IMP THEOREMS CLASS 12 PDF

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🎯 HERE YOU GET ALL 12TH & OTHER ENTRACE STUDY MATERIAL 🎯 Join Telegram ;- @NOTESPROVIDER12TH_BOARD AND DAILY QUIZZES 😊 JOIN TELEGRAM 😊 ✴️ KEEP SHARE THIS LINKS ✴️ 👇👇👇👇👇👇👇👇 JOIN TELEGRAM ALL LINKS FOR NOTES Click👉➡️@SCIENCE11MAHARASTRA Click➡@HSCNOTESPROVIDER_12TH_BOARD Click➡ @HSCNOTESPROVIDER12TH Click 👉➡ @NOTESPROVIDER12TH ❇️ ALL HSC MHT CET NEET JEE NOTES BOTS ❇️ 👇👇👇👇👇👇👇👇 Click👉➡ @NOTESPROVIDER12TH_BOT ➡️ Join Telegram:- @NOTESPROVIDER12TH_BOARD ☝☝☝☝☝☝ click@MHTCET_NOTESPROVIDER_BOT ❇️ALL HSC MHT CET NEET JEE NOTES ❇️ ENTRANCE PREPARATION Click➡ @HSCNEETCETQUIZCOLLECTION Click➡ @NEET_JEE_TOPPERS_2021 Click➡️ @MHTCET_NOTESPROVIDER ✳️ KEEP SHARE THIS LINK ✳️ ✴️ JOIN BACKUP CHANNEL 👇 ✴️ Click👉@NOTESPROVIDER12TH_BOARD 👇👇👇👇👇👇👇 Join telegram for more notes Click➡️@NOTESPROVIDER12TH Click➡️ SCIENCE BUDDYS JOIN TELEGRAM CHANNEL 👉click link @NOTESPROVIDER12TH SEARCH TELEGRAM👉click link @NOTESPROVIDER12TH USE THIS BOT FOR NOTES TELEGRAM click👉 @NOTESPROVIDERTH_BOT  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Join Telegram:- @NOTESPROVIDER12TH_BOARD  Join Telegram ;- @NOTESPROVIDER12TH_BOARD The Sine Rule : a b c In ∆ABC, sin A = sin B = sin C = 2R, where R is the circumradius of ∆ABC. Proof :Let AD be perpendicular to BC. 𝐴 AD = b sin C 𝑏 1 ∴ A(∆ABC) = 2 BC × AD 1 = 2 a × b sin C 𝐵 𝐷 𝐶 1 ∴ A(∆ABC) = 2 ab sin C ∴ 2A(∆ABC) = ab sin C Similarly 2A(∆ABC) = ac sin B and 2A(∆ABC) = bc sin A ∴ bc sin A = ac sin B = ab sin C 𝐴 Divide by abc, bc sin A ac sin B ab sin C ∴ = = 𝑏 abc abc abc 2𝑅 sin A sin B sin C 𝑂 ∴ = = a b c Join Telegram:- @NOTESPROVIDER12TH_BOARD 𝐵 𝐶 a b c ∴ sin A = sin B = sin C … (1) 𝑃 To prove that each ratio is equal to 2R. As the sum of three angles of triangle is 1800 , at least one of the angle of the triangle is not right angle. Suppose A is not right angle. Draw diameter through A. Let it meet circle in P. ∴ AP = 2R and ∆ACP is a right angled triangle. ∠ABC and ∠APC are inscribed in the same arc. ∴ m∠ABC = m∠APC b b ∴ sin B = sin P = AP = 2R b ∴ sin B = 2R b ∴ sin B = 2R … (2) From (1) and (2) , we get a b c = sin B = sin C = 2R sin A 1  Join Telegram ;- @NOTESPROVIDER12TH_BOARD The Cosine Rule : In ∆ABC, (i) a2 = b2 + c 2 − 2bc cos A (ii) b2 = c 2 + a2 − 2ca cos B (iii) c 2 = a2 + b2 − 2ab cos C Proof :Take A as the origin, X − axis along AB and the line perpendicular to AB through A as the Y − axis. 𝑌 𝐶(𝑏 cos 𝐴 , 𝑏 sin 𝐴) The co-ordinates of 𝑏 A(0, 0) 𝑎 B(c, 0) and C(b cos A , b sin A) 𝐴 𝑐 𝐵 𝑋 2 2 2 To prove that a = b + c − 2bc cos A BC = √(c − b cos A)2 + (0 − b sin A)2 (by distance formula) BC 2 = (c − b cos A)2 + (0 − b sin A)2 a2 = c 2 + b2 cos 2 A − 2 bc cos A + b2 sin2 A a2 = c 2 + b2 cos 2 A b2 sin2 A@NOTESPROVIDER12TH_BOARD +Telegram:- Join − 2 bc cos A a2 = c 2 + b2 (cos 2 A + sin2 A) − 2 bc cos A a2 = c 2 + b2 − 2 bc cos A ∴ a2 = b2 + c 2 − 2bc cos A b2 +c2 −a2 cosA = 2bc Similarly, we can prove that b2 = c 2 + a2 − 2ca cos B c 2 = a2 + b2 − 2ab cos C The Projection Rule: In ∆ABC, (i) a = b cos C + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos A Proof: (i) a = b cos C + c cos B By cosine rule we have, 2  Join Telegram ;- @NOTESPROVIDER12TH_BOARD a2 +c2 −b2 a2 +b2 −c2 cosB = and cosC = 2ac 2ab Consider, a2 +b2 −c2 a2 +c2 −b2 b cos C + c cos B = b. + c. 2ab 2ac a2 +b2 −c2 a2 +c2 −b2 b cos C + c cos B = + 2a 2a a2 +b2 −c2 +a2 +c2 −b2 b cos C + c cos B = 2a 2a2 b cos C + c cos B = 2a b cos C + c cos B = a