{"id":35251,"date":"2022-08-25T11:44:07","date_gmt":"2022-08-25T06:14:07","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35251"},"modified":"2022-08-25T11:44:07","modified_gmt":"2022-08-25T06:14:07","slug":"inter-2nd-year-maths-2b-definite-integrals-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-definite-integrals-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2B Definite Integrals Important Questions"},"content":{"rendered":"

Students get through Maths 2B Important Questions<\/a> Inter 2nd Year Maths 2B Definite Integrals Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2B Definite Integrals Important Questions<\/h2>\n

Question 1.
\n\\(\\int_{2}^{3} \\frac{2 x}{1+x^{2}} d x\\) [T.S. Mar. 16; May 06]
\nSolution:
\nI = \\(\\left[\\ln \\left|1+x^{2}\\right|\\right]_{2}^{3}\\)
\n= ln 10 – ln 5
\n= ln (10\/5)
\n= ln 2<\/p>\n

\"Inter<\/p>\n

Question 2.
\n\\( \\int_{0}^{\\pi} \\sqrt{2} \\cdot \\sqrt{2} \\sqrt{\\cos ^{2} \\frac{\\theta}{2} d \\theta}\\) [A.P. Mar. 16; Mar. 05]
\nSolution:
\nI = \\( \\int_{0}^{\\pi} \\sqrt{2} \\cdot \\sqrt{2} \\sqrt{\\cos ^{2} \\frac{\\theta}{2} d \\theta}\\)
\n= \\(\\int_{0}^{\\pi} 2 \\cdot \\cos \\theta \/ 2 d \\theta\\)
\n= \\([4 \\sin \\theta \/ 2]_{0}^{\\pi}\\)
\n= 4 (sin \\(\\frac{\\pi}{2}\\) – sin 0)
\n= 4<\/p>\n

Question 3.
\n\\(\\int_{0}^{2}|1-x| d x\\) [A.P. Mar. 15; May 11]
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nI = \\(\\int_{1}^{5} \\frac{d x}{\\sqrt{2 x-1}}\\) [T.S. Mar. 15]
\nSolution:
\nLet 2x – 1 = t2<\/sup>
\n2 dx = 2t dt
\ndx = t dt
\nUL : t = 3
\nLL : t = 1
\nI = \\(\\int_{1}^{3} \\frac{t d t}{t}\\)
\n= \\(\\int_{1}^{3} d t\\)
\n= \\([\\mathrm{t}]_{1}^{3}\\) = 3 – 1
\n= 2<\/p>\n

\"Inter<\/p>\n

Question 5.
\nI = \\(\\int_{0}^{1} \\frac{x^{2}}{x^{2}+1} d x\\) [Mar. 11]
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\n\\(\\int_{0}^{2 \\pi}\\) sin2<\/sup>x cos4<\/sup>; x dx [T.S. Mar. 15; Mar 14]
\nSolution:
\nsin2<\/sup>x cos4<\/sup>x is even function.
\n\"Inter<\/p>\n

Question 7.
\nEvaluate \\(\\int_{0}^{a} \\sqrt{a^{2}-x^{2}} d x\\) [T.S. Mar. 16]
\nSolution:
\nPut x = a sin \u03b8 \u21d2 dx = a cos \u03b8 . d\u03b8
\n\u03b8 = 0 \u21d2 x = 0, x = a \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n\"Inter<\/p>\n

Question 8.
\nFind \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin2<\/sup> x cos4<\/sup> x dx [A.P. Mar. 16]
\nSolution:
\nf(x) is even
\n\u2234 \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin2<\/sup> x cos4<\/sup> x dx = 2 \\(\\int_{0}^{\\pi \/ 2} \\sin ^{2} x \\cdot \\cos ^{4} x d x\\)
\n= 2 . \\(\\frac{3}{6}\\) . \\(\\frac{1}{4}\\) . \\(\\frac{1}{4}\\) . \\(\\frac{\\pi}{4}\\) = \\(\\frac{\\pi}{16}\\)<\/p>\n

\"Inter<\/p>\n

Question 9.
\n\\(\\int_{0}^{\\pi \/ 2} \\cdot \\frac{\\sin ^{5} x}{\\sin ^{5} x+\\cos ^{5} x} d x\\) [Mar. 14, 08]
\nSolution:
\n\"Inter<\/p>\n

