1 Rahman Sir DEFINITE INTEGRAL Properties of definite integral π π 1. Property I. ∫π π(π)π π = ∫π π(π)π π 2 Property II. ∫π π(π)π π = − ∫π π(π)π π π π π π π 3. Property III. ∫π π(π)π π = ∫π π(π)π π + ∫π π(π)π π. πππππ π < π < π π 4. π Property IV. ∫π π(π)π π = ∫π π(π + π − π)π x π π Special case ∫π π(π)π π = ∫π π(π − π)π π 5. π Property V. ∫−π π(π)π π = π. ππ π(π)ππ ππ π ππππππππ π = 2∫π π(π)π π if f(x) is even function π π π Proof : ∫−π π(π)π π = ∫−π π(π)π π + ∫π π(π)π π Let π = −π, ππππ π π = −π π πππππ, π = −π, π = π π π Now ∫–π π(π)π π = ∫π π(−π)(−π π) π = − ∫ π(−π)π π π π ∫π π(−π)π π by property II π = ∫π π(−π)π π by property I π π π … (1) π»πππ , ∫−π π(π)π π = ∫π π(π)π π + ∫π π(−π)π π This concept is used when f(x) is neither odd nor even. Case I. when f(x) is odd function, then f(-x) = -f(x) π π π From (1) ∫−π π(π)π π = ∫π π(π)π π + ∫π π(−π)π π π π π π = ∫π π(π)π π − ∫π π(π)π π as π(−π) = −π(π) =0 Case II. when f(x) is even function, then f(-x) = f(x) π π π From (1) ∫−π π(π)π π = ∫π π(π)π π + ∫π π(−π)π π = ∫π π(π)π π + ∫π π(π)π π as π(−π) = π(π) π ππ = 2 ∫π π(π)π π π 6. Property VI: ∫π π(π)π π = π ∫π π(π) π π ππ π(ππ − π) = π(π) = 0 if π(ππ − π) = −π(π) π π π Type I. Problems based on the property : ∫π π(π₯) ππ₯ = ∫π π(π₯) ππ₯ + ∫π π(π₯) This property is used when (i) Integrand f(x) is different in different part of αΎπ, παΏ in which the function is to be integrated. 2 Rahman Sir DEFINITE INTEGRAL (ii) (iii) Integrand f(x) is a modulus function for this put modulus part equal to zero and find the values of x which is between lower and upper limits. Integrand f(x) is Greatest Integer function. 1 − 2π₯, π₯ ≤ 0 1 Evaluate ∫−1 π(π₯) ππ₯ π€βπππ π(π₯) = ΰ΅ 1 + 2π₯, π₯ ≥ 0 1 0 1 ∫−1 π(π₯) ππ₯ = ∫−1(1 − 2π₯) ππ₯ + ∫0 (1 + 2π₯) ππ₯ = αΎπ₯ − π₯ 2 αΏ0−1 + αΎπ₯ + π₯ 2 αΏ10 = αΎ(0) − (−1 − 1)αΏ + αΎ1 + 1αΏ= 2 + 2 = 4 3 2 Find ∫0 Θπ₯πππ ππ₯Θ ππ₯ Solution: Put π₯πππ ππ₯ = 0 ⇒ π₯ = 0 ππ πππ ππ₯ = 0 π₯ = 0 ππ ππ₯ = ππ π₯ = (2π+1) 2 (2π+1) π₯= π where n = 0,1, 2 ,3.. 