{"id":38403,"date":"2022-12-03T16:05:14","date_gmt":"2022-12-03T10:35:14","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38403"},"modified":"2022-12-03T16:05:14","modified_gmt":"2022-12-03T10:35:14","slug":"maths-2b-definite-integrals-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2b-definite-integrals-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type"},"content":{"rendered":"
Students must practice these Maths 2B Important Questions<\/a> TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. <\/p>\n Question 7. Question 8. Question 9. Question 10. <\/p>\n Question 11. Question 12. Question 13. Question 14. Question 15. <\/p>\n Question 16. Question 17. Question 18. Question 19. Question 20. <\/p>\n Question 21. Question 22. Question 23. <\/p>\n Question 24. Question 25. <\/p>\n Question 26. Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type Question 1. Evaluate . [(TS) May ’18; (AP) Mar. ’16, ’15] Solution: Put tan … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type<\/h2>\n
\nEvaluate \\(\\int_0^{\\pi \/ 2} \\frac{d x}{4+5 \\cos x}\\). [(TS) May ’18; (AP) Mar. ’16, ’15]
\nSolution:
\nPut tan \\(\\frac{x}{2}\\) = t then \\(\\sec ^2 \\frac{x}{2} \\cdot \\frac{1}{2} \\mathrm{~d} x=\\mathrm{dt}\\)
\n<\/p>\n
\nEvaluate \\(\\int_0^{\\pi \/ 4} \\frac{\\sin x+\\cos x}{9+16 \\sin 2 x} d x\\). [(AP) Mar. ’17; (TS) May ’15]
\nSolution:
\nPut sin x – cos x = t
\n\u21d2 (sin x – cos x)2<\/sup> = t2<\/sup>
\n(cos x + sin x ) dx = dt
\nsin2<\/sup>x + cos2<\/sup>x – 2 sin x cos x = t2<\/sup>
\n1 – sin 2x = t2<\/sup>
\nsin 2x = 1 – t2<\/sup>
\nLower limit: x = 0 \u21d2 t = -1
\nUpper limit: x = \\(\\frac{\\pi}{4}\\) \u21d2 t = 0
\n
\n<\/p>\n
\nEvaluate \\(\\int_0^1 \\frac{\\log (1+x)}{1+x^2} d x\\). [(TS) Mar. ’20; (TS) May ’19, ’17; ’12]
\nSolution:
\nPut x = tan \u03b8
\nthen dx = sec2<\/sup>\u03b8 d\u03b8
\nLower limit: x = 0 \u21d2 \u03b8 = 0
\nUpper limit: x = 1 \u21d2 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n
\n
\n<\/p>\n
\nShow that \\(\\int_0^{\\pi \/ 2} \\frac{x}{\\sin x+\\cos x} d x=\\frac{\\pi}{2 \\sqrt{2}} \\log (\\sqrt{2}+1)\\). [(AP) Mar. ’20, ’18; ’17 (TS)]
\nSolution:
\n
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_0^{\\pi \/ 2} \\frac{\\sin ^2 x}{\\cos x+\\sin x} d x\\). [(TS) May ’15]
\nSolution:
\n
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_0^\\pi \\frac{x}{1+\\sin x} d x\\)
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_0^\\pi \\frac{x \\sin x}{1+\\sin x} d x\\). [Mar. ’16 (TS); May; Mar. ’15 (AP); March ’13]
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_0^\\pi \\frac{x \\sin x}{1+\\cos ^2 x} d x\\). [(AP) May ’18, ’16, ’14]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\frac{x \\sin ^3 x}{1+\\cos ^2 x} d x\\). [(TS) May ’18; Mar. ’15]