Similarly, we can prove (i) b = c cos A + a cos C (iii) c = a cos B + b cos A B−C (b−c) A Napier’s Analogy: In ∆ABC, tan ( ) = (b+c) cot 2 2 Proof: By sine rule b = k sin B and c = k sinC (b−c) k sin B −k sinC ∴ (b+c) = k sin B+k sinC Join Telegram:- @NOTESPROVIDER12TH_BOARD (b−c) sin B − sinC ∴ (b+c) = sin B+ sinC B−C B+c (b−c) 2 sin( ).cos( 2 ) 2 ∴ (b+c) = B+C B−c 2 sin( 2 ).cos( 2 ) (b−c) B+C B−C ∴ (b+c) = cot ( ) tan ( ) 2 2 (b−c) π A B−C ∴ (b+c) = cot (2 − 2 ) tan ( ) 2 (b−c) A B−C ∴ (b+c) = tan (2 ) tan ( ) 2 B−C (b−c) A ∴ tan ( ) = (b+c) cot 2 2 Similarly we can prove that C−A (c−a) B tan ( ) = (c+a) cot 2 2 3  Join Telegram ;- @NOTESPROVIDER12TH_BOARD A−B (a−b) C tan ( ) = (a+b) cot 2 2 RECALL A homogeneous equation of degree two in x and y, ax 2 + 2hxy + by 2 = 0 represents a pair of lines passing through the origin if h2 − ab ≥ 0. −h−√h2 −ab −h+√h2 −ab Slopes of these lines are m1 = and m2 = b b 2h The sum of slope is, m1 + m2 = − b a Product of slope is m1 m2 = b 2√h2 −ab The difference of slope is |m1 − m2 | = b Theorem: HomogeneousJoin equation degree two in 𝐱 and y, 𝐚𝐱𝟐 + 𝟐𝐡𝐱𝐲 + 𝐛𝐲 𝟐 = 𝟎 of@NOTESPROVIDER12TH_BOARD Telegram:- represents a pair of lines passing through the origin if 𝐡𝟐 − 𝐚𝐛 ≥ 𝟎. Proof : Consider the homogeneous equation of degree two in x and y, ax 2 + 2hxy + by 2 = 0 ⋯(I) Here a, b and h are constants not all zero. Consider two cases (ii) b = 0 ( or a = 0 ) (iii) b ≠ 0 ( or a ≠ 0 ) Case ( i ) 𝐛=𝟎 Equation ( I ) becomes ax 2 + 2hxy = 0 ∴ x(ax + 2hy) = 0 ∴ x = 0 or ax + 2hy = 0. 4  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Above are the combined equation of line y- axis and ax + 2hy = 0 both passes through origin. Case ( ii ) 𝐛≠𝟎 ax 2 + 2hxy + by 2 = 0 by 2 + 2hxy = −ax 2 Multiplying both side by b b2 y 2 + 2hbxy = −abx 2 To make L.H.S. complete square we add h2 x 2 to both sides. b2 y 2 + 2hbxy + h2 x 2 = h2 x 2 − abx 2 (by + hx)2 = (h2 − ab)x 2 Taking square roots on both side by + hx = ±√h2 − ab x Join Telegram:- @NOTESPROVIDER12TH_BOARD by = −hx ± √h2 − ab x by = (−h ± √h2 − ab )x (−h±√h2 −ab ) y= x as b ≠ 0 b (−h−√h2 −ab ) (−h+√h2 −ab ) y= x and y = x h h Above are the two equations of the lines in the form y = mx both passes through origin. Therefore the equation ax 2 + 2hxy + by 2 = 0 represents a pair of lines passing through the origin if h2 − ab ≥ 0. Theorem: The acute angle 𝛉 between the lines represented by 𝐚𝐱𝟐 + 𝟐𝐡𝐱𝐲 + 𝐛𝐲𝟐 = 𝟎 is 𝟐√𝐡𝟐 −𝐚𝐛 given by 𝐭𝐚𝐧 𝛉 = | |. 𝐚+𝐛 Proof: Let m1 and m2 be slopes of lines represented by the equation 5  Join Telegram ;- @NOTESPROVIDER12TH_BOARD ax 2 + 2hxy + by 2 = 0. −h−√h2 −ab −h+√h2 −ab ∴ m1 = and m2 = b b As θ is the acute angle between the lines, 1m −m2 tanθ = |1+m | 1 m2 −h−√h2 −ab −h+√h2 −ab |m1 −m2 | = | − | b b −h−√h2 −ab+h−√h2 −ab =| b | −2√h2 −ab =| | b 2√h2 −ab =| | b 2√h2 −ab b ∴ tan θ = | a | 1+b Join Telegram:- @NOTESPROVIDER12TH_BOARD 2√h2 −ab b ∴ tan θ = | b+a | b 2√h2 −ab ∴ tan θ = | |. a + b ≠ 0 a+b Remark: 1) Lines represented by ax 2 + 2hxy + by 2 = 0 are coincident if and only if m1 = m2 ∴ m1 − m2 = 0 2√h2 −ab ∴ =0 b ∴ h2 − ab = 0 ∴ h2 = ab Thus lines represented by ax 2 + 2hxy + by 2 = 0 are coincident if and only if 𝐡𝟐 = 𝐚𝐛. 6  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 2) Lines represented by ax 2 + 2hxy + by 2 = 0 are perpendicular if and only if m1 m2 = −1. a ∴ b = −1 ∴ a = −b ∴a+b= 0 Thus lines represented by ax 2 + 2hxy + by 2 = 0 are perpendicular to each other if and only if 𝐚 + 𝐛 = 𝟎. Theorem: Let 𝐚̅ and 𝐛̅ be non-collinear vectors. A vector 𝐫̅ is coplanar with 𝐚̅ and 𝐛̅ if and only if there exist unique scalars 𝐭 𝟏 and 𝐭 𝟐 such that 𝐫̅ = 𝐭 𝟏 𝐚̅ + 𝐭 𝟐 𝐛̅. Proof: Only If Part : Given: Suppose r̅ is coplanar a̅ and b̅. with@NOTESPROVIDER12TH_BOARD Join Telegram:- To show: There exist unique scalars t1 and t 2 such that r̅ = t1 a̅ + t 2 b̅. Let a̅ be along OA and b̅ be along OB. Let ̅̅̅̅ OP = r̅ , Draw lines parallel to OB, meeting OA in M and parallel to OA, meeting OB in N. ON = t 2 b̅ and ̅̅̅̅̅ Then ̅̅̅̅ OM = t1 a̅ for some t1 , t 2 ∈ R By triangle law or parallelogram law We have r̅ = t1 a̅ + t 2 b̅ If – Part: Given: Suppose r̅ = t1 a̅ + t 2 b̅ To show: r̅ , a̅ and b̅ are coplanar. 7  Join Telegram ;- @NOTESPROVIDER12TH_BOARD As a̅ and b̅ are coplanar t1 a̅ , t 2 b̅ are also coplanar. Therefore t1 a̅ + t 2 b̅ ,a̅ , b̅ are coplanar. Therefore a̅ , b̅ and r̅ are coplanar. Uniqueness : Suppose vector r̅ = t1 a̅ + t 2 b̅ - - - - - - - (1) Can also be written as r̅ = S1 a̅ + S2 b̅ - - - - - - - - (2) Subtracting (2) from (1) we get ̅ = (t1 − S1 )a̅ + (t 2 − S2 )b̅ 0 But a̅ and b̅ are non-collinear vectors ∴ t1 − S1 = t 2 − S2 = 0 ∴ t1 = S1 and t 2 = S2 Therefore the uniqueness follows. Join Telegram:- @NOTESPROVIDER12TH_BOARD Theorem: Three vectors 𝐚̅ , 𝐛̅ and 𝐜̅ are coplanar if and only if there exists a non- zero linear combination 𝐱 𝐚̅ + 𝐲𝐛̅ + 𝐳𝐜̅ = ̅ 𝟎 with (𝐱, 𝐲, 𝐳) ≠ (𝟎, 𝟎, 𝟎) Proof: Only If – Part: Assume a̅ , b̅ and c̅ are coplanar. Case – 1: Suppose that any two of a̅ , b̅ and c̅ are collinear vectors, say a̅ and b̅. ∴ There exist scalars x & y at least one of which is non zero such that xa̅ + yb̅ = 0 ̅ i.e. xa̅ + yb̅ + 0c̅ = 0 ̅ and (x , y, 0) is the required Solution for xa̅ + yb̅ + zc̅ = 0̅. Case – 2 :No two vectors a̅ , b̅ and c̅ are collinear. As c̅ is coplanar with a̅ and b̅. ∴ We have scalars x , y such that c̅ = xa̅ + yb̅. ∴ xa̅ + yb̅ − c̅ = 0 ̅ and (x, y, −1) is the required Solution for xa̅ + yb̅ + zc̅ = 0 ̅. 8  Join Telegram ;- @NOTESPROVIDER12TH_BOARD If – Part: Conversely suppose that xa̅ + yb̅ + zc̅ = 0 ̅ where one of x , y , z is non zero, say x y z ≠ 0. ∴ c̅ = (− z) a̅ + (− z) b̅ ∴ c̅ is coplanar with a̅ and b̅. ∴ a̅ , b̅ and c̅ are coplanar vectors. Section Formula: Theorem: (Section formula for internal division) Let A(a̅) and B(b̅) be any two points in the space and R(r̅ ) be a point on the line segment AB ̅ +na̅ mb dividing it internally in the ratio m ∶ n then r̅ = m+n Proof :As R is a point on the line segment AB (A-R-B) and ̅̅̅̅ AR and ̅̅̅̅ RB are in same direction. AR m = n, RB Join Telegram:- @NOTESPROVIDER12TH_BOARD n(AR) = m(RB) ̅̅̅̅ have same direction and magnitude, ̅̅̅̅ and nAR As mRB ̅̅̅̅ ̅̅̅̅ = nAR ∴ mRB ̅̅̅̅ − OR ∴ m(OB ̅̅̅̅) = n(OR ̅̅̅̅ − OA ̅̅̅̅) ∴ m(b̅ − r̅ ) = n(r̅ − a̅) ∴ mb̅ − mr̅ = nr̅ − na̅ ∴ mb̅ + na̅ = mr̅ + nr̅ = (m + n)r̅ ̅ +na̅ mb ∴ r̅ = m+n Theorem: (Section formula for External Division) Let 𝐀(𝐚̅) and 𝐁(𝐛̅) be any two points in the space and 𝐑(𝐫̅ ) be a point on the line 𝐦𝐛̅−𝐧𝐚̅ segment 𝐀𝐁 dividing it externally in the ratio 𝐦 ∶ 𝐧 then 𝐫̅ = 𝐦−𝐧 9  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Proof: As the point R divides the line segment AB externally, we have either A − B − R or R−A−B Assume that A − B − R AR ∶ BR = m ∶ n AR m ∴ BR = n n(AR) = m(BR) ̅̅̅̅) and m(BR As n(AR ̅̅̅̅) have same magnitude and direction ∴ n ̅̅̅̅ ̅̅̅̅ AR = mBR ∴ n(r̅ − a̅) = m(r̅ − b̅Join ) Telegram:- @NOTESPROVIDER12TH_BOARD ∴ nr̅ − na̅ = mr̅ − mb̅ ∴ mb̅ − na̅ = mr̅ − nr̅ = (m − n)r̅ ̅ −na̅ mb ∴ r̅ = m−n Theorem: Prove that medians of a triangle are concurrent. Solution: Let A, B and C be the vertices of the triangle ABC. Let D, E and F be the mid-points of the sides BC, AC and AB respectively. Let a̅, b̅, c̅, d ̅ , e̅ and f̅ be the position vectors of the points A, B, C, D, E and F respectively. ∴ By mid-point formula, 10  ̅ ̅ Join Telegram ;- @NOTESPROVIDER12TH_BOARD ̅ = b+c̅ , e̅ = a̅+c̅ , f̅ = a̅+b ∴ d 2 2 2 ̅ = b̅ + c̅ , 2e̅ = a̅ + ̅c , ∴ 2d 2f̅ = a̅ + b̅ ̅ + a̅ = a̅ + b̅ + c̅ , 2e̅ + b̅ = a̅ + b̅ + ̅c , 2f̅ + ̅c = a̅ + b̅ + ̅c ∴ 2d ̅ +a̅ 2d ̅ 2e̅+b 2f̅+ ̅c ̅ +c̅ a̅+b ∴ = = = = g̅ (say) 3 3 3 3 ̅ +c̅ a̅+b ̅ +(1)a̅ (2)d (2)e ̅ ̅ +(1)b (2)f̅+(1) ̅c Then we have g̅ = = = = 3 2+1 2+1 2+1 If G is the point whose position vector is g̅ , then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE , CF internally in the ratio 2 ∶ 1. Therefore the medians of a triangle are concurrent. Prove that the angle bisectors of a triangle are concurrent. Solution: Join Telegram:- @NOTESPROVIDER12TH_BOARD Let A, B and C be vertices of triangle. Let AD, BE and CF be the angle bisectors of the triangle ABC. Also AB = z , BC = x , AC = y. Now the angle bisector AD meets the side BC at the point D. Here point D divides the line segment BC internally in the ratio AB : AC i.e. z : y. Let a̅ , b̅ , c̅, d ̅ , e̅ and f̅ be the position vectors of the points A, B, C, D, E and F respectively. Hence by section formula for internal division, ̅ ̅ +xa̅ ̅ = zc̅+yb Similarly we get , e̅ = xa̅+zc̅ We have d and f̅ = yb z+y x+z y+x ̅ = zc̅ + yb̅ ∴ (z + y)d 11  Join Telegram ;- @NOTESPROVIDER12TH_BOARD ̅ + xa̅ = xa̅ + zc̅ + yb̅ ∴ (z + y)d Similarly (x + z)e̅ + yb̅ = xa̅ + zc̅ + yb̅ And (x + y)f̅ + xc̅ = xa̅ + zc̅ + yb̅ ̅ + xa̅ = (x + z)e̅ + yb̅ = (x + y)f̅ + xc̅ = xa̅ + zc̅ + yb̅ ∴ (z + y)d ̅ +xa (z+y)d ̅ (x+z)e ̅ ̅ +yb (x+y)f̅+xc̅ ̅ xa̅+zc̅+yb ∴ = = = = h̅ (say) x+y+z x+y+z x+y+z x+y+z (z+y)d+xa ̅ ̅ (x+z)e ̅ +yb (x+y)f+xc̅ ̅ ̅ Then we have h̅ = (z+y)+x = (x+z)+y = (x+y)+z That is point H(h̅) divides AD in the ratio (y + z): x, BE in the ratio (x + z): y and CF in the ratio (x + y): z. This shows that the point H is the point of concurrence of the angle bisectors AD, BE and CF of the triangle ABC. Thus, the angle bisectors of a triangle are concurrent. Prove that the median of a trapezium is parallel to sides of the trapezium and its length is Join Telegram:- @NOTESPROVIDER12TH_BOARD half the sum of parallel sides. Solution: Suppose that the ABCD is trapezium with AD and BC are parallel sides. A B Let E be the mid-point of side AD and F be the mid-point of side BC. ̅EF ̅̅̅ = ̅EF ̅̅̅ = 1 ̅̅̅̅ + (mDC 1 ̅̅̅̅ + (mDC 2 2F Here side EF is the median of the trapezium ABCD. E ̅̅̅̅ DC ) = ̅̅̅̅ DC̅EF̅̅̅)== ̅̅̅̅ 1=(m + EF 1 1(m + ̅̅̅̅ + Also ̅̅̅̅ 𝐴𝐵 is parallel to ̅̅̅̅ 𝐷𝐶. 1 2 (mDC ̅̅̅̅ ̅̅̅̅+ 2 (mDC 2 D2 1)DC = 1)DC ̅̅̅̅ DC ) = ̅̅̅̅C = ̅̅̅̅kDC)̅̅̅̅ ̅̅̅DC̅ = kDC1̅̅̅̅ ̅EF ̅̅̅ = Let a̅ , b̅, c̅, d ̅ , e̅ and f̅ be the position vectors of the points A, B, C, D, E and EF F. 1 = 1 (m , + (m + , 2 1 (mDC 2(mDC ̅̅̅̅ ̅̅̅̅ 2 ̅̅̅̅ w + w1)DC 2̅̅̅̅ = We have to show that the median EF is parallel to the sides AB and DC. ̅̅̅̅ )h= = DC 1)DC ̅̅̅̅ DC ) = ̅̅̅̅ ̅̅̅̅ hkDC 1 kDC 1 2, (me+ e, 2 (m + But it is sufficient to show that EF is parallel to AB or DC. r w̅̅̅̅ = 1)DC r w1)DC ̅̅̅̅ = e h ̅̅̅̅ kDCh̅̅̅̅ e kDC e, As ̅̅̅̅ AB is parallel to ̅̅̅̅ DC. ,e k kr w wr eh For any scalar m we have he i i e e s sk r rk e e 12 s s i i c cs k ks a a  Join Telegram ;- @NOTESPROVIDER12TH_BOARD ̅̅̅̅ AB = m ̅̅̅̅ DC ….. (i) ̅ a̅+d ̅ b+c̅ Now by mid-point formula e̅ = and f̅ = 2 2 ̅̅ = f̅ − e̅ Therefore, ̅̅ EF ̅ +c̅ b ̅ a̅+d = − 2 2 1 = 2 (b̅ + c̅ − a̅ − d ̅) 1 = 2 (b̅ − a̅ + c̅ − d ̅) ̅̅ = 1 (AB ̅̅ EF ̅̅̅̅ + ̅̅̅̅ DC ) …..