2 1 3 ; π€βππ π = 0, πππ π₯ = π€βππ π = 1 2 2 1 2 3 2 1 2 I= ∫0 π₯πππ ππ₯ ππ₯ - ∫ π₯πππ ππ₯ ππ₯ 1 Now, ∫ π₯πππ ππ₯ ππ₯ = π₯ ∫ πππ ππ₯ ππ₯ − π ∫ π ππππ₯ ππ₯ π₯ 1 = π π ππππ₯ + π2 πππ ππ₯ 1 2 3 2 1 2 Hence, I= ∫0 π₯πππ ππ₯ ππ₯ - ∫ π₯πππ ππ₯ ππ₯ π₯ 1 = απ π ππππ₯ + π2 πππ ππ₯α 1 1 1 2 π₯ − απ π ππππ₯ + π2 πππ ππ₯α 0 3 1 α2π − π2 α − α2π (−1) − 2πα = 5 1 1+1+3 2π 1 = 2π − π2 3 2 ∫0 αΎπ₯αΏ ππ₯ where αΎπ₯αΏ is greatest integer function. Now, f(x) = 0. When π₯ ∈ (0,1) 1 − π2 3 2 1 2 3 Rahman Sir DEFINITE INTEGRAL 3 f(x) = 1. When π₯ ∈ (1, 2 ) 3 2 3 1 3 3 1 ∫0 αΎπ₯αΏ ππ₯ = ∫0 0 ππ₯ + ∫12 1 ππ₯ = 0 + αΎπ₯αΏ12 = 2 − 1 = 2 π π₯ π₯ Example: The integral of ∫0 ΰΆ§1 + 4 sin2 2 − 4 sin 2 is 2π (π΄) π − 4 (B) 3 − 4 − 4ΰΆ₯3 π (C) 4ΰΆ₯3 − 4 π π₯ (D) 4ΰΆ₯3 − 4 − 3 π₯ Solution: Let I= ∫0 ΰΆ§1 + 4 sin2 2 − 4 sin 2 dx π 2 π₯ = ∫0 ΰΆ§α2 sin 2 − 1α dx π π₯ β΅ ΰΆ₯(π₯)2 = Θπ₯Θ = ∫0 α2 sin 2 − 1α dx π₯ π₯ 1 π₯ π π Put 2 sin 2 − 1 = 0 ππ sin 2 = 2 or 2 = 6 ππ π₯ = 3 π π₯ π π₯ When π₯ ∈ α0, 3 α π‘βππ 2 sin 2 − 1 ≤ 0 and When π₯ ∈ α 3 , πα π‘βππ 2 sin 2 − 1 ≥ 0 π π π₯ π₯ πΌ = − ∫03 (2 sin 2 − 1) dx + ∫π (2 sin 2 − 1) dx 3 π 3 π₯ π₯ π = − α −4 cos 2 − π₯α + α −4 cos 2 − π₯απ 0 ξ3 3 π ξ3 π = α4 × 2 + 3 − 4α + α 0 − π + 4 × 2 + 3 α 2π π = 4ΰΆ₯3 − 4 + 3 − π = 4ΰΆ₯3 − 4 − 3 π π»ππππ, the correct option is (D) 4ΰΆ₯3 − 4 − 3 Type II. Problems based on the property: π π ∫π π(π₯) ππ₯ = ∫π π(π + π − π₯) ππ₯ π π or ∫0 π(π₯) ππ₯ = ∫0 π(π − π₯) ππ₯ special case πβππ property is used when π(π₯) + π(π + π − π₯) becomes πππ‘πππππππ function of x i.e., (π) f(x) is trigonometric function of complementary angles. (ii) Integrand is of the form x f(x) so that x is eliminated and the function becomes πππ‘πππππππ function of x. (iii) Integrand is a logarithmic function. 4 Rahman Sir DEFINITE INTEGRAL Working rule: First we assume the Integration as π I = ∫π π(π₯) ππ₯ … (1) π I = ∫π π(π + π − π₯) ππ₯…. (2) Adding (1) and (2) we get π ∫π αΎπ(π₯) + π(π + π − π₯)αΏ ππ₯ which becomes πππ‘πππππππ function of x. π π₯π πππ₯ Example: Evaluate: ∫0 1+cos2 π₯ ππ₯ π Let πΌ = ∫0 1+cos2 π₯ ππ₯ π₯π πππ₯ … (1) π (π−π₯)sin(π−π₯) π ππ πππ₯ π −π₯π πππ₯ Or πΌ = ∫0 1+cos2 (π− π₯) dx = ∫0 1+cos2 π₯ dx+ ∫0 1+cos2 π₯ dx … (2) Adding (1) and (2) we get π ππ πππ₯ 2I= ∫0 1+cos2 π₯dx πππ‘ πππ π₯ = π‘ , π‘βππ − π πππ₯ππ₯ = ππ‘ , π€βππ π₯ = 0 π‘ = πππ 0 = 1 πππ π€βππ π₯ = π π‘ = −1 −1 −ππ‘ 1+π‘ 2 1 2 1 1 πΌ = 2 π ∫1 ππ‘ = 2 π ∫−1 1+π‘ 2 1 ππ‘ = 2 π ∫0 1+π‘ 2 π π β΅ ∫π π(π)π π = − ∫π π(π)π π β΅ π(π‘)ππ ππ£ππ ππ’πππ‘πππ π π2 παΎtan−1 π₯αΏ10 = πΰ΅£tan−1(1 ) − tan−1 0ࡧ = π α 4 α = 4 π ππ₯ Evaluate: ∫π3 1+ π‘πππ₯ ξ 6 π ππ₯ 3 π 1+ ξπ‘πππ₯ 6 Let πΌ = ∫ π = ∫ ξπππ π₯ππ₯ ξπ πππ₯ Or πΌ = ∫π3 πππ π₯+ ξ 6 π ξπππ π₯ππ₯ 3 π πππ π₯+ ξπ πππ₯ ξ 6 … (1) π π π ΰΆ§cos( + −π₯)ππ₯ 3 6 3 π π π π π 6 ΰΆ§πππ ( + −π₯)+ ΰΆ§π ππ( + −π₯) 3 6 3 6 ππ πΌ = ∫ π ξπ πππ₯ππ₯ ππ πΌ = ∫π3 πππ π₯+ ξπ πππ₯ ξ … (2) 6 Adding (1) and (2) we get 5 Rahman Sir DEFINITE INTEGRAL π 3 π 6 2I = ∫ ππ₯ π 3 π 6 π π π = αΎπ₯αΏ = 3 − 6 = 6 Hence, I = π 12 π 4 Evaluate ∫0 log(1 + π‘πππ₯ ) ππ₯ π 4 Solution: let I = ∫0 log(1 + π‘πππ₯ ) ππ₯ … (1) π 4 π Or I = ∫0 log 1 + tan( 4 − π₯) ππ₯ π 4 π 4 = I = ∫0 log(1 + π 4 = ∫ log( 0 tan −π‘πππ₯ π 4 1+tan .π‘πππ₯ ) ππ₯ 1 + π‘πππ₯ + 1 −π‘πππ₯ ) ππ₯ 1 + π‘πππ₯ π 4 2 = ∫0 log(1+π‘πππ₯) ππ₯ π 4 π 4 Or I = ∫0 πππ2ππ₯ − ∫0 log(1 + π‘πππ₯ ) ππ₯ …. (2) Adding (1) and (2) we get π 4 π 4 π 2 I = ∫0 πππ2ππ₯ = πππ2αΎπ₯αΏ0 = 4 log 2 1 π Or I = 8 log 2 2−π₯ π¬πππππππ: ∫−1 πππ α2+π₯α ππ₯ 1 2−π₯ Solution: Let I = ∫−1 πππ α2+π₯α ππ₯ … (1) 1 2−(1−1−π₯) π Or I= ∫−1 πππ ΰΈ¬2+(1−1−π₯)ΰΈ¬ ππ₯ 1 2−(0−π₯) = ∫−1 πππ ΰΈ¬2+(0−π₯)ΰΈ¬ ππ₯ 1 2+π₯ … Or I = ∫−1 πππ α2−π₯α ππ₯ (2) Adding (1) and (2) we get 1 2−π₯ 2+π₯ π β΅ ∫π π(π₯) ππ₯ = ∫π π(π + π − π₯) ππ₯ 2I = ∫−1 πππ α2+π₯α + π ππ α2−π₯α ππ₯ 6 Rahman Sir DEFINITE INTEGRAL 1 1 2I = ∫−1 πππ1dx =∫−1 0 ππ₯ = 0 Or I = 0 1+cos2 π₯ πΌ1 πΉπππ . πΌπ πΌ1 = ∫sin2 π₯ πΌ 2 1+cos2 π₯ Solution: πΌ1 = ∫sin2 π₯ π₯παΌπ₯(2 − π₯)α½ππ₯ and 1+cos2 π₯ πΌ2 = ∫sin2 π₯ παΌπ₯(2 − π₯)α½ππ₯ π₯π(2 − π₯)ππ₯ Let F(x) = π₯παΌπ₯(2 − π₯)α½ Now, π + π − π₯ = (1 + cos2 π₯ + sin2 π₯ − π₯) = (2 − π₯) ∴ πΉ(π + π − π₯) = (2 − π₯)παΌ(2 − π₯)π₯α½ 1+cos2 π₯ Hence, ∫sin2 π₯ 1+cos2 π₯ πΌ1 = ∫sin2 π₯ 1+cos2 π₯ π₯παΌπ₯(2 − π₯)α½ππ₯ = ∫sin2 π₯ 1+cos2 π₯ 2παΌπ₯(2 − π₯)α½ππ₯ − ∫sin2 π₯ (π₯ − 2)παΌπ₯(2 − π₯)α½ππ₯ π₯παΌπ₯(2 − π₯)α½ππ₯ 1+cos2 π₯ ππ πΌ1 = ∫ 2παΌπ₯(2 − π₯)α½ππ₯ − πΌ1 sin2 π₯ 1+cos2 π₯ Or 2πΌ1 = 2∫sin2 π₯ παΌπ₯(2 − π₯)α½ππ₯ Or 2 πΌ1 = 2 πΌ2 ⇒ πΌ1 = πΌ2 πΌ or πΌ1 = 1 2 ΰ΅£π₯ 2 ࡧ 10 Example: The Value of integral ∫4 (A) 6 αΎπ₯ 2 −28π₯+196αΏ+ αΎπ₯ 2 αΏ (B) 3 ΰ΅£π₯ 2 ࡧ 10 Solution: Let I = ∫4 (C ) 7 αΎπ₯ 2 −28π₯+196αΏ+ αΎπ₯ 2 αΏ ΰ΅£π₯ 2 ࡧ 10 Or I = ∫4 αΎ(π₯−14)2 αΏ+ αΎπ₯ 2 αΏ ππ₯ is (D) 1 3 ππ₯ ππ₯ … (1) ΰ΅£π₯ 2 ࡧ Let π(π₯) = αΎ(π₯−14)2 αΏ+ αΎπ₯ 2 αΏ Now, π(π + π − π₯) = π(10 + 4 − π₯) = π(14 − π₯) ΰ΅£(14−π₯)2 ࡧ Therefore, π(14 − π₯) = αΎ(π₯)2 αΏ+ ΰ΅£(14−π₯)2 ࡧ 10 Or I = ∫4 ΰ΅£(14−π₯)2 ࡧ αΎ(π₯)2 αΏ+ ΰ΅£(14−π₯)2 ࡧ Adding (1) and (2) we get, π π ππ₯ … (2) using the property ∫π π(π₯) ππ₯ = ∫π π(π + π − π₯) ππ₯ 