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_3^7 \\sqrt{\\frac{7-x}{x-3}} d x\\)
\nSolution:
\nPut x = 3 cos2<\/sup>\u03b8 + 7 sin2<\/sup>\u03b8 then
\ndx = (3 . 2 cos \u03b8 (-sin \u03b8) + 7 . 2 sin \u03b8 cos \u03b8) d\u03b8
\n= (-3 . 2 sin \u03b8 . cos \u03b8 + 7 . 2 sin \u03b8 cos \u03b8) d\u03b8
\n= 4 . 2 sin \u03b8 cos \u03b8 d\u03b8
\n= 4 sin 2\u03b8 d\u03b8
\n7 – x = 7 – 3 cos2<\/sup>\u03b8 – 7 sin2<\/sup>\u03b8
\n= 7(1- sin2<\/sup>\u03b8) – 3 cos2<\/sup>\u03b8
\n= 7 cos2<\/sup>\u03b8 – 3 cos2<\/sup>\u03b8
\n= 4 cos2<\/sup>\u03b8
\nx – 3 = 3 cos2<\/sup>\u03b8 + 7 sin2<\/sup>\u03b8 – 3
\n= -3(1 – cos2<\/sup>\u03b8) + 7 sin2<\/sup>\u03b8
\n= -3 sin2<\/sup>\u03b8 + 7 sin2<\/sup>\u03b8
\n= 4 sin2<\/sup>\u03b8
\nLower limit: x = 3 \u21d2 \u03b8 = 0
\nUpper limit: x = 7 \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n<\/p>\n
\nShow that the area enclosed between the curves y2<\/sup> = 12(x + 3) and y2<\/sup> = 20(5 – x) is \\(64 \\sqrt{\\frac{5}{3}}\\).
\nSolution:
\nGiven curves are y2<\/sup> = 12(x + 3)
\ny = \\(2 \\sqrt{3} \\sqrt{x+3}\\) ……(1)
\ny2<\/sup> = 20(5 – x)
\ny = \\(2 \\sqrt{5} \\sqrt{5-x}\\) ……..(2)
\nSolving (1) and (2)
\n\\(2 \\sqrt{3} \\sqrt{x+3}\\) = \\(2 \\sqrt{5} \\sqrt{5-x}\\)
\nsquaring on both sides
\n12(x + 3) = 20(5 – x)
\n\u21d2 3x + 9 = 25 – 5x
\n\u21d2 8x = 16
\n\u21d2 x = 2
\n
\n<\/p>\n
\nShow that the area of the region bounded by \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (ellipse) is \u03c0ab. Also, deduce the area of the circle x2<\/sup> + y2<\/sup> = a2<\/sup>. [(AP) May ’17; Mar. ’14]
\nSolution:
\nGiven ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n
\n
\n\u2234 Required area = \u03c0ab sq. units
\nIf b = a, the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) becomes circle as, x2<\/sup> + y2<\/sup> = a2<\/sup>.
\nArea of circle = \u03c0a(a) = \u03c0a2<\/sup> (since b = a).<\/p>\n
\nLet AOB be the positive quadrant of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) with OA = a, OB = b. Then show that the area bounded between the chord AB and the arc AB of the ellipse is \\(\\frac{(\\pi-2) a b}{4}\\).
\nSolution:
\nGiven equation of ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nSince A = (a, 0) and B = (0, b).
\nWe have the equation of chord AB is \\(\\frac{x}{a}+\\frac{y}{b}=1\\)
\n
\n<\/p>\n
\nm Find the area bounded between the curves y2<\/sup> = 4ax, x2<\/sup> = 4by. [(AP) May ’19]
\nSolution:
\nGiven curves are
\ny2<\/sup> = 4ax ………(1)
\n\u21d2 y = \\(\\sqrt{a} \\sqrt{x} 2\\)
\nx2<\/sup> = 4by ………(2)
\n\u21d2 y = \\(\\frac{x^2}{4 b}\\)
\n
\n<\/p>\n
\nFind the area bounded between the curves y2<\/sup> = 4ax, x2<\/sup> = 4ay.