Question 10.
\nI = \\(\\int_{0}^{\\pi \/ 4} \\frac{\\sin x+\\cos x}{9+16 \\sin 2 x} d x\\) [Mar. 08]
\nSolution:
\n\"Inter
\n= –\\(\\frac{1}{40} \\ln \\left[\\frac{1 \/ 4}{9 \/ 4}\\right]\\) = \\(\\frac{1}{40}\\) . 2ln . 3 = \\(\\frac{1}{20}\\) ln 3<\/p>\n

Question 11.
\ny = x3<\/sup> + 3, y = 0, x = -1, x = 2 [Mar. 05]
\nSolution:
\nRequired area PABQ
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 12.
\nx = 4 – y2<\/sup>, x = 0. [Mar. 11]
\nSolution:
\nThe given parabola x = 4 – y2<\/sup> meets, the x – axis at A (4, 0) and Y – axis at P(0, 2) and Q(6, -2).
\nThe parabola is symmetrical about X – axis
\n\"Inter
\nRequired area = 2 Area of OAP
\n\"Inter<\/p>\n

Question 13.
\nEvaluate the following definite integrals.
\n(i) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{4} x \\cos ^{5} x d x\\)
\n(ii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{5} x \\cos ^{4} x d x\\)
\n(iii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{6} x \\cos ^{4} x d x\\)
\nSolution:
\n(i) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{4} x \\cos ^{5} x d x\\)
\n= \\(\\frac{4}{9}\\) . \\(\\frac{2}{7}\\) . \\(\\frac{1}{5}\\) = \\(\\frac{8}{315}\\)<\/p>\n

(ii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{5} x \\cos ^{4} x d x\\)
\n= \\(\\frac{3}{9}\\) . \\(\\frac{1}{7}\\) . \\(\\frac{4}{5}\\) . \\(\\frac{2}{3}\\) = \\(\\frac{8}{315}\\)<\/p>\n

(iii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{6} x \\cos ^{4} x d x\\)
\n= \\(\\frac{3}{10}\\) . \\(\\frac{1}{8}\\) . \\(\\frac{5}{6}\\) . \\(\\frac{3}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\)
\n= \\(\\frac{3}{512}\\) \u03c0<\/p>\n

Question 14.
\n\\(\\int_{0}^{\\pi \/ 2} \\frac{d x}{4+5 \\cos x}\\) [A.P. Mar. 16, 15]
\nSolution:
\n\"Inter
\n= \\(\\frac{1}{3}\\left[\\ln \\frac{4}{2}\\right]=\\frac{1}{3} \\ln 2\\)<\/p>\n

\"Inter<\/p>\n

Question 15.
\n\\(\\int_{0}^{\\pi} \\frac{x}{1+\\sin x} d x\\) [May 11]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 16.
\n\\(\\int_{0}^{\\pi} \\frac{x \\sin ^{3} x}{1+\\cos ^{2} x} d x\\) [T.S. Mar. 15; Mar. 11]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 17.
\n\\(\\int_{0}^{1} \\frac{\\log (1+x)}{1+x^{2}} d x\\) [Mar. 07, 05]
\nSolution:
\nPut x = tan \u03b8
\ndx = sec2<\/sup> \u03b8 d\u03b8
\nx = 0 \u21d2 \u03b8 = 0
\nx = 1 \u21d2 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 18.
\n\\(\\int_{0}^{\\pi \/ 4} \\log (1+\\tan x) d x\\) [A.P. Mar. 16]
\nSolution:
\nI = \\(\\int_{0}^{\\pi \/ 4} \\log \\left[1+\\tan \\left(\\frac{\\pi}{4}-x\\right)\\right] d x\\)
\n\"Inter<\/p>\n

Question 19.
\ny = 4x – x2<\/sup>, y = 5 – 2x. [T.S. Mar. 16]
\nSolution:
\n\"Inter
\ny = 4x – x2<\/sup> …………….. (i)
\ny = 5 – 2x ……………….. (ii)
\ny = -([x – 2]2<\/sup>) + 4
\ny – 4 = (x – 2)2<\/sup>
\nSolving equations (i) and (ii) we get
\n4x – x2<\/sup> = 5 – 2x
\nx2<\/sup> – 6x + 5 = 0
\n(x – 5) (x – 1) = 0
\nx = 1, 5
\nRequired area = \\(=\\int_{1}^{5}\\left(4 x-x^{2}-5+2 x\\right) d x\\)
\n= \\(\\int_{-1}^{5}\\left(6 x-x^{2}-5\\right) d x\\)
\n= \\(\\left(3 x^{2}-\\frac{x^{3}}{3}-5 x\\right)_{1}^{5}\\)
\n= (75 – \\(\\frac{125}{3}\\) – 25) – (3 – \\(\\frac{1}{3}\\) – 5)
\n= 50 – \\(\\frac{125}{3}\\) + 2 + \\(\\frac{1}{3}\\)
\n= \\(\\frac{150-125+6+1}{3}\\) = \\(\\frac{32}{3}\\) sq. units.<\/p>\n