7 Rahman Sir DEFINITE INTEGRAL 10 2I = ∫4 1 ππ₯ = αΎπ₯αΏ10 4 = 10 − 4 = 6 6 Or I = 2 = 3 Hence, the correct option is (B) π 2 πππ π₯−π πππ₯π₯ 0 10−π₯ 2 +ππ₯ 2 Example: The Value of integral ∫ π (B) π (A) 2 π 2 πππ π₯−π πππ₯ 0 10−π₯ 2 +ππ₯ 2 Solution: Let I = ∫ (C ) 0 =∫ = ∫0 π πππ₯−πππ π₯ π2 ππ₯ π2 10−࡬ −ππ₯+π₯ 2 ΰ΅°− + 4 2 4 π 2 π πππ₯−πππ π₯ 0 10−π₯ 2 +ππ₯ 2 ππ₯ ππ₯ … (2) ππ₯ =∫ Or I (D) 4π …(1) ππ₯ π π π cos( −π₯)−sin( −π₯) 2 2 2 π 0 πα −π₯α π 10−( −π₯)2 + 2 2 2 π 2 ππ₯ is Adding (1) and (2) we get, π 2I = ∫02 0 10−π₯ 2 + ππ₯ 2 ππ₯ = 0 Or I = 0 Type III. Problems based on the property. π (i) ∫−π π(π₯) ππ₯ = 0, if f(x) is odd function. i.e., π(−π₯) = −π(π₯) (ii) ∫−π π(π₯) ππ₯ = 2 ∫0 π(π₯) ππ₯ if f(x) is even function. i.e., π(−π₯) = π(π₯) (iii) ∫−π π(π₯) ππ₯ = ∫0 π(π₯) ππ₯ + ∫0 π(−π₯) ππ₯ when f(x) is neither odd nor even function i.e., π π π π π some part is odd and some part is even 1 π¬πππππππ ∫−1 sin5 π₯ cos4 π₯ ππ₯ 1 Solution: Let I = ∫−1 sin5 π₯ cos 4 π₯ ππ₯. Let π(π₯) = sin5 π₯ πππ 4 π₯ π(−π₯) = sin5(−π₯) cos 4 (−π₯) = − sin5 π₯ cos4 π₯ = −π(π₯) π πΌ = 0 , β΅ ∫−π π(π₯) ππ₯ = 0, if f(x) is odd function. i.e., π(−π₯) = −π(π₯) π π−π₯ π π−π₯ π¬πππππππ ∫−π ΰΆ§π+π₯ dx Let I = ∫−π ΰΆ§π+π₯ dx DEFINITE INTEGRAL π π−π₯ ππ₯ = ∫−π 2 ΰΆ₯ π −π₯2 π π π = ∫−π 2 ΰΆ₯ π −π₯ 2 8 Rahman Sir ππ₯ − ∫−π −π₯ ΰΆ₯π2 −π₯ 2 ππ₯ = πΌ1 − πΌ2 πΌ2 = 0 ππ π I= ∫−π −π₯ π ππ₯ ΰΆ₯π2 −π₯ 2 π = 2 ∫0 is odd function ΰΆ₯π2 −π₯ 2 π ΰΆ₯π2 −π₯ 2 ππ₯ as π ΰΆ₯π2 −π₯ 2 is even function π₯ π π = 2a αsin−1 πα = 2π sin−1 1 = 2π. 2 = ππ 0 π ππ₯ Evaluate ∫ 2π 1+π π πππ₯ − 2 This function is neither odd nor even π ππ₯ 2 π π πππ₯ 1+π − 2 Solution: Let I = ∫ π 1 1 = ∫02 α1+π π πππ₯ + α ππ₯ 1+π sin(−π₯) π π π ∫−π π(π₯) ππ₯ = ∫0 π(π₯) ππ₯ + ∫0 π(−π₯) ππ₯ when f(x) is neither odd nor even function i.e., some part is odd and some part is even π 2 1 = ∫0 α1+π π πππ₯ + π 2 1 = ∫0 ΰ΅€1+π π πππ₯ + π 2 1 1+π −sin(π₯) π π πππ₯ 1+π sin(π₯) π α ππ₯ ࡨ ππ₯ π = ∫0 ππ₯ = 2 − 0 = 2 π 2 πππ π₯ π − 1+π π₯ 2 Evaluate ∫ ππ₯ [CBSE(F)2015] This function is neither odd nor even π 2 πππ π₯ π − 1+π π₯ 2 ∴ ∫ π π 2 πππ π₯ cos(−π₯) ππ₯ = ∫0 α1+π π₯ + 1+π −π₯ α ππ₯ π π ∫−π π(π₯) ππ₯ = ∫0 π(π₯) ππ₯ + ∫0 π(−π₯) ππ₯ when f(x) is neither odd nor even function i.e., some part is odd and some part is even 9 Rahman Sir DEFINITE INTEGRAL π 2 πππ π₯ = ∫0 α1+π π₯ + π πππ π₯ = ∫02 α1+π π₯ + π = ∫02 cos(π₯) 1 π 1+ π₯ ex cos(π₯) 1+π π₯ (1+π π₯ )πππ π₯ 1+π π₯ α ππ₯ α ππ₯ ππ₯ π π π = ∫02 πππ π₯ ππ₯ = αΎπ πππ₯αΏ02 = sin α2 α − π ππ0 = 1 π N.C.E.R.T. Example 36: Evaluate ∫02 ππππ πππ₯ ππ₯ π Solution: Let I= ∫02 ππππ πππ₯ ππ₯ … (1) π π Or I = ∫02 logsin α 2 − π₯α ππ₯ π Or I = ∫02 ππππππ π₯ ππ₯ … (2) Adding (1) and (2) π π 2I = ∫02 ππππ πππ₯ ππ₯ + ∫02 ππππππ π₯ ππ₯ π Or 2I = ∫02 ππππ πππ₯. πππ π₯ ππ₯ β΅ ππππ΄ + ππππ΅ = ππππ΄π΅ π 2 Or 2I = ∫0 log (2π πππ₯.πππ π₯) 2 ππ₯ π π Or 2I = ∫02 ππππ ππ2π₯ ππ₯ − πππ2 ∫02 ππ₯ π 2I = πΌ1 − 2 πππ2 π Now πΌ1 = ∫02 ππππ ππ2π₯ ππ₯ π Put 2x = t , then 2 dx = ππ‘ , π€βππ π₯ = 0 π‘ = 0 πππ π€βππ π₯ = 2 , π‘ = π 1 π πΌ1 = 2 ∫0 ππππ πππ‘ ππ‘ 2 π = 2 ∫02 ππππ πππ‘ ππ‘ π ππ π = ∫02 ππππ πππ‘ ππ‘ = ∫02 ππππ πππ₯ ππ₯ π Hence, 2I = I− 2 πππ2 π Or I = − 2 πππ2 π since, ∫π π(π)π π = π ∫π π(π) π π ππ π(ππ − π) = π(π) π π [ β΅ ∫π π(π)π π = ∫π π(π)π π] 10 Rahman Sir DEFINITE INTEGRAL EXERCISE 7.11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 1. π 2 ∫0 cos2 π₯ dx π Let I = ∫02 cos2 π₯ dx … (1) Solution: π π π Or I = ∫02 cos 2 ( 2 − π₯) dx = ∫02 sin2 π₯ dx .. (2) Adding (1) and (2) we get π π 2 I = ∫02 (cos2 π₯ + sin2 π₯) dx = ∫02 ππ₯ π π π Or I = αΎπ₯αΏ02 = α2 − 0 α = 2 2. π 2 ΰΆ₯π πππ₯ ∫0 ΰΆ₯π πππ₯+ dx ξπππ π₯ π ΰΆ₯π πππ₯ Solution: Let I = ∫02 ΰΆ₯π πππ₯+ π dx … (1) π π 2 Or I = ∫0 ξπππ π₯ ΰΆ§sin( 2 −π₯) π dx π ΰΆ§sin( 2 −π₯)+ ΰΆ§cosα 2 − π₯α π π π β΅ ∫π π(π)π π = ∫π π(π − π)π π πππ πππ α π − πα = ππππ and πππ α π − πα = ππππ π 2 Or I = ∫0 ξπππ π₯ dx ΰΆ₯π πππ₯ πππ π₯+ ξ … (2) Adding (1) and (2) we get π π 2I = ∫02 ππ₯ = α2 − 0α π Or I = 4 3. π 2 ∫0 3 sin2 π₯ 3 dx 3 sin2 π₯+cos2 π₯ 3 π 2 Solution: Let I = ∫0 sin2 π₯ 3 3 π 2 Or I = ∫0 π π 3 … (1) dx sin2 π₯+cos2 π₯ π 2 sin2 ( − π₯) 3 π 2 3 π 2 ) sin2 α −π₯α+cos2 ( − π₯ π π β΅ ∫π π(π)π π = ∫π π(π − π)π π πππ πππ α π − πα = ππππ and πππ α π − πα = ππππ 11 Rahman Sir DEFINITE INTEGRAL 3 π 2 = ∫0 Or I cos2 π₯ 3 …. (2) 3 cos2 π₯+sin2 π₯ Adding (1) and (2) we get π 2 π 2I = ∫0 ππ₯ = α2 − 0α π Or I = 4 π 2 4. cos5 π₯ ∫0 sin5 π₯ +cos5 π₯ dx π 2 cos5 π₯ … (1) Let I = ∫0 sin5 π₯ +cos5 π₯ dx Solution: π Or I = ∫02 π π 2 cos5 ( − π₯) π 2 π 2 sin5 ( − π₯ )+cos5 ( − π₯) ππ₯ π π π β΅ ∫π π(π)π π = ∫π π(π − π)π π πππ πππ α π − πα = ππππ and πππ α π − πα = ππππ π sin5 π₯ … (2) Or I = ∫02 cos5 π₯ +sin5 π₯ dx Adding (1) and (2) we get π 2 π 2I = ∫0 ππ₯ = α − 0α 2 π Or I = 4 5. 