\nSolution:
\nGiven curves are y2<\/sup> = 4ax
\n\u21d2 y = \\(2 \\sqrt{a} \\sqrt{x}\\) …….(1)
\nx2<\/sup> = 4ay
\n\u21d2 y = \\(\\frac{x^2}{4 a}\\) ………(2)
\n
\nSolving (1) & (2)
\n\\(2 \\sqrt{a} \\sqrt{x}\\) = \\(\\frac{x^2}{4 a}\\)
\nsquaring on both sides
\n4ax = \\(\\frac{x^4}{16 a^2}\\)
\n\u21d2 x4<\/sup> = 64a3<\/sup>x
\n\u21d2 x4<\/sup> – 64a3<\/sup>x = 0
\n\u21d2 x(x3<\/sup> – 64a3<\/sup>) = 0
\n\u21d2 x = 0 (or) x3<\/sup> – 64a3<\/sup> = 0
\n\u21d2 x3<\/sup> = 64a3<\/sup>
\n\u21d2 x = 4a
\n\u2234 x = 0, 4a
\n<\/p>\n
\nEvaluate \\(\\int_a^b \\sqrt{(\\mathbf{x}-\\mathbf{a})(\\mathbf{b}-\\mathbf{x})} \\mathbf{d x}\\). [Mar. ’18 (TS)]
\nSolution:
\nPut x = a cos2<\/sup>\u03b8 + b sin2<\/sup>\u03b8
\ndx = [a(2 cos \u03b8) (-sin \u03b8) + b 2 sin \u03b8 cos \u03b8] d\u03b8
\n= [-a sin 2\u03b8 + b sin 2\u03b8] d\u03b8
\n= (b – a) sin 2\u03b8 d\u03b8
\nNow, x – a = a cos2<\/sup>\u03b8 + b sin2<\/sup>\u03b8 – a
\n= -a(1 – cos2<\/sup>\u03b8) + b sin2<\/sup>\u03b8
\n= -a sin2<\/sup>\u03b8 + b sin2<\/sup>\u03b8
\n= (b – a) sin2<\/sup>\u03b8
\nb – x = b – a cos2<\/sup>\u03b8 – b sin2<\/sup>\u03b8
\n= b(1 – sin2<\/sup>\u03b8) – a cos2<\/sup>\u03b8
\n= b cos2<\/sup>\u03b8 – a cos2<\/sup>\u03b8
\n= (b – a) cos2<\/sup>\u03b8
\nLower limit: x = a \u21d2 \u03b8 = 0
\nUpper limit: x = b \u21d2 \u03b8 = \u03c0\/2
\n
\n<\/p>\n
\nFind \\(\\int_{-a}^a\\left(x^2+\\sqrt{a^2-x^2}\\right) d x\\)
\nSolution:
\n<\/p>\n
\nEvaluate \\(\\int_0^\\pi x \\sin ^3 x d x\\)
\nSolution:
\n
\n<\/p>\n
\nFind \\(\\int_0^\\pi x \\sin ^7 x \\cos ^6 x d x\\). [Mar. ’19 (TS)]
\nSolution:
\n
\n<\/p>\n
\nIf In<\/sub> = \\(\\int_0^{\\pi \/ 2} \\sin ^n x d x\\), then show that In<\/sub> = \\(\\frac{n-1}{n} I_{n-2}\\). [Mar. ’15 (TS)]
\nSolution:
\n<\/p>\n
\nIf In<\/sub> = \\(\\int_0^{\\pi \/ 2} \\cos ^n x d x\\), then show that In<\/sub> = \\(\\frac{\\mathbf{n}-1}{n} I_{n-2}\\)
\nSolution:
\n
\n<\/p>\n
\nThe circle x2<\/sup> + y2<\/sup> = 8 is divided into two parts by the parabola 2y = x2<\/sup>. Find the area of both parts.