Question 20.
\ny2<\/sup> = 4x, y2<\/sup> = 4(4 – x) [May 11]
\nSolution:
\nEquations of the curves are y2<\/sup> = 4x ………………… (1)
\ny2<\/sup> = 4(4 – x) …………………. (2)
\nEliminating y, we get
\n4x = 4 (4 – x)
\n2x = 4 \u21d2 x = 2
\nSubstituting in equation (1), y2<\/sup> = 8
\n\"Inter
\n= 2[\\(\\frac{4}{3}\\)(2\\(\\sqrt{2}\\)) – \\(\\frac{4}{3}\\)(-2\\(\\sqrt{2}\\))]
\n= 2(\\(\\frac{8 \\sqrt{2}}{3}\\) + \\(\\frac{8 \\sqrt{2}}{3}\\))
\n= 2(\\(\\frac{16 \\sqrt{2}}{3}\\)) = \\(\\frac{32 \\sqrt{2}}{3}\\) sq. units<\/p>\n

\"Inter<\/p>\n

Question 21.
\nShow that the area of the region bounded by \\(\\frac{x^{2}}{a^{2}}\\) + \\(\\frac{y^{2}}{b^{2}}\\) = 1 (ellipse) is it ab. also deduce the area of the circle x2<\/sup> + y2<\/sup> = a2<\/sup>. [Mar. 14, May 05]
\nSolution:
\n\"Inter
\nThe ellipse is symmetrical about X and Y axis
\nArea of the ellipse = 4 Area of CAB
\n= 4 . \\(\\frac{\\pi}{4}\\) ab
\n\"Inter
\n(from Prob. 8 in ex 10(a))
\n= \u03c0ab
\nSubstituting b = a, we get the circle
\nx2<\/sup> + y2<\/sup> = a2<\/sup>
\nArea of the circle = \u03c0a(a) = \u03c0a2<\/sup> sq. units.<\/p>\n

Question 22.
\nEvaluate \\(\\int_{\\pi \/ 6}^{\\pi \/ 3} \\frac{\\sqrt{\\sin x}}{\\sqrt{\\sin x}+\\sqrt{\\cos x}} d x\\) [Mar. 14]
\nSolution:
\nLet A = \\(\\int_{\\pi \/ 6}^{\\pi \/ 3} \\frac{\\sqrt{\\sin x}}{\\sqrt{\\sin x}+\\sqrt{\\cos x}} d x\\)
\nPut x = \\(\\frac{\\pi}{2}\\) – t, dx = – dt
\n\"Inter
\n= \\(\\int_{\\pi \/ 6}^{\\pi \/ 3} d x=(x)_{\\pi \/ 6}^{\\pi \/ 3}\\)
\n= \\(\\frac{\\pi}{3}-\\frac{\\pi}{6}=\\frac{\\pi}{6}\\)
\nA = \\(\\frac{\\pi}{12}\\)<\/p>\n

\"Inter<\/p>\n

Question 23.
\nFind \\(\\int_{0}^{\\pi} \\frac{x \\cdot \\sin x}{1+\\sin x} d x\\) [T.S. Mar. 16] [A.P. Mar. 15]
\nSolution:
\n\"Inter
\n\"Inter
\n= \\(\\left(2 \\tan \\frac{x}{2}\\right)_{0}^{\\pi \/ 2}\\)
\n= 2 . tan \\(\\frac{\\pi}{2}\\) – 2 . 0
\n= 2 – 1 = 2
\n2A = \\(\\pi(x)_{0}^{\\pi}\\) – 2\u03c0 = \u03c0(\u03c0) – 2 = \u03c02<\/sup> – 2\u03c0
\nA = \\(\\frac{\\pi^{2}}{2}\\) – \u03c0<\/p>\n