5 ∫−5Θπ₯ + 2Θ dx Solution: put π₯ + 2 = 0 , ⇒ π₯ = −2 −2 5 I= ∫−5 −(π₯ + 2) ππ₯ + ∫−2(π₯ + 2)ππ₯ −2 π₯2 = − α 2 + 2π₯α −5 5 π₯2 + α 2 + 2π₯α −2 25 25 = α−(2 − 4) + α 2 − 10αα + αα 2 + 10α − (2 − 4)α 25 25 = α 2 − 8α + α12 + 2 α = 29 8 ∫2 Θπ₯ − 5Θ dx Solution: let π₯ − 5 = 0 , ππ π₯ = 5 It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8]. 6. 5 8 I= ∫2 −(π₯ − 5) ππ₯ + ∫5 (π₯ −)ππ₯ 5 π₯2 π₯2 = − α 2 − 5π₯α + α 2 − 5π₯α 2 25 8 5 25 = − αα 2 − 25α − (2 − 10)α + α(32 − 40) − α 2 − 25αα 12 Rahman Sir DEFINITE INTEGRAL 25 25 = − 2 + 25 − 8 −8 − 2 + 25 = −25 + 25 + 25 − 16 = 9 7. 1 ∫0 π₯(1 − π₯)π dx Solution: 1 ∫0 π₯(1 − π₯)π dx Let I = 1 π I= ∫0 (1 − π₯)(π₯)π dx π β΅ ∫π π(π)π π = ∫π π(π − π)π π 1 = ∫0 (π₯ π − π₯ π+1 )ππ₯ π₯ π+1 1 π₯ π+2 = α π+1 − π+2 α 1 0 (π+2)−(π+1) 1 = απ+1 − π+2α = (π+1)(π+2) 1 = (π+1)(π+2) 8. π 4 ∫0 log(1 + π‘πππ₯) dx π 4 Solution: Let I = ∫0 log(1 + π‘πππ₯) dx … (1) π 4 π = ∫0 log α1 + tan β« Ϋβ¬− π₯β«Ϋβ¬α dx 4 π 4 π 4 tan −π‘πππ₯ =∫0 log α1 + α αα dx π 1+tan π‘πππ₯ 4 π 4 1−π‘πππ₯ = ∫0 log α1 + 1+π‘πππ₯α ππ₯ π = ∫04 log α 1+π‘πππ₯+1−π‘πππ₯ π 1+π‘πππ₯ α ππ₯ 2 = ∫04 log α1+π‘πππ₯α ππ₯ π π Or I = log2 ∫04 ππ₯ − ∫04 log(1 + π‘πππ₯) dx … (2) Adding (1) and (2) we get, π π 2I = log2 ∫04 ππ₯ = 4 πππ2 π Or I = 8 πππ2 9. 2 ∫0 π₯ΰΆ₯2 − π₯ dx Solution: 2 Let I = ∫0 π₯ΰΆ₯2 − π₯ dx DEFINITE INTEGRAL 2 = ∫0 (2 − π₯)ΰΆ₯2 − (2 − π₯) dx 2 = ∫0 (2 − π₯)ξπ₯ 2 3 2 = 2∫0 ξπ₯ ππ₯ − ∫0 π₯ 2 ππ₯ 3 4 5 2 = ΰ΅€3 π₯ 2 − 5 π₯ 2 ࡨ 4 2 0 2 = α3 × 2ΰΆ₯2 − 5 × 4 × ΰΆ₯2α 8ξ 2 8ξ2 = 3 − 5 = 10. 40ξ2−24ξ2 16ξ2 = 15 15 π 2 ∫0 (2ππππ πππ₯ − ππππ ππ2π₯) dx π 2 Let I = ∫0 (2ππππ πππ₯ − ππππ ππ2π₯) dx Solution: π 2 π π I = ∫0 (2ππππ ππ( 2 − π₯) − ππππ ππ2 α2 − π₯α ) dx Or π π β΅ ∫ π(π)π π = ∫ π(π − π)π π π π π 2 … (2) πΌ = ∫0 (2ππππππ π₯ − ππππ ππ2π₯) ππππππ (1)πππ (2) we get π 2 2πΌ = ∫0 (2ππππ πππ₯ + 2ππππππ π₯ − 2ππππ ππ2π₯) π 2 π 2 2πΌ = 2 ∫0 ππππ πππ₯πππ π₯ππ₯ − 2 ∫0 ππππ ππ2π₯ ππ₯ π 2 = 2 ∫0 log (2π πππ₯πππ π₯ππ₯) 2 π 2 − 2 ∫0 ππππ ππ2π₯ ππ₯ π 2 π 2 π 2 = 2 ∫0 ππππ ππ2π₯ ππ₯ − 2 πππ2 ∫0 ππ₯ − 2 ∫0 ππππ ππ2π₯ ππ₯ π 2 = −2 πππ2 ∫0 ππ₯ π 1 = −2 α2 α πππ2 = −ππππ2 = π log 2 π 1 ππ πΌ = 2 log 2 π 2 π − 2 11. ∫ sin2 π₯ dx π Solution: Let As sin 2 (−π₯) I= ∫ 2π sin2 π₯ dx − 2 = sin2 π₯ , π‘βπππππππ, sin2 π₯ is an even