(ii) 2 ̅̅ = 1 (mDC ̅̅ EF ̅̅̅̅ + ̅̅̅̅ 1 ̅̅̅̅ = kDC DC ) = 2 (m + 1)DC ̅̅̅̅̅, where k is scalar 2 ̅̅ is scalar multiple of ̅̅̅̅ ∴ ̅̅ EF DC ̅̅̅̅. ̅̅̅̅ is parallel to DC ∴ EF 1 Join Telegram:- @NOTESPROVIDER12TH_BOARD ̅̅̅̅ + DC ̅̅̅̅ = (AB Also, (ii) → EF ̅̅̅̅ ) 2 ̅̅̅̅| = 1 |AB |EF ̅̅̅̅ + DC ̅̅̅̅| 2 1 ∴ l(median) = [ sum of length of sode AB and DC] 2 Prove by vector method that the angle subtended on semicircle is a right angle. Solution: Consider a circle with center Origin O and radios r. Let A and B be the end points of the diameter AB of the circle. Let C be any point on the circumference. Let a̅, b̅ and c̅ be the position vectors of the points A, B and C respectively. ∴ From figure,|a̅ | = |b̅ | = |c̅ | = r 13  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Also, b̅ = −a̅ Consider, ̅̅̅̅ CB = (a̅ − c̅) ∙ (b̅ − c̅) CA ∙ ̅̅̅̅ = (a̅ − c̅) ∙ (−a̅ − c̅) = −(a̅ − c̅) ∙ (a̅ + c̅) = −[ |a̅|2 − |c̅|2 ] = −[r 2 − r 2 ] As |a̅| = |c̅| = r =0 ̅̅̅̅ and CB ∴ CA ̅̅̅̅ are perpendicular to each other. ∴ m∠ACB = 90o ∴ Angle subtended on semicircle is a right angle. Join Telegram:- @NOTESPROVIDER12TH_BOARD 14  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Theorem: If y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x such that the composite function 𝐲 = 𝐟[𝐠(𝐱)]is a differentiable function of x 𝐝𝐲 𝐝𝐲 𝐝𝐮 then 𝐝𝐱 = 𝐝𝐮 × 𝐝𝐱 Proof: Let δx be the small increment in x and δu and δy are the corresponding increments in u and y respectively. As y = f (u) is a differentiable function of u. dy δy ∴ du = lim …..(i) δu→0 δu Also, u = g (x) is a differentiable function of x du δu ∴ dx = lim …..(ii) δx→0 δx Consider, δy δy δu = δu × δx δx Taking the limit as δx → 0JoinonTelegram:- both sides @NOTESPROVIDER12TH_BOARD We get, δy δy δu lim = lim (δu × δx ) δx→0 δx δx→0 δy δy δu lim = lim (δu ) × lim (δx ) δx→0 δx δx→0 δx→0 δy δy du lim = lim (δu ) × dx by (ii) δx→0 δx δx→0 As δx → 0, we get, δu → 0 ( ∵ u is a continuous function of x) δy δy du lim = lim × dx δx→0 δx δu→0 δu δy dy du lim = dx × dx (1) by (i) δx→0 δx ∴RHS of (1) is differentiable implies LHS of (1) is also differentiable. δy dy lim = dx δx→0 δx Then equation (1) becomes, dy dy du = du × dx dx 15  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Theorem: Suppose 𝐲 = 𝐟(𝐱)is a differentiable function of 𝐱 on an interval I and y is One- 𝐝𝐲 𝐝𝐱 𝟏 𝐝𝐲 one, onto and 𝐝𝐱 on I. Also if 𝐟 −𝟏 (𝐲) is differentiable on 𝐟(𝐈) then𝐝𝐲 = 𝐝𝐲 , 𝐝𝐱 ≠ 𝟎 𝐝𝐱 Proof: Let δx be the small increment in x and δy is the corresponding increments y. As y = f(x) is a differentiable function of x. dy δy ∴ dx = lim …..(i) δx→0 δx Consider, δx  y  =1 y x x 1 y  = , where  0 y y x x Taking the limit as δx → 0 on both sides We get, Join Telegram:- @NOTESPROVIDER12TH_BOARD δx 1 lim = lim ( δy ) δx→0 δy δx→0 δx δx 1 lim =( δy ) δx→0 δy lim δx→0 δx δx 1 dy lim = ( dy ), dx ≠ 0 by (i) δx→0 δy dx As δx → 0 implies that δy → 0 δx 1 lim = ( dy ) …..(1) δy→0 δy dx As RHS of (1) is differentiable implies LHS of (1) is also differentiable. dx 1 dy ∴ = where  0 dy dy dx dx 16  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Theorem: If x = f(t) and y = g(t) are differentiable functions of t so that y is a differentiable 𝐝𝐲 𝐝𝐱 𝐝𝐲 𝐝𝐭 function of x and if 𝐝𝐭 ≠ 𝟎 𝐭𝐡𝐞𝐧 = 𝐝𝐱. 