Question 24.
\nFind \\(\\int^{\\pi} x \\sin ^{7} x \\cos ^{6} x d x .\\) [May 05] [T.S. Mar. 19]
\nSolution:
\n\"Inter
\nA = \\(\\pi \\int_{0}^{\\pi \/ 2} \\sin ^{7} x \\cos ^{6} x d x\\)
\n= \u03c0 . \\(\\frac{6}{17}\\) . \\(\\frac{3}{11}\\) . \\(\\frac{1}{9}\\) . \\(\\frac{6}{7}\\) . \\(\\frac{4}{5}\\) . \\(\\frac{2}{3}\\)
\n= \u03c0 \\(\\frac{16}{3003}\\)<\/p>\n

Question 25.
\nEvaluate \\(\\int_{1}^{2} x^{5} d x\\) dx
\nSolution:
\n\\(\\int_{1}^{2} x^{5} \\cdot d x=\\left[\\frac{x^{6}}{6}\\right]_{1}^{2}\\)
\n= \\(\\frac{2^{6}}{6}-\\frac{1}{6}=\\frac{63}{6}=\\frac{21}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 26.
\nEvaluate \\(\\int_{0}^{\\pi} \\sin x d x\\)
\nSolution:
\n\\(\\int_{0}^{\\pi} \\sin x d x=[-\\cos x]_{0}^{\\pi}\\)
\n= – cos \u03c0 – (- cos 0)
\n= + 1 + 1= 2<\/p>\n

Question 27.
\nEvaluate \\(\\int_{0}^{a} \\frac{d x}{x^{2}+a^{2}}\\)
\nSolution:
\n\\(\\int_{0}^{a} \\frac{d x}{x^{2}+a^{2}}=\\left[\\frac{1}{a} \\tan ^{-1}\\left(\\frac{x}{a}\\right)\\right]_{0}^{a}\\)
\n= \\(\\frac{1}{a}\\) [tan-1<\/sup> (1) – tan-1<\/sup> (0)]
\n= \\(\\frac{1}{a}\\) (\\(\\frac{\\pi}{4}\\) – 0) = \\(\\frac{\\pi}{4 a}\\)<\/p>\n

Question 28.
\nEvaluate \\(\\int_{1}^{4} x \\sqrt{x^{2}-1} d x\\)
\nSolution:
\ng(x) = x2<\/sup> – 1
\nf(t) = \\(\\sqrt{t}\\)
\ng'(x) = 2x
\n\"Inter<\/p>\n

Question 29.
\nEvaluate \\(\\int_{0}^{2} \\sqrt{4-x^{2}} d x\\)
\nSolution:
\nLet g(\u03b8) = 2 sin \u03b8 \u21d2 g'(\u03b8) = 2 cos \u03b8
\nf(x) = \\(\\sqrt{4-x^{2}}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 30.
\nEvaluate \\(\\int_{0}^{16} \\frac{x^{1 \/ 4}}{1+x^{1 \/ 2}} d x\\)
\nSolution:
\nPut t4<\/sup> = x \u21d2 dx = 4t3<\/sup> . dt
\n\"Inter<\/p>\n

Question 31.
\nEvaluate \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2} \\sin |x| d x\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 32.
\nShow that \\(\\int_{0}^{\\pi \/ 2} \\sin ^{n} x d x=\\int_{0}^{\\pi \/ 2} \\cos ^{n} x d x\\)
\nSolution:
\nf(x) = sinn<\/sup>x.
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 33.
\nEvaluate \\(\\int_{0}^{\\pi \/ 2} \\frac{\\cos ^{5 \/ 2} x}{\\sin ^{5 \/ 2} x+\\cos ^{5 \/ 2} x} d x\\)
\nSolution:
\n\"Inter
\n= \\(\\int_{0}^{\\pi \/ 2} d x=(x)_{0}^{\\pi \/ 2}=\\frac{\\pi}{2}\\)
\nA = \\(\\int_{0}^{\\pi \/ 2} \\frac{\\cos ^{\\frac{5}{2}} x}{\\sin ^{\\frac{5}{2}} x+\\cos ^{\\frac{5}{2}} x} d x=\\frac{\\pi}{4}\\)<\/p>\n

Question 34.
\nShow that \\(\\int_{0}^{\\pi \/ 2} \\frac{x}{\\sin x+\\cos x} d x=\\frac{\\pi}{2 \\sqrt{2}} \\log (\\sqrt{2}+1)\\)
\nSolution:
\n\"Inter
\n\"Inter
\n= \\(\\frac{\\pi}{4 \\sqrt{2}}\\) log (\\(\\sqrt{2}\\) + 1)2<\/sup>
\n= \\(\\frac{\\pi}{4 \\sqrt{2}}\\) 2 log (\\(\\sqrt{2}\\) + 1)
\n= \\(\\frac{\\pi}{4 \\sqrt{2}}\\) log (\\(\\sqrt{2}\\) + 1)<\/p>\n

