Shazia Abbas

Assignment 1 Evaluate the integration: Question: ≠ 𝟏 𝟏 𝒙𝟑 − 𝒙 + 𝟐𝟎 ∫ 𝒅𝒙 𝟑 − 𝟓𝒙 𝒙 𝟎 Solution: 1 =∫0 [1 + 1 =∫0 1𝑑𝑥 −6𝑥+20 𝑥 3 +5𝑥 + ]dx 1 −6𝑥+20 ∫0 𝑥 3 +5𝑥 dx 1 −6𝑥+20 = |𝑥 |10 +∫0 𝑥 3 +5𝑥 dx 1 −6𝑥+20 =[1 − 0]+∫0 𝑥 3 +5𝑥 1 −6𝑥+20 I=1+∫0 𝑥 3 +5𝑥 dx dx→ (𝐴) We resolve it into partial fractions −6𝑥+20 −6𝑥+20 = 𝑥 3 +5𝑥 𝑥(𝑥 2 +5) −6𝑥+20 𝐴 𝐵𝑥+𝐶 = + 𝑥(𝑥 2 +5) 𝑥 𝑥 2 +5 → (𝐵 ) Multiplying both sides by x(𝑥 2 + 5) −6x+20=A(𝑥 2 + 5)+(𝐵𝑥 + 𝑐 )x→ (𝐶 ) For A, put x=0 in (𝐶 ) 20=5A→ 𝑨=𝟒 Now we find values of B and C From (𝐶 ) −6x+20=A𝑥 2 +5A+B𝑥 2 +Cx Equating coefficients of 𝑥 2 and x 𝑥 2 , 𝐴 + 𝐵 = 0 → (𝑖 ) X, C=−6→ (𝑖𝑖 ) Put A=4 B=−4 So, −6𝑥+20 4 −4𝑥−6 = + 𝑥(𝑥 2 +5) 𝑥 𝑥 2 +5 Put in (𝐴) in (𝑖 ) C=−𝟔 1 4 4𝑥+6 𝑥 𝑥 2 +5 I=1+∫0 [ − 14 =1+∫0 𝑥 − ]dx 1 4𝑥+6 ∫0 [𝑥 2 +5]dx 1 2(2𝑥+3) =1+4|ln 𝑥 |10 − ∫0 dx 𝑥 2 +5 1 2𝑥+3 =1+4[ln(1) − 𝑙𝑛(0)] − 2 ∫0 1 2𝑥 =1+4[0 − (−∞)] − 2 ∫0 2 =1−2 ln|𝑥 + 5|10 − 𝑥 2 +5 dx 𝑥 2 +5 √5 3 𝑥 2 +(√5) 1 √5 √5 =1−2[ln(6) − 𝑙𝑛(5)] − 6[10.776 − 0]+C =1−2[0.0792] − 6[10.776]+C 𝐼 ′ =63.8144 1 tan−1 ( ) − 0 I+C=63.8144 dx 𝑥 1 −1 6 | 𝑡𝑎𝑛 ( )| +C √5 √5 0 tan−1 ( )]+C =−63.8144 + 𝐶 2 1 =1−2[ln(1 + 5) − 𝑙𝑛(0 + 5)] − 6 [ 1 1 dx−2 ∫2 √5 Question:≠ 𝟐 ∫ √𝟑 + 𝟐𝒙 − 𝒙𝟐 dx Solution: =∫ √3 + 2𝑥 − 𝑥 2 dx =∫ √𝑥 2 − 2𝑥 − 3 𝑑𝑥 =∫ √𝑥 2 − 2𝑥 + 1 − 2 𝑑𝑥 ∵ (𝑎 − 𝑏)2 =𝑎2 − 2𝑎𝑏 + 𝑏 2 2 2 =∫ √(𝑥 − 1)2 − (√2) dx ∵ ∫ √𝑢2 (𝑥−1) = 2 − ∵ (√2) =2 𝑢 𝑎2 du= √𝑢2 2 − 2 𝑎2 √(𝑥 − 1)2 − (√2) − 2 √(𝑥 − 1)2 − (√2) |+C − 𝑎2 2 ln|𝑢 + √𝑢2 − 𝑎2 |+C 2 (√2) 2 ln |(𝑥 − 1) +