𝐝𝐱 𝐝𝐭 Proof: Let δt be the small increment in t say then δx and δy are corresponding increments in the x and y respectively. As x and y are differentiable functions of t, We have, dx lim δx ∴ = ----- (i) dt δt → 0 δt dy lim δy = ----- (ii) dt δt → 0 δt Consider, δy δy δt δx Join Telegram:- @NOTESPROVIDER12TH_BOARD = δx ∵ ≠ 0. δx δt δt Taking the limit δt → 0 on both sides we get, lim δy lim δy δt→0δt = lim δx δt → 0 δx δt δt→0 dy lim δy dt = dx -----(1) by (i) and (ii) δt → 0 δx dt As RHS of (1) is differentiable implies LHS of (1) is also differentiable. dy lim δy = dx δt → 0 δx Thus the equation (1) becomes, dy dy dx dx = dt dx Where dt ≠ 0 dt 17  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 𝒇 ′ (𝒙) 𝐟 ′ (𝐱) Theorem: ∫ 𝒅𝒙 = 2 √𝒇(𝒙) + c √𝒇(𝒙) Theorem: ∫ 𝐝𝐱 = log |𝐟(𝐱)| + 𝐜 𝐟(𝐱) Proof: Consider, Proof: Consider, d 1 d 1 (2 √f(x) + c) = 2 ∙ ∙ f ′ (x) (log |f(x)| + c ) = ∙ f′(x) dx 2f(x) dx f(x) ∴ By definition of integral ∴ By definition of integral 1 1 ∫ ∙ f ′ (x) dx = log |f(x)| + c ∫2∙ ∙ f ′ (x)dx = 2 √f(x) + c f(x) 2f(x) f ′ (x) f ′ (x) ∫ dx = 2 √f(x) + c ∫ dx = log |f(x)| + c f(x) f(x) 𝟏 𝟏 𝐱 Theorem: ∫ 𝐝𝐱 = 𝐭𝐚𝐧−𝟏 ( ) + 𝐜 𝐱 𝟐 +𝐚𝟐 𝐚 𝐚 Proof: Consider, Join Telegram:- @NOTESPROVIDER12TH_BOARD 𝑑 1 𝑥 𝑑 1 𝑥 𝑑 [ tan−1 (𝑎) + 𝑐] = [ tan−1 (𝑎)] + 𝑐 𝑑𝑥 𝑎 𝑑𝑥 𝑎 𝑑𝑥 𝑑 1 𝑥 1 1 𝑑 𝑥 [ tan−1 ( ) + 𝑐] = ( )+0 𝑑𝑥 𝑎 𝑎 𝑎 𝑥 2 𝑑𝑥 𝑎 1 + (𝑎 ) 𝑑 1 𝑥 1 1 1 [ tan−1 (𝑎) + 𝑐] = 𝑥2 (𝑎 ) 𝑑𝑥 𝑎 𝑎 1+ 2 𝑎 𝑑 1 𝑥 1 1 [ tan−1 ( ) + 𝑐] = 2 2 𝑑𝑥 𝑎 𝑎 𝑎 𝑎 + 𝑥2 𝑎2 𝑑 1 𝑥 1 [ tan−1 ( ) + 𝑐] = 2 𝑑𝑥 𝑎 𝑎 𝑥 + 𝑎2 ∴ By definition of integration, 1 1 −1 𝑥 ∴∫ 𝑑𝑥 = tan ( )+𝑐 𝑥 2 + 𝑎2 𝑎 𝑎 18  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 𝟏 𝟏 𝐱−𝐚 Theorem:∫ 𝐱 𝟐 −𝐚𝟐 𝐝𝐱 = 𝐥𝐨𝐠 (𝐱+𝐚) + 𝐜 𝟐𝐚 1 Proof: Let , I = ∫ x2 −a2 dx 1 I= ∫ dx (x + a)(x − a) 1 A B Consider, (x+a)(x−a) = (x+a) + (x−a) ∴ 1 = A(x − a) + B((x + a)) Put x = −a then we get, 1 = A(−a − a) 1 A = − 2a Put x = a then we get, 1 = B(a + a) 1 B = 2a Join Telegram:- @NOTESPROVIDER12TH_BOARD 1 1 1 A B − 2a ∫ dx = ∫ ( + ) dx = ∫ ( + 2a ) dx (x + a)(x − a) (x + a) (x − a) (x + a) (x − a) 1 1 1 1 1 ∫ dx = ∫ ( − ) dx = [log|x − a| − log|x + a|] + c (x + a)(x − a) 2a (x − a) (x + a) 2a 1 1 x−a ∫ dx = log | |+c (x + a)(x − a) 2a x+a 1 1 x−a ∴∫ dx = log ( )+c x 2 − a2 2a x+a 𝟏 𝟏 𝐚+𝐱 Theorem:∫ 𝐚𝟐 −𝐱 𝟐 𝐝𝐱 = 𝐥𝐨𝐠 (𝐚−𝐱) + 𝐜 𝟐𝐚 1 Proof: Let , I = ∫ a2 −x2 dx 1 I= ∫ dx (a + x)(a − x) 19  1 A B Join Telegram ;- @NOTESPROVIDER12TH_BOARD Consider, (a+x)(a−x) = (a+x) + (a−x) ∴ 1 = A(a − x) + B(a + x) Put x = −a then we get, 1 = A(a + a) 1 A = 2a Put x = a then we get, 1 = B(a + a) 1 B = 2a 1 1 1 A B 2a 2a ∫ dx = ∫ ( + ) dx = ∫ ( + ) dx (a + x)(a − x) (a + x) (a − x) (a + x) (a − x) 1 1 1 1 1 ∫ dx = ∫ ( + ) dx = [log|𝑎 + 𝑥| − log|a − x|] + c (a + x)(a − x) 2a (𝑎 + 𝑥) (a − x) 2a 1 1 a+x ∫ dx = log | |+c (a + x)(a − x) 2a a−x Join Telegram:- @NOTESPROVIDER12TH_BOARD 1 1 a+x ∴∫ dx = log ( )+c a2 − x 2 2a a−x 𝟏 Theorem: ∫ 𝐝𝐱 = 𝐥𝐨𝐠(𝐱 + √𝐱𝟐 − 𝐚𝟐 ) + 𝐜 √𝐱 𝟐 −𝐚𝟐 1 Proof:Let I = ∫ 2 2 dx √x −a Put x = a sec θ x ie θ = sec −1 ( ) a ∴ dx = asecθ ∙ tanθ ∙ dθ 1 I= ∫ a ∙ secθ ∙ tanθ ∙ dθ √a2 sec 2 θ − a2 asecθ ∙ tanθ = ∫ dθ √a2 (sec 2 θ − 1) 20  Join Telegram ;- @NOTESPROVIDER12TH_BOARD a ∙ secθ ∙ tanθ = ∫ dθ √a2 tan2 θ a ∙ secθ ∙ tanθ = ∫ dθ a ∙ tan θ = ∫ secθ dθ = log ( sec θ + tan θ ) + C = log ( sec θ + √tan2 θ ) + C = log ( sec θ + √sec 2 θ − 1) + C x x2 √ = log ( + 2 − 1) + C a a x + √x 2 − a2 = log ( )+C a Join Telegram:- @NOTESPROVIDER12TH_BOARD = log (x + √x 2 − a2 ) − log a + C = log (x + √x 2 − a2 ) + c Where C − loga = c 1 ∴∫ dx = log (x + √x 2 − a2 ) + c √x 2 − a2 𝟏 Theorem: ∫ 𝐝𝐱 = 𝐥𝐨𝐠(𝐱 + √𝐱𝟐 + 𝐚𝟐 ) + 𝐜 √𝐱 𝟐 +𝐚𝟐 1 Proof:Let I = ∫ 2 2 dx √x +a Put x = a tan θ x ie θ = tan−1 ( ) a ∴ dx = a ∙ sec 2 θ ∙ dθ 21  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 1 2 I= ∫ a ∙ sec θ ∙ dθ √a2 tan2 θ + a2 a sec 2 θ = ∫ dθ √a2 (tan2 θ + 1) a ∙ secθ ∙ tanθ = ∫ dθ √a2 tan2 θ a ∙ sec 2 θ = ∫ dθ a ∙ sec θ = ∫ secθ dθ = log ( sec θ + tan θ ) + C = log ( tan θ + √sec 2 θ ) + C = log ( tan θ + √tan2 θ + 1) + C x x 2 Join Telegram:- @NOTESPROVIDER12TH_BOARD √ = log ( + 2 + 1) + C a a x + √x 2 + a2 = log ( )+C a = log (x + √x 2 + a2 ) − log a + C = log (x + √x 2 + a2 ) + c Where C − loga = c 1 ∴∫ dx = log (x + √x 2 + a2 ) + c √x 2 + a2 22  Join Telegram ;- @NOTESPROVIDER12TH_BOARD Theorem: If u and v two differentiable functions of x then 𝐝𝐮 ∫ 𝐮 𝐯 𝐝𝐱 = 𝐮 ∫ 𝐯 𝐝𝐱 − ∫ [𝐝𝐱 ∫ 𝐯 𝐝𝐱] 𝐝𝐱 Proof: let ∫ v dx = w … (i) dw ∴v= … (ii) dx Consider, d d d (u. w) = u w + w dx u dx dx du = uv + w dx By definition of integration du (u. w) = ∫ [uv + w ] dx dx du = ∫ uv dx + ∫ w dx dx Join Telegram:- du @NOTESPROVIDER12TH_BOARD u ∫ v dx = ∫ uv dx + ∫ dx w dx du ∫ u v dx = u ∫ v dx − ∫ [dx ∫ v dx] dx 𝐱 𝐚𝟐 𝐱 Theorem: ∫ √𝐚𝟐 − 𝐱𝟐 𝐝𝐱 = √𝐚𝟐 − 𝐱 𝟐 + 𝐬𝐢𝐧−𝟏 (𝐚) + 𝐜 𝟐 𝟐 Proof: Let I =  √a2 − x 2 ∙ 1dx d(√a2 −x2 ) = √a2 − x 2  1 dx −  [ ∙ ∫ 1𝑑𝑥] 𝑑𝑥 dx 1 = √a2 − x 2 x − ∫ (−2x)(x)dx 2√a2 −x2 x2 = √a2 − x 2 x +  dx √a2 −x2 a2 − ( a2 −x2) = √a2 − x 2 x +  dx √a2 −x2 a2 ( a2 −x2 ) = √a2 − x 2 x + ∫ [√a2 −x2 − ] dx √a2 −x2 23  1 Join Telegram ;- @NOTESPROVIDER12TH_BOARD = x √a2 − x 2 + a2  dx − √a2 − x 2 dx √a2 −x2 1 I = x √a2 − x 2 + a2  dx – I √a2 −x2 x I+I= x √a2 − x 2 + a2 sin−1 (a) + c x a2 x I= √a2 − x 2 + sin−1 (a) + c 2 2 x a2 x   √a2 − x 2 dx = √a2 − x 2 + sin−1 (a) + c 2 2 𝐱 𝐚𝟐 Theorem: ∫ √𝐱𝟐 − 𝐚𝟐 𝒅𝒙 = √𝐱𝟐 − 𝐚𝟐 − 𝐥𝐨𝐠|𝒙 + √𝐱𝟐 − 𝐚𝟐 | + 𝐜 𝟐 𝟐 Proof: Let I =  √x 2 − a2 ∙ 1dx d(√x2 −a2 ) = √x 2 − a2  1 dx −  [ ∙ ∫ 1dx] dx dx 1 = √x 2 − a2 x − ∫ (2x)(x)dx 2√x2 −a2 Join Telegram:- x2 @NOTESPROVIDER12TH_BOARD = √x 2 − a2 x − ∫ dx √x2 −a2 x2 −a2 +a2 = √x 2 − a2 x − ∫ dx √x2 −a2 x2 −a2 a2 = √x 2 − a2 x − ∫ [ + ] dx √x2 −a 2 √x2 −a2 1 = √x 2 − a2 x − ∫ √x 2 − a2 dx − a2  dx √x2 −a2 I = x √x 2 − a2 − I − a2 log|x + √x 2 − a2 | I + I = x√x 2 − a2 − a2 log|x + √x 2 − a2 | 2I = x√x 2 − a2 − a2 log|x + √x 2 − a2 | x a2  I= √x 2 − a2 − log|x + √x 2 − a2 | + c 2 2 24  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 𝐱 𝐚𝟐 Theorem: ∫ √𝐱𝟐 + 𝐚𝟐 𝒅𝒙 = √𝐱𝟐 + 𝐚𝟐 + 𝟐 𝟐 𝐥𝐨𝐠|𝒙 + √𝐱 + 𝐚 | + 𝐜 𝟐 𝟐 Proof: Let I =  √x 2 + a2 ∙ 1dx d(√x2 +a2 ) = √x 2 + a2  1 dx −  [ ∙ ∫ 1dx] dx dx 1 = √x 2 + a2 x − ∫ (2x)(x)dx 2√x2 +a2 x2 = √x 2 + a2 x − ∫ dx √x2 +a2 x2 +a2 −a2 = √x 2 + a2 x − ∫ dx √x2 −a2 x2 +a2 a2 = √x 2 + a2 x − ∫ [ − ] dx √x2 +a 2 √x2 +a2 1 = √x 2 + a2 x − ∫ √x 2 + a2 dx + a2  dx √x2 +a2 I = x √x 2 + a2 − I + a2 log|x + √x 2 + a2 | I + I = x√x 2 + a2 Join + aTelegram:- 2 √x 2 + a2 | log|x +@NOTESPROVIDER12TH_BOARD 2I = x√x 2 + a2 + a2 log|x + √x 2 + a2 | x a2 I= √x 2 + a2 + log|x + √x 2 + a2 | + c 2 2 Theorem: ∫ 𝒆𝒙 ∙ [𝒇(𝒙) + 𝒇′ (𝒙)]𝒅𝒙 = 𝒆𝒙 𝒇(𝒙) + 𝒄 Proof: Consider, d (ex ∙ f(x) + c) = ex ∙ f ′ (x) + ex ∙ f(x) dx d (ex ∙ f(x) + c) = ex ∙ [f(x) + f ′ (x) ] dx ∴By definition of integral ∫ ex ∙ [f(x) + f ′ (x)]dx = ex ∙ f(x) + c 25  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 𝐚 𝐛 𝐚 Property I: ∫𝐚 𝐟(𝐱)𝐝𝐱 = 𝟎 Property II:∫𝐚 𝐟(𝐱)𝐝𝐱 = − ∫𝐛 𝐟(𝐱)𝐝𝐱 Proof: Let ∫ f(x)dx = g(x) + c Proof: If ∫ f(x)dx = g(x) + c a b ∴ ∫a f(x) dx = [g(x) + c]aa ∴ ∫a f(x)dx = g(b) − g(a) = [ g(a) − g(a)] = −[g(a) − g(b)] a =0 = − ∫b f(x)dx b a Thus ∫a f(x)dx = − ∫b f(x)dx 𝐛 𝐛 Property III: ∫𝐚 𝐟(𝐱)𝐝𝐱 = ∫𝐚 𝐟(𝐭)𝐝𝐭 Property IV: 𝐛 𝐜 𝐛 Proof: Let ∫ f(x)dx = g(x) + c ∫𝐚 𝐟(𝐱)𝐝𝐱 = ∫𝐚 𝐟(𝐱)𝐝𝐱 + ∫𝐜 𝐟(𝐱)𝐝𝐱 b Join Telegram:- @NOTESPROVIDER12TH_BOARD ∫ f(x)dx = g(b) − g(a) Where 𝐚 < 𝐜 < 𝐛 i.e. 