𝑜𝑔|(4)^2/3|−𝑙𝑜𝑔|(3)^2/2| =𝑙𝑜𝑔|(4^2/3)/(3^2/2)| =𝑙𝑜𝑔|4^2/3 × 2/3^2 | =𝑙𝑜𝑔|16/3 × 2/9| =𝑙𝑜𝑔|32/27 | =𝐥𝐨𝐠⁡(𝟑𝟐/𝟐𝟕) (log⁡A−log⁡B=𝑙𝑜𝑔 A/𝐵) (log⁡A−log⁡B=𝑙𝑜𝑔 A/𝐵) Example 27 Evaluate the following integrals: (iv) ∫_0^(𝜋/4)▒〖sin^3⁡2𝑡 cos⁡2 𝑡〗 𝑑𝑡 Let F(𝑥)=∫1▒〖𝑠𝑖𝑛^3 2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡〗 Let s𝑖𝑛 2𝑡=𝑢 Differentiating w.r.t.𝑥 (𝑑(sin⁡2𝑡))/𝑑𝑡=𝑑𝑢/𝑑𝑡 2c𝑜𝑠 2𝑡 =𝑑𝑢/𝑑𝑡 𝑑𝑡=𝑑𝑢/(2 𝑐𝑜𝑠 2𝑡) Putting value of u and du in our integral ∫1▒〖𝑠𝑖𝑛^3 2𝑡 𝑐𝑜𝑠 2𝑡 𝑑𝑡〗=∫1▒〖𝑢^3 𝑐𝑜𝑠 2𝑡 × 𝑑𝑢/(2 𝑐𝑜𝑠 2𝑡)〗 =1/2 ∫1▒〖𝑢^3 𝑑𝑢〗 =1/2 𝑢^(3+1)/(3+1)=1/2 𝑢^4/4= 𝑢^4/8 Putting back 𝑢=𝑠𝑖𝑛 2𝑡 =1/8 𝑠𝑖𝑛^4 2𝑡 Hence, F(𝑡)=1/8 𝑠𝑖𝑛^4 2𝑡 Now, ∫_0^(𝜋/4)▒〖𝑠𝑖𝑛^3 2𝑡 𝑐𝑜𝑠 2𝑡=𝐹(𝜋/4)−𝐹(0) 〗 =1/8 𝑠𝑖𝑛^4 2(𝜋/4)−1/8 𝑠𝑖𝑛^4 2(0) =1/8 𝑠𝑖𝑛^4 𝜋/2−1/8 𝑠𝑖𝑛^4 (0) =1/8 ×1^4−1/8 ×0^4 =1/8 ×1−0 =𝟏/𝟖