\"Inter<\/p>\n

Question 35.
\nEvaluate \\(\\int_{\\pi \/ 6}^{\\pi \/ 3} \\frac{\\sqrt{\\sin x}}{\\sqrt{\\sin x}+\\sqrt{\\cos x}} d x\\) [Mar 14]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 36.
\nFind \\(\\int_{-a}^{a}\\left(x^{2}+\\sqrt{a^{2}-x^{2}}\\right) d x\\)
\nSolution:
\n\"Inter
\n= 2(\\(\\frac{a^{3}}{3}\\) – 0) + 2(0 + \\(\\frac{a^{2}}{3}\\) sin-1<\/sup> (1) – 0 – 0)
\n= \\(\\frac{2a^{3}}{3}\\) + a2<\/sup> . \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 37.
\nFind \\(\\int_{0}^{\\pi} \\frac{x \\cdot \\sin x}{1+\\sin x} d x\\) [T.S. Mar. 16] [A.P. Mar. 15]
\nSolution:
\n\"Inter
\n\"Inter
\n= \\(\\left(2 \\tan \\frac{x}{2}\\right)_{0}^{\\pi \/ 2}\\)
\n= 2 . tan \\(\\frac{\\pi}{2}\\) – 2 . 0
\n= 2 – 1 = 2
\n2A = \\(\\pi(x)_{0}^{\\pi}\\) – 2\u03c0 = \u03c0(\u03c0) – 2 = \u03c02<\/sup> – 2\u03c0
\nA = \\(\\frac{\\pi^{2}}{2}\\) – \u03c0<\/p>\n

\"Inter<\/p>\n

Question 38.
\nEvaluate \\(\\int_{0}^{\\pi \/ 2} x \\sin x d x\\)
\nSolution:
\n\\(\\int_{0}^{\\pi \/ 2} x \\cdot \\sin x d x=(-x \\cdot \\cos x)_{0}^{\\pi \/ 2}+\\int_{0}^{\\pi \/ 2} \\cos x d x\\)
\n= (0 – 0) + \\((\\sin x)_{0}^{\\pi \/ 2}\\)
\n= sin \\(\\frac{\\pi}{2}\\) – sin 0 = 1 – 0 = 1<\/p>\n

Question 39.
\nEvaluate \\(\\lim _{n \\rightarrow \\infty} \\sum_{i=1}^{n} \\frac{1}{n}\\left[\\frac{n-i}{n+i}\\right]\\) by using the method of finding definite integral as the limit of a sum.
\nSolution:
\n\"Inter<\/p>\n

Question 40.
\nEvaluate \\(\\lim _{n \\rightarrow \\infty} \\frac{2^{k}+4^{k}+6^{k}+\\ldots .+(2 n)^{k}}{n^{k+1}}\\) by using the method of finding definite integral as the limit of a sum.
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 41.
\nEvaluate \\(\\lim _{n \\rightarrow \\infty}\\left[\\left(1+\\frac{1}{n}\\right)\\left(1+\\frac{2}{n}\\right) \\cdots\\left(1+\\frac{n}{n}\\right)\\right]^{\\frac{1}{n}}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 42.
\nLet f: R \u2192 R be a continuous periodic function and T be the period of it. Then prove that for any positive integer n,
\n\\(\\int_{0}^{n T} f(x) d x=n \\int_{0}^{T} f(x) d x\\) ………………. (1)
\nSolution:
\nLet k be an integer arid define
\ng : [kT, (k + 1)T] \u2192 [0, T] as g(t) = t – kT.
\nThen g'(t) = 1 for all t \u2208 [kT, (k + 1)T].
\nHence by \\(\\int_{g(c)}^{g(d)} f(t) d t=\\int_{c}^{d} f(g(x)) g^{\\prime}(x) d x,(f \\circ g)\\)
\ng’ is integrable on [kT, (k + 1 )T] and
\n\\(\\int_{k T}^{(k+1) T} f(g(t)) g^{\\prime}(t) d t=\\int_{0}^{T} f(x) d x\\) ………… (2)
\nWe have f(g(t)) g'(t) = f(t – kT), 1 = f(t),
\nsince f is periodic with T as the period.
\nHence \\(\\int_{k T}^{(k+1) T} f(g(t)) g^{\\prime}(t) d t=\\int_{k T}^{(k+1) T} f(t) d t\\) ………….. (3)
\nThus from (2) and (3),
\n\\(\\int_{k T}^{(k+1) T} f(t) d t=\\int_{0}^{T} f(t) d t\\) ………………. (4)
\nLet us now prove eq. (1) by using the principle of mathematical induction.
\nFor n = 1, clearly (1) is true.
\nAssume (1) is true for a positive integer m.
\n\"Inter
\nHence eq. (1) is true for n = m + 1
\nThus, eq. (1) \u00a1s true for any positive integer n, by the principle of mathematical induction.<\/p>\n