𝐜 ∈ [𝐚, 𝐛] a = [(g(b) + c) − ( g(a) + c] Proof: Let ∫ f(x)dx = g(x) + c c b ∫a f(x)dx + ∫c f(x)dx = [g(x) + c]ca + [g(x) + c]bc = g(b) − g(a) --- (i) b = [(g(c) + c) − (g(a) + c)] + ∫ f(t)dt = [g(t) + c]ba [(g(b) + c) − (g(c) + c)] a = [(g(b) + c) − ( g(a) + c] = g(c) + c − g(a) − c + g(b) + c − g(c) − c = g(b) − g(a) --- (ii) = g(b) − g(a) From (i) and (ii) = [g(x) + c]ba b b b ∫a f(x)dx = ∫a f(t)dt = ∫a f(x)dx b c b Thus∫a f(x)dx = ∫a f(x)dx + ∫c f(x)dx where a < c < b 26  Join Telegram ;- @NOTESPROVIDER12TH_BOARD 𝐛 𝐛 Property V: ∫𝐚 𝐟(𝐱) 𝐝𝐱 = ∫𝐚 𝐟(𝐚 + 𝐛 − 𝐱) 𝐝𝐱 Proof: Let ∫ f(x) dx = g(x) + c Consider b R. H. S. = ∫a f(a + b − x) dx Put a + b − x = t i.e. x = a + b − t ∴ −dx = dt ⟹ dx = −dt When x = a ⟹ t = b When x = b ⟹ t = a a ∴ R. H. S. = ∫b f(t) (−dt) a = − ∫b f(t) dt b b a = ∫a f(t) dt - - - - - - - (∵ ∫a f(x) dx = − ∫b f(x)dx) Join Telegram:- @NOTESPROVIDER12TH_BOARD b b b = ∫a f(x) dx - - - - - - - (∵ ∫a f(x) dx = ∫a f(t)dt) = L. H. S. b b Thus, ∫a f(x) dx = ∫a f(a + b − x) dx 𝐚 𝐚 Property VI: ∫𝟎 𝐟(𝐱) 𝐝𝐱 = ∫𝟎 𝐟(𝐚 − 𝐱) 𝐝𝐱 Proof: Let ∫ f(x) dx = g(x) + c Consider, a R. H. S. = ∫0 f(a − x) dx Put a − x = t i.e. x = a − t ∴ −dx = dt ⟹ dx = −dt When x = 0 ⟹ t = a 27  Join Telegram ;- @NOTESPROVIDER12TH_BOARD When x = a ⟹ t = 0 0 ∴ R. H. S. = ∫a f(t) (−dt) 0 = − ∫a f(t) dt a b a = ∫0 f(t) dt - - - - - - - (∵ ∫a f(x) dx = − ∫b f(x)dx) a b b = ∫0 f(x) dx - - - - - - - (∵ ∫a f(x) dx = ∫a f(t)dt) = L. H. S. a a Thus, ∫0 f(x) dx = ∫0 f(a − x) dx 𝟐𝐚 𝐚 𝐚 Property VII: ∫𝟎 𝐟(𝐱) 𝐝𝐱 = ∫𝟎 𝐟(𝐱) 𝐝𝐱 + ∫𝟎 𝐟(𝟐𝐚 − 𝐱) 𝐝𝐱 Proof: Consider a a R. H. S. = ∫0 f(x) dx + ∫0 f(2a − x) dx R. H. S. = I1 + I2 -Join - - -Telegram:- - - - - - -@NOTESPROVIDER12TH_BOARD - - - - - - -- - - - - - - - (i) a Consider I2 = ∫0 f(2a − x) dx Put 2a − x = t i.e. x = 2a − t ∴ −dx = dt ⟹ dx = −dt When x = 0 ⟹ t = 2a When x = a ⟹ t = a a ∴ I2 = ∫2a f(t) (−dt) a ∴ I2 = − ∫2a f(t) dt 2a b a = ∫a f(t) dt - - - - - - - (∵ ∫a f(x) dx = − ∫b f(x)dx) 2a b b = ∫a f(x) dx - - - - - - - (∵ ∫a f(x) dx = ∫a f(t)dt) From equation (i) 28  a a a Join 2a Telegram ;- @NOTESPROVIDER12TH_BOARD ∫ f(x) dx + ∫ f(2a − x) dx = ∫ f(x) dx + ∫ f(x) dx 0 0 0 a 2a b c b = ∫0 f(x) dx ------ By ∫a f(x)dx = ∫a f(x)dx + ∫c f(x)dx 2a a a Thus, ∫0 f(x) dx = ∫0 f(x) dx + ∫0 f(2a − x) dx 𝐚 𝐚 Property VIII: ∫−𝐚 𝐟(𝐱) 𝐝𝐱 = 𝟐 ∫𝟎 𝐟(𝐱) 𝐝𝐱 , if 𝐟(𝐱) is even function. =𝟎 , if 𝐟(𝐱) is odd function. Proof: f(x) even function if f(−x) = f(x) And f(x) odd function if f(−x) = −f(x) a 0 a ∫−a f(x) dx = ∫−a f(x) dx + ∫0 f(x) dx - - - - - - - - - - - - - - - (i) 0 Consider, I = ∫−a f(x) dx Put x = −t ∴ dx = −dt Join Telegram:- @NOTESPROVIDER12TH_BOARD When x = −a ⟹ t = a When x = 0 ⟹ t = 0 0 I = ∫a f(−t)(−dt) 0 = − ∫a f(−t)dt a b a = ∫0 f(−t) dt - - - - - - (∫a f(x) dx = − ∫b f(x) dx) a b b = ∫0 f(−x) dx - - - - - - (∫a f(x) dx = ∫a f(t) dt) Equation (i) becomes a a a ∫−a f(x) dx = ∫0 f(−x) dx + ∫0 f(x) dx a a ∫−a f(x) dx = ∫0 [f(−x) + f(x)] dx a If f(x) is odd function then f(−x) = −f(x), hence ∫−a f(x) = 0 29  Join Telegram ;-a@NOTESPROVIDER12TH_BOARD a If f(x) is even function then f(−x) = f(x), hence ∫−a f(x) = 2 ∫0 f(x) dx Hence, a a ∫−a f(x) dx = 2 ∫0 f(x) dx , if f(x) is even function. =0 , if f(x) is odd function. Join Telegram:- @NOTESPROVIDER12TH_BOARD 30