![INDEFINITE INTEGRATIONS | d Jftx)dx Jxdx INDIRECT FORMULAE [f()]"+1 J f'() [f(x)]"dx 1 +C 3 " n+1 (f(x))2 f'(x) - f(x) dx 2 +C ೆಕ 2 2 f'(x)V f(x) dx [f(x)]3/ 3 +C 3 _f'() V f(x) +C 4 dx 2 Vf(x) MATHS TUITIONS ೫ WARASIGUDA, SECUNDERABAD f@) 5 log | f(x) | + C dx F f'(x) ] ] dx +C 6 F f(x))2 f(x) 0 ex ( f(x) + f'(x) ) dx ex flx) + c = (< f'() + f(x) ) dx 8 f(x) + C 3 . = INDEFINITE INTEGRATIONS | d Jftx)dx Jxdx INDIRECT FORMULAE [f()]"+1 J f'() [f(x)]"dx 1 +C 3 n+1 (f(x))2 f'(x) - f(x) dx 2 +C ೆಕ 2 2 f'(x)V f(x) dx [f(x)]3/ 3 +C 3 _f'() V f(x) +C 4 dx 2 Vf(x) MATHS TUITIONS ೫ WARASIGUDA, SECUNDERABAD f@) 5 log | f(x) | + C dx F f'(x) ] ] dx +C 6 F f(x))2 f(x) 0 ex ( f(x) + f'(x) ) dx ex flx) + c = (< f'() + f(x) ) dx 8 f(x) + C 3 . =](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_378415_3150939f_1785234377154_sc.jpg?tenant=sc&referrer=pwa-sharechat-v2&f=154_sc.jpg)
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P RAVINDRA#➗10th గణితం #📘ఇంటర్ సైన్స్ & మ్యాథ్స్ #📖ఎడ్యుకేషన్✍ #🏆పోటీ పరీక్షల స్పెషల్ #👩💻టెట్/DSC ప్రత్యేకం![LOGARITHMS Synopsis and Formulae logx N 0>0,1>0,1#1 a = a = xN ; logm logm logm 2 (ab) b a+ S ()= logm] logm logm bl 0 logb logb 111 m a Q ಕ 1 logb logb" | b >0,b#1, 5 0 m 0>0, i # 0 ) m logb" logb 0 n loga 0 1 ೆಕ logb 8 b = 1 PRAVINDRA LOGARITHMS Synopsis and Formulae logx N 0>0,1>0,1#1 a = a = xN ; logm logm logm 2 (ab) b a+ S ()= logm] logm logm bl 0 logb logb 111 m a Q ಕ 1 logb logb" | b >0,b#1, 5 0 m 0>0, i # 0 ) m logb" logb 0 n loga 0 1 ೆಕ logb 8 b = 1 PRAVINDRA](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_569656_1324445d_1785001568444_sc.jpg?tenant=sc&referrer=feed-service&f=444_sc.jpg)
1410- P RAVINDRA#➗10th గణితం #🏆పోటీ పరీక్షల స్పెషల్ #🤔Guess the Answer❓ #📘ఇంటర్ సైన్స్ & మ్యాథ్స్ #📖ఎడ్యుకేషన్✍
![MATHS TUITIONS 1th MATHS WARASIGUDA, SECUNDERABAD MENSURATION SYNOPSIS CUBOID Lateral Surface Area of the Cuboid (L.S.A) = 2h(l + b) here [ = length Total Surface Area of the Cuboid (T.S.A) २(Ib + bh + hl) b = breadth h = height V = Ibh Volume of the Cuboid 2) CUBE : 4$2 Lateral Surface Area of the Cube (L.S.A) here side or) edge 6$2 Total Surface Area of the Cube (T.S.A) of the cube $3 Volume of the Cube V 3) REGULAR CIRCULAR CYLINDER : Curved Surface Area of the Cylinder (C.S.A) = 2nrh here 1 radius of the base Total Surface Area of the Cylinder (TSA) = Zir(h + r) h = height TT2h Volume of the Cylinder; V SPHERE 4TT2 Surface Area of the Sphere; A= here ٣ = radius 3 Volume of the Sphere, V T1*3 5) HEMISPHERE Hemisphere (C.S.A) 2112' Curved Surface Area of the hereT = rudius 31T2 Total Surface Area of the Hemisphere (T S.A) = 2 Volume of the Hemisphere, V Tr3 3 6) RIGHt CIRCULAR CONE here r = radius Curved Surface Area of the Cone (C.S.A) = nrl of the base Total Surface Area of the Cone (T.S.A) = nr(r + 4) height h = TT2h Volume of the Cone, V = slant height] 3 Also [2 = 12 + h2 Practice Today, Success Tomorrow! PRAVINDRA, MSc Dedicated to Your Success in MATHS MATHS TUITIONS 