Question 43.
\nFind
\n(i) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{4} x d x\\)
\n(ii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{7} x d x\\)
\n(iii) \\(\\int_{0}^{\\pi \/ 2} \\cos ^{8} x d x\\)
\nSolution:
\n(i) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{4} x d x\\)
\n= \\(\\frac{3}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\) = \\(\\frac{3\\pi}{16}\\)<\/p>\n

(ii) \\(\\int_{0}^{\\pi \/ 2} \\sin ^{7} x d x\\)
\n= \\(\\frac{6}{7}\\) . \\(\\frac{4}{5}\\) . \\(\\frac{2}{3}\\) = \\(\\frac{16}{35}\\)<\/p>\n

(iii) \\(\\int_{0}^{\\pi \/ 2} \\cos ^{8} x d x\\)
\n= \\(\\frac{7}{8}\\) . \\(\\frac{5}{6}\\) . \\(\\frac{3}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\) = \\(\\frac{35\\pi}{256}\\)<\/p>\n

\"Inter<\/p>\n

Question 44.
\nEvaluate \\(\\int_{0}^{a} \\sqrt{a^{2}-x^{2}} d x\\) [T.S. Mar. 16]
\nSolution:
\nPut x = a sin \u03b8 \u21d2 dx = a cos \u03b8 . d\u03b8
\n\u03b8 = 0 \u21d2 x = 0, x = a \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n\\(\\int_{0}^{a} \\sqrt{a^{2}-x^{2}} d x=\\int_{0}^{\\pi \/ 2}(a \\cos \\theta)(a \\cos \\theta) d \\theta\\)
\n= a2<\/sup> \\(\\int_{0}^{\\pi \/ 2}\\) cos2<\/sup> \u03b8 d\u03b8
\n= a2<\/sup> . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\) = \\(\\frac{\\pi a^{2}}{4}\\)<\/p>\n

Question 45.
\nEvaluate the following definite integrals.
\n(i) \\(\\int_{0}^{\\pi \/ 2}\\) sin4<\/sup> x . cos5<\/sup> x dx
\n(ii) \\(\\int_{0}^{\\pi \/ 2}\\) sin5<\/sup> x . cos4<\/sup> x dx
\n(iii) \\(\\int_{0}^{\\pi \/ 2}\\) sin6<\/sup> x . cos4<\/sup> x dx
\nSolution:
\n(i) \\(\\int_{0}^{\\pi \/ 2}\\) sin4<\/sup> x . cos5<\/sup> x dx
\n= \\(\\frac{4}{9}\\) . \\(\\frac{2}{7}\\) . \\(\\frac{1}{5}\\) = \\(\\frac{8}{315}\\)<\/p>\n

(ii) \\(\\int_{0}^{\\pi \/ 2}\\) sin5<\/sup> x . cos4<\/sup> x dx
\n= \\(\\frac{3}{9}\\) . \\(\\frac{1}{7}\\) . \\(\\frac{4}{5}\\) . \\(\\frac{2}{3}\\) = \\(\\frac{8}{315}\\)<\/p>\n

(iii) \\(\\int_{0}^{\\pi \/ 2}\\) sin6<\/sup> x . cos4<\/sup> x dx
\n= \\(\\frac{3}{10}\\) . \\(\\frac{1}{8}\\) . \\(\\frac{5}{6}\\) . \\(\\frac{3}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\)
\n= \\(\\frac{3}{512}\\) \u03c0<\/p>\n