function 13 Rahman Sir 14 Rahman Sir DEFINITE INTEGRAL π π It is known that if f(x) is an even function, then ∫−π π(π)π π = 2∫π π(π)π π π ∴ πΌ = 2 ∫02 sin2 π₯ ππ₯ π 1 = ∫02 (1 − πππ 2π₯) ππ₯ = απ₯ − 2 π ππ2π₯α π π 2 0 π = α 2 − 0 − (0 − 0)α = 2 π π₯ 12. ∫0 1+π πππ₯ dx π π₯ … (1) I = ∫0 1+π πππ₯ dx Solution: Let π (π−π₯) I = ∫0 1+sin(π−π₯) dx π (π−π₯ =∫0 1+π πππ₯) dx Or I … (2) Adding (1) and (2) we get, π 1 2I = π ∫0 1+π πππ₯ dx π 1−π πππ₯ 2I = π ∫0 (1+π πππ₯)(1−π πππ₯) dx π 1−π πππ₯ = π ∫0 (1−sin2 π₯ ) dx π 1−π πππ₯ = π ∫0 cos2 π₯ dx π = π ∫0 (sec 2 π₯ − π‘πππ₯π πππ₯) ππ₯ = π αΎπ‘πππ₯ − π πππ₯αΏπ0 = παΎπ‘πππ − π πππαΏ − παΎπ‘ππ0 − π ππ0αΏ = π(0 + 1) − π(0 − 1) = 2π ππ πΌ = π π 13. ∫−2π sin7 π₯ dx 2 Solution: Let π 2 π − 2 I = ∫ sin7 π₯ dx As sin7 (−π₯) = (− sin π₯ )7 = − sin7 π₯ , therefore, sin7 π₯ is an odd function. π It is known that if f(x) is an odd function, then ∫−π π(π)π π =0 15 Rahman Sir DEFINITE INTEGRAL π 2 π − 2 ∴ I = ∫ sin7 π₯ dx = 0 2π 14. ∫0 cos5 π₯ dx Solution: Let 2π I = ∫0 cos 5 π₯ dx … (1) πππ 5 (2π − π₯) = cos5 π₯ ππ π It is known that ∫π π(π)π π = π ∫π π(π) π π ππ π(ππ − π) = π(π) π Or I = 2 ∫0 cos5 π₯ dx π ⇒ πΌ = 2(0) β΅ cos(π − π₯) = −πππ π₯ and ∫π π(π) π π = π, ππ π(π − π) = −π(π π π πππ₯−πππ π₯ 15. ∫02 1+π πππ₯πππ π₯ dx π Solution: : Let π π πππ₯−πππ π₯ I = ∫02 1+π πππ₯πππ π₯ … (1) π β΅ ∫π π(π)π π = ∫π π(π − π)π π π πΌ = ∫02 π 2 π 2 π π 1+π ππ( −π₯)πππ ( −π₯) 2 2 sin( −π₯)−πππ ( −π₯) π 2 πππ π₯−π πππ₯ … (2) πΌ = ∫0 1+π πππ₯πππ π₯ Adding (1) and (2) we get, 2I = 0 Or I = 0 π 16. ∫0 log(1 + πππ π₯) dx π let I =∫0 log(1 + πππ π₯) Solution: Or I = π ∫0 log(1 + cos(π − π₯)) … (1) π π = ∫0 log(1 − πππ π₯) dx … (2) β΅ cos(π − π₯) = −πππ π₯ Adding (1) and (2) we get π 2πΌ = ∫0 log(1 − cos2 π₯) dx π π Or 2I = ∫0 ππππ ππ2 π₯ππ₯ = 2 ∫0 ππππ πππ₯ ππ₯ π Or I = ∫0 ππππ πππ₯ ππ₯ … (3) π 2 Or I = 2 ∫0 ππππ πππ₯ ππ₯ as β΅ sin(π − π₯) = π πππ₯ ππ π ∫ π(π)π π = π ∫ π(π) π π ππ π(ππ − π) = π(π) π π π 2 Or I = 2 ∫0 ππππ πππ₯ ππ₯ π β΅ ∫π π(π)π π = ∫π π(π − π)π π … (4) 16 Rahman Sir DEFINITE INTEGRAL π 2 π π Or I = 2 ∫0 ππππ ππ( 2 − π₯) ππ₯ π 2 Or I = 2 ∫0 ππππππ π₯ ππ₯ π β΅ ∫π π(π)π π = ∫π π(π − π)π π … (5) Adding (4) and (5) we get, π 2 2I = 2 ∫0 ππππ πππ₯ πππ π₯ ππ₯ π 2 1 Or 2I =2 ∫0 log ( 2π πππ₯πππ π₯ ) 2 ππ₯ π 2 π 2 2I = 2 ∫0 ππππ ππ2π₯ ππ₯ − 2 πππ2 ∫0 ππ₯ 2I = πΌ1 − ππππ2 