1th MATHS WARASIGUDA, SECUNDERABAD MENSURATION SYNOPSIS CUBOID Lateral Surface Area of the Cuboid (L.S.A) = 2h(l + b) here [ = length Total Surface Area of the Cuboid (T.S.A) २(Ib + bh + hl) b = breadth h = height V = Ibh Volume of the Cuboid 2) CUBE : 4$2 Lateral Surface Area of the Cube (L.S.A) here side or) edge 6$2 Total Surface Area of the Cube (T.S.A) of the cube $3 Volume of the Cube V 3) REGULAR CIRCULAR CYLINDER : Curved Surface Area of the Cylinder (C.S.A) = 2nrh here 1 radius of the base Total Surface Area of the Cylinder (TSA) = Zir(h + r) h = height TT2h Volume of the Cylinder; V SPHERE 4TT2 Surface Area of the Sphere; A= here ٣ = radius 3 Volume of the Sphere, V T1*3 5) HEMISPHERE Hemisphere (C.S.A) 2112' Curved Surface Area of the hereT = rudius 31T2 Total Surface Area of the Hemisphere (T S.A) = 2 Volume of the Hemisphere, V Tr3 3 6) RIGHt CIRCULAR CONE here r = radius Curved Surface Area of the Cone (C.S.A) = nrl of the base Total Surface Area of the Cone (T.S.A) = nr(r + 4) height h = TT2h Volume of the Cone, V = slant height] 3 Also [2 = 12 + h2 Practice Today, Success Tomorrow! PRAVINDRA, MSc Dedicated to Your Success in MATHS](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_391106_b3181a5_1785168389873_sc.jpg?tenant=sc&referrer=feed-service&f=873_sc.jpg)
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MATH Solutions By PHANI#🏆పోటీ పరీక్షల స్పెషల్ #👩💻టెట్/DSC ప్రత్యేకం #➗10th గణితం #📖ఎడ్యుకేషన్✍ #📘ఇంటర్ సైన్స్ & మ్యాథ్స్![ALGEBRA FORMULA HANDBOOK PART1: 30 Essentiol Alebraic_Identities for Students 1. BASIC IDENTITIES 2. CUBIC IDENTITIES (4 + b)? = 42 + 24b + b2 (a+ 6)3 342b + 34b? + b3 a3 + (4 - b) 2 a2 _ Zab (a+6)3 b2 43 + b3 + 34b (4+ b) 42 + b2 = (4 + b) 2' (4 - b)3 43 _ 342b + 34b2 ' 03 Zab a2 _ 62 = (a+b) (a-b) (-b)3 43 _ b3 _ 3.b (4 - b) (4 - b) 2 (4 + b) 2 43 + b3 = (4+b) ( 42 Aab 0b + 02) ( - b)2 02 + 02 = + 2ab (a-b) (a? + ab + B2) - b3 0 2(02 + b2) . (a + b)२ + (a - b)2 43 + b3 = (4 + b)3 3ab (a+b) (a+6)2 (a-6)2 44b _ b3 (a-b)3 + 3ab (a-b) (4x೬)? - (42೬)2 ab 3. IDENTITIES INVOLVING THREE VARIABLES (a+ b+c)2 42 + b2 + c2 + 2 (4b + bc + c4 ) = a? + b2+c2 = (a+b+c)2 _ 2 (ab a bc + ca)| a (a+b+c)2 _ (a2+ 62+c2) 2(ab bc + ca) (a+b+c)3 43 + b3 + c3 + 3 (4 + b) (b + c) (c +4) 43 + b3 + c3 ' (a+b+c) (a?| ४२ + ८२ 3abc ab - bc ca) ३ (a + b + c) [ (a - b)२ ৮3 + c3 (b - c)२ (c-a)2] a 3 3abc 4. PRODUCT IDENTITIES 5. SPECIAL RESULTS (x+a)(r+b) = x2 (a+b)x + ab bc(b-c)+ ca(c-a)+ ab(a-b) (b - c) (c - &) (4 - b) (xta) (x-b) = x2 + (a-b)x-abl a२ (b - c) + b२ (c - a ) + c२ (a - b) (x-a) (x-b) = x2 (a+b)x+ab (b - c) (c - &) (4 - b) (X+p) (X+0) (X+r) c2) + b(c? _ 42) + c (4? _ b2) (P+9+r)x2 0(62 (Pq (b - c) (c - &) (4 - b) 4r + rP)x + Pqr Practice more Score higher Understand better ALGEBRA FORMULA HANDBOOK PART1: 30 Essentiol Alebraic_Identities for Students 1. BASIC IDENTITIES 2. CUBIC IDENTITIES (4 + b)? = 42 + 24b + b2 (a+ 6)3 342b + 34b? + b3 a3 + (4 - b) 2 a2 _ Zab (a+6)3 b2 43 + b3 + 34b (4+ b) 42 + b2 = (4 + b) 2' (4 - b)3 43 _ 342b + 34b2 ' 03 Zab a2 _ 62 = (a+b) (a-b) (-b)3 43 _ b3 _ 3.b (4 - b) (4 - b) 2 (4 + b) 2 43 + b3 = (4+b) ( 42 Aab 0b + 02) ( - b)2 02 + 02 = + 2ab (a-b) (a? + ab + B2) - b3 0 2(02 + b2) . (a + b)२ + (a - b)2 43 + b3 = (4 + b)3 3ab (a+b) (a+6)2 (a-6)2 44b _ b3 (a-b)3 + 3ab (a-b) (4x೬)? - (42೬)2 ab 3. IDENTITIES INVOLVING THREE VARIABLES (a+ b+c)2 42 + b2 + c2 + 2 (4b + bc + c4 ) = a? + b2+c2 = (a+b+c)2 _ 2 (ab a bc + ca)| a (a+b+c)2 _ (a2+ 62+c2) 2(ab bc + ca) (a+b+c)3 43 + b3 + c3 + 3 (4 + b) (b + c) (c +4) 43 + b3 + c3 ' (a+b+c) (a?| ४२ + ८२ 3abc ab - bc ca) ३ (a + b + c) [ (a - b)२ ৮3 + c3 (b - c)२ (c-a)2] a 3 3abc 4. PRODUCT IDENTITIES 5. SPECIAL RESULTS (x+a)(r+b) = x2 (a+b)x + ab bc(b-c)+ ca(c-a)+ ab(a-b) (b - c) (c - &) (4 - b) (xta) (x-b) = x2 + (a-b)x-abl a२ (b - c) + b२ (c - a ) + c२ (a - b) (x-a) (x-b) = x2 (a+b)x+ab (b - c) (c - &) (4 - b) (X+p) (X+0) (X+r) c2) + b(c? _ 42) + c (4? _ b2) (P+9+r)x2 0(62 (Pq (b - c) (c - &) (4 - b) 4r + rP)x + Pqr Practice more Score higher Understand better](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_750173_1953fd9e_1781934716122_sc.jpg?tenant=sc&referrer=feed-service&f=122_sc.jpg)
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![INTEGRATION BY PARTS u [ (u' | v dx) dx Jvdx V dx u = ILATE RULE (Choose u) 8 where Inverse functions 1, 1 are functions of X Logarithmic functions Algebraic functions A according to u and v must be chosen Trigonometric functions ] the ILATE rule, where E Exponential functions MATHS TUITIONS WARASIGUDA, SECUNDERABADK Algebraic Logarithmic 1 Trigonometric Exponential Inverse functions functions functions functions functions 0 EXAMPLE Remember! J x log dx Example: 1 dx Choose u first log x Solution: Let du (as per ILATE rule) 2 Jx dx 0 dx dv > V Find du, V By intesration by parts, Apply the formula Jvdx - J (u Jvdx) dx J x 108 > dx carefully-| (2) - [ (T . 2 ) dx Simplify the result (log x) 12 J (2 log dx 2 22 log Y ೩2 + C २ 2 2 2 د2 ) x 108 > log X dx +C INTEGRATION BY PARTS u [ (u' | v dx) dx Jvdx V dx u = ILATE RULE (Choose u) 8 where Inverse functions 1, 1 are functions of X Logarithmic functions Algebraic functions A according to u and v must be chosen Trigonometric functions ] the ILATE rule, where E Exponential functions MATHS TUITIONS WARASIGUDA, SECUNDERABADK Algebraic Logarithmic 1 Trigonometric Exponential Inverse functions functions functions functions functions 0 EXAMPLE Remember! J x log dx Example: 1 dx Choose u first log x Solution: Let du (as per ILATE rule) 2 Jx dx 0 dx dv > V Find du, V By intesration by parts, Apply the formula Jvdx - J (u Jvdx) dx J x 108 > dx carefully-| (2) - [ (T . 2 ) dx Simplify the result (log x) 12 J (2 log dx 2 22 log Y ೩2 + C २ 2 2 2 د2 ) x 108 > log X dx +C](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_409393_1a35223f_1785234408045_sc.jpg?tenant=sc&referrer=feed-service&f=045_sc.jpg)
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