Question 46.
\nFind \\(\\int_{0}^{2 \\pi}\\) sin4<\/sup> x . cos6<\/sup> x dx [T.S. Mar. 19]
\nSolution:
\nf(x) = sin4<\/sup> x . cos6<\/sup> x dx
\nf(2\u03c0 – x) = f(\u03c0 – x) = f(x)
\n\\(\\int_{0}^{2 \\pi}\\) sin4<\/sup> x . cos6<\/sup> x dx = 2 \\(\\int_{0}^{2 \\pi}\\) sin4<\/sup> x . cos6<\/sup> x dx
\n= 4 \\(\\int_{0}^{2 \\pi}\\) sin4<\/sup> x . cos6<\/sup> x dx
\n= 4 . \\(\\frac{5}{10}\\) . \\(\\frac{3}{8}\\) . \\(\\frac{1}{6}\\) . \\(\\frac{3}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\)
\n= \\(\\frac{3}{128}\\) \u03c0<\/p>\n

\"Inter<\/p>\n

Question 47.
\nFind \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin2<\/sup> x . cos4<\/sup> x dx [A.P. Mar. 16, 19]
\nSolution:
\nf(x) is even
\n\u2234 \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin2<\/sup> x . cos4<\/sup> x dx = 2 \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin2<\/sup> x . cos4<\/sup> x dx
\n= 2 . \\(\\frac{3}{6}\\) . \\(\\frac{1}{4}\\) . \\(\\frac{1}{4}\\) . \\(\\frac{\\pi}{4}\\)
\n= \\(\\frac{\\pi}{16}\\)<\/p>\n

Question 48.
\nFind \\(\\int_{0}^{\\pi}\\) x sin7<\/sup> x . cos6<\/sup> x dx [May 05]
\nSolution:
\n\"Inter
\n\\(\\int_{0}^{\\pi}\\) x sin7<\/sup> x . cos6<\/sup> x dx = 2 \\(\\int_{0}^{\\pi}\\) sin7<\/sup> x . cos6<\/sup> x dx
\nA = \u03c0 \\(\\int_{0}^{\\pi}\\) x sin7<\/sup> x . cos6<\/sup> x dx
\n= \u03c0 . \\(\\frac{6}{17}\\) . \\(\\frac{3}{11}\\) . \\(\\frac{1}{9}\\) . \\(\\frac{6}{7}\\) . \\(\\frac{4}{5}\\) . \\(\\frac{2}{3}\\)
\n= \u03c0 . \\(\\frac{16}{3003}\\)<\/p>\n

Question 49.
\nFind \\(\\int_{-a}^{a}\\) a2<\/sup> (a2<\/sup> – x2<\/sup>)3\/2<\/sup> dx
\nSolution:
\nf(x) = x2<\/sup> (a2<\/sup> – x2<\/sup>)
\nf(x) is even
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 50.
\nFind \\(\\int_{0}^{1} x^{3 \/ 2} \\sqrt{1-x} d x\\)
\nSolution:
\nPut x = sin2<\/sup> \u03b8
\ndx = 2 sin \u03b8 . cos \u03b8 . d\u03b8
\nx = 0 \u21d2 \u03b8 = 0, x = 1 \u21d2 \u03b8 = \\(\\frac{\\pi}{2}\\)
\n\\(\\int_{0}^{1} x^{3 \/ 2} \\sqrt{1-x} d x\\)
\n= \\(\\int_{0}^{\\pi \/ 2} \\sin ^{3} \\theta \\cdot \\cos \\theta .2 \\sin \\theta \\cos \\theta d \\theta\\)
\n= \\(2 \\int_{0}^{\\pi \/ 2} \\sin ^{4} \\theta \\cdot \\cos ^{2} \\theta d \\theta\\)
\n= 2 . \\(\\frac{3}{6}\\) . \\(\\frac{1}{4}\\) . \\(\\frac{1}{2}\\) . \\(\\frac{\\pi}{2}\\) = \\(\\frac{\\pi}{16}\\)<\/p>\n

Question 51.
\nFind the area under the curve f(x) = sin x in [0, 2\u03c0].
\nSolution:
\n\"Inter
\nf(x) = sin x,
\nWe know that in [0, \u03c0], sin x \u2265 0 and [\u03c0, 2\u03c0], sin x \u2264 0
\nRequired area = \\(\\int_{1}^{\\pi}\\) sinx dx + \\(\\int_{\\pi}^{2 \\pi}\\) (-sinx) dx
\n= \\((-\\cos x)_{0}^{\\pi}[\\cos x]_{\\pi}^{2 \\pi}\\)
\n= – cos \u03c0 + cos 0 + cos 2\u03c0 – cos \u03c0
\n= -(-1) + 1 + 1-(-1) = 1 + 1 + 1 + 1
\n= 4.<\/p>\n