\nSolution:
\nGiven, the circle x2<\/sup> + y2<\/sup> = 8 ……..(1)
\n\u21d2 y = \\(\\sqrt{8-x^2}\\)
\nThe parabola is 2y = x2<\/sup> ……..(2)
\n\u21d2 y = \\(\\frac{x^2}{2}\\)
\nSolving (1) and (2)
\n2y + y2<\/sup> = 8
\n\u21d2 y2<\/sup> + 2y – 8 = 0
\n\u21d2 (y + 4)(y – 2) = 0
\n\u21d2 y = -4 or y = 2
\nFrom (1) \u21d2 x2<\/sup> + (-4)2<\/sup> = 8
\nx2<\/sup> = 8 – 16
\n-8 \u2260 8
\nFrom (2) \u21d2 x2<\/sup> = 2(2)
\n\u2234 x = \u00b12
\n
\n
\n<\/p>\n
\nIf In<\/sub> = \\(\\int_0^{\\pi \/ 4} \\tan ^n x d x\\), then show that In<\/sub>+ In-2<\/sub> = \\(\\frac{1}{n-1}\\). [Mar. ’06, ’99, ’98]
\nSolution:
\n
\nPut tan x = t
\nsec2<\/sup>x dx = dt
\nLower limit: x = 0 \u21d2 t = 0
\nUpper limit: x = \\(\\frac{\\pi}{4}\\) \u21d2 t = 1
\n
\n\u2234 In<\/sub> + In-2<\/sub> = \\(\\frac{1}{n-1}\\)<\/p>\n
\nIf \\(I_{m, n}=\\int_0^{\\pi \/ 2} \\sin ^m x \\cos ^n x d x\\), then show that \\(I_{m, n}=\\frac{m-1}{m+n} I_{m-2, n}\\). [May ’99]
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int_2^6 \\sqrt{(6-x)(x-2)} d x\\)
\nSolution:
\nPut x = 2cos2<\/sup>\u03b8 + 6sin2<\/sup>\u03b8 then
\ndx = [2(2) cos \u03b8 . (-sin \u03b8) + 6(2) sin \u03b8 cos \u03b8] d\u03b8
\n= [-(2) . 2 sin \u03b8 cos \u03b8 + 6(2) sin \u03b8 cos \u03b8] d\u03b8
\n= [-2 . sin 2\u03b8 + 6 sin 2\u03b8] d\u03b8
\n= 4 sin 2\u03b8 d\u03b8
\n6 – x = 6 – 2 cos2<\/sup>\u03b8 – 6 sin2<\/sup>\u03b8
\n= 6(1 – sin2<\/sup>\u03b8) – 2 cos2<\/sup>\u03b8
\n= 6 cos2<\/sup>\u03b8 – 2 cos2<\/sup>\u03b8
\n= 4 cos2<\/sup>\u03b8
\nx – 2 = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8 – 2
\n= -2(1 – cos2<\/sup>\u03b8) + 6 sin2<\/sup>\u03b8
\n= -2 sin2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\n= 4 sin2<\/sup>\u03b8
\nLower limit: x = 2 \u21d2 \u03b8 = 0
\nUpper limit: x = 6 \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n
\n<\/p>\n
\nEvaluate \\(\\int_4^9 \\frac{d x}{\\sqrt{(9-x)(x-4)}}\\)
\nSolution:
\nPut x = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8 then
\ndx = [4(2) cos \u03b8 (-sin \u03b8) + 9(2) sin \u03b8 . cos \u03b8] d\u03b8 = 0
\n= (-4 sin 2\u03b8 + 9 sin 2\u03b8) d\u03b8
\n= 5 sin 2\u03b8 d\u03b8
\n9 – x = 9 – 4 cos2<\/sup>\u03b8 – 9 sin2<\/sup>\u03b8
\n= 9(1 – sin2<\/sup>\u03b8) – 4 cos2<\/sup>\u03b8
\n= 9 cos2<\/sup>\u03b8 – 4 cos2<\/sup>\u03b8
\n= 5 cos2<\/sup>\u03b8
\nx – 4 = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8 – 4
\n= -4 (1 – cos2<\/sup>\u03b8) + 9 sin2<\/sup>\u03b8
\n= -4 sin2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\n= 5 sin2<\/sup>\u03b8
\nLower limit: x = 4 \u21d2 \u03b8 = 0
\nUpper limit: x = 9 \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n<\/p>\n","protected":false},"excerpt":{"rendered":"