π 2 πΌ1 =2 ∫0 ππππ ππ2π₯ ππ₯ let 2π₯ = π‘ π‘βππ 2ππ₯ = ππ‘ , π€βππ π₯ = 0 π‘ = 0, π€βππ π₯ = π 2 π π‘βππ π‘ = π 2 π πΌ1 = 2 ∫0 ππππ πππ‘ ππ‘ = ∫0 ππππ πππ₯ ππ₯ π = πΌ ππππ (3) or 2I =πΌ − π πππ2 Or I = −π log 2 π π₯ 17. ∫0 π₯+ξ π−π₯ dx ξ ξ π let I = ∫0 Solution: π I = ∫0 ξπ₯ dx ξπ₯+ ξπ−π₯ ξπ−π₯ ξπ−π₯+ ΰΆ₯π−(π−π₯) π Or I = ∫0 Adding (1) and (2) we get, π 2I = ∫0 ππ₯ = αΎπ₯αΏπ0 = π π Or I = 2 4 18. ∫0 Θπ₯ − 1Θ dx …. (1) dx ξπ−π₯ dx ξπ−π₯+ ξπ₯ π as ∫π π(π₯)ππ₯ = ∫π πΰ΅«π¦ΰ΅―ππ¦ … (2) 17 Rahman Sir DEFINITE INTEGRAL 4 Let I = ∫0 Θπ₯ − 1Θ dx Solution: Put π₯ − 1 = 0 π‘βππ π₯ = 1 It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4 1 4 ∴ πΌ = ∫0 (1 − π₯) ππ₯ + ∫1 (π₯ − 1) ππ₯ π₯2 1 π₯2 4 = απ₯ − 2 α + α 2 − π₯α 0 1 1 1 = αα1 − 2α − (0)α + α(8 − 4) − α2 − 1αα 1 1 =2 +4+2=5 π π 19. ∫0 π(π₯) π(π₯) dx = 2 ∫0 π(π₯) ππ₯ , ππ π πππ π πππ πππππππ ππ π(π₯) = π(π − π₯) and π(π₯) + π(π − π₯) = 4 π … (1) Let I= ∫0 π(π₯) π(π₯) dx Solution: π π β΅ ∫π π(π)π π = ∫π π(π − π)π π π π ππ πΌ = ∫0 π(π − π₯) π(π − π₯)dx = ∫0 π(π₯) π(π − π₯) dx π ππ πΌ = ∫0 π(π₯) π(π − π₯) dx π΄πππππ (1)πππ (2) we get, β΅ π(π₯) = π(π − π₯) … (2) π π 2πΌ = ∫0 ΰ΅£π(π₯)π(π₯) + π(π₯)π(π − π₯)ࡧ dx = ∫0 ΰ΅£π(π₯) + π(π − π₯)ࡧ π(π₯)ππ₯ π 2I = ∫0 4 π(π₯)ππ₯ β΅ π(π₯) + π(π − π₯) = 4 π Or I = 2 ∫0 π(π₯) ππ₯ Choose the correct answer in Exercises 20 and 21. π 2 π − 2 20. ∫ (π₯ 3 + π₯πππ π₯ + tan5 π₯ + 1) dx (A) 0 (C ) π (B) 2 π 2 π − 2 Solution: ∫ (π₯ 3 + π₯πππ π₯ + tan5 π₯ + 1) dx π 2 π − 2 3 5 π 2 π − 2 = ∫ (π₯ + π₯πππ π₯ + tan π₯) ππ₯ + ∫ (1)dx Let f(x) = π₯ 3 + π₯πππ π₯ + tan5 π₯. F(-x)= (−π₯)3 + (−π₯) cos(−π₯) + tan5(−π₯) = −π₯ 3 − π₯πππ π₯ − tan5 π₯ = −(π₯ 3 + π₯πππ π₯ + tan5 π₯) = −π(π₯) ∴ π(π₯)ππ πππ ππ’πππ‘πππ (D) 1 18 Rahman Sir DEFINITE INTEGRAL π 2 π − 2 ππ ∫ (π₯ 3 + π₯πππ π₯ + tan5 π₯) dx = 0 π 2 π − 2 π π π»ππππ, πΌ = ∫ (1)dx = α 2 α − (− 2 ) = π Hence, the correct option is (C) π 4+3π πππ₯ 21. ∫02 πππ α4+3πππ π₯α dx 3 (A) 2 Solution: (B) 4 π 2 (D) −2 (C ) 0 4+3π πππ₯ Let I = ∫0 πππ α4+3πππ π₯α dx … (1) π 2 Or I = ∫0 πππ α€ π 2 π 2 π 4+3cos( −π₯) 2 4+3sin( −π₯) 4+3πππ π₯ Or I = ∫0 πππ α4+3π πππ₯ α dx α€ dx …. (2) Adding (1) and (2) we get, π 2I = ∫02 πππ1 ππ₯ = 0 Or, I = 0 Hence, the correct option is (C) π π β΅ ∫π π(π)π π = ∫π π(π − π)π π DEFINITE INTEGRAL 19 Rahman Sir
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