Question 52.
\nFind the area under the curve f(x) = cos x in [0, 2\u03c0].
\nSolution:
\nWe know that cos x \u2265 0 in (0, \\(\\frac{\\pi}{2}\\)) \u222a (\\(\\frac{3\\pi}{2}\\), \u03c0) and \u2264 0 in \\(\\left(\\frac{\\pi}{2}, \\frac{3 \\pi}{2}\\right)\\)
\n\"Inter
\n= sin \\(\\frac{\\pi}{2}\\) – sin 0 – sin \\(\\frac{3\\pi}{2}\\) + sin \\(\\frac{\\pi}{2}\\) + sin 2\u03c0 – sin \\(\\frac{3\\pi}{2}\\)
\n= 1 – 0 – (-1) + 1 + 0 – (-1)
\n= 1 + 1 + 1 + 1 = 4.<\/p>\n

\"Inter<\/p>\n

Question 53.
\nFind the area bounded by the parabola y = x2<\/sup>, the X-axis and the lines x = -1, x = 2.
\nSolution:
\n\"Inter<\/p>\n

Question 54.
\nFind the area cut off between the line y = 0 and the parabola y = x2<\/sup> – 4x + 3.
\nSolution:
\n\"Inter
\nEquation of the parabola is
\ny = x2<\/sup> – 4x + 3
\nEquation of the line is y = 0
\nx2<\/sup> – 4x + 3 = 0
\n(x – 1) (x – 3) = 0
\nx = 1, 3
\nThe curve takes negative values for the values of x between 1 and 3.
\nRequired area = \\(\\int_{1}^{3}\\) -(x2<\/sup> – 4x + 3)dx
\n= \\(\\int_{1}^{3}\\) (-x2<\/sup> + 4x – 3) dx
\n= \\(\\left(-\\frac{x^{3}}{3}+2 x^{2}-3 x\\right)_{1}^{3}\\)
\n= (-9 + 18 – 9) – (-\\(\\frac{1}{3}\\) + 2 – 3)
\n= \\(\\frac{1}{3}\\) – 2 + 3 = \\(\\frac{4}{3}\\)<\/p>\n

\"Inter<\/p>\n

Question 55.
\nFind the area bounded by y = sin x and y = cos x between any two consecutive points of intersection.
\nSolution:
\n\"Inter
\n= \\(\\frac{1}{\\sqrt{2}}\\) + \\(\\frac{1}{\\sqrt{2}}\\) + \\(\\frac{1}{\\sqrt{2}}\\) + \\(\\frac{1}{\\sqrt{2}}\\) = 4\\(\\frac{1}{\\sqrt{2}}\\) = 2\\(\\sqrt{2}\\)<\/p>\n

Question 56.
\nFind the area of one of the curvilinear triangles bounded by y = sin x, y = cos x and X – axis.
\nSolution:
\n\"Inter
\nln (0, \\(\\frac{\\pi}{4}\\)) cos x \u2265 sin x and (\\(\\frac{\\pi}{4}\\), \\(\\frac{\\pi}{2}\\)), cos x \u2264 sin x.
\nRequired area = \\(\\int_{0}^{\\pi \/ 4}\\) sin x dx + \\(\\int_{\\pi \/ 4}^{\\pi \/ 2}\\) cos x dx
\n= \\((-\\cos x)_{0}^{\\pi \/ 4}+(\\sin x)_{\\pi \/ 4}^{\\pi \/ 2}\\)
\n= – cos \\(\\frac{\\pi}{4}\\) + cos 0 + sin \\(\\frac{\\pi}{2}\\) – sin \\(\\frac{\\pi}{4}\\)
\n= – \\(\\frac{1}{\\sqrt{2}}\\) + 1 + 1 – \\(\\frac{1}{\\sqrt{2}}\\)
\n= 2(1 – \\(\\frac{1}{\\sqrt{2}}\\)) = 2 – \\(\\sqrt{2}\\)<\/p>\n

Question 57.
\nFind the area of the right angled triangle with base b and altitude h, using the fundamental theorem of integral calculus.
\nSolution:
\n\"Inter
\nOAB is a right angled triangle and \u2220B = 90\u00b0 take ‘O’ as the origin and OB as positive X-axis
\nIf OB = band AB = h, then Co-ordinates of A are (b, h)
\nEquation of OA is y = \\(\\frac{h}{b}\\) x
\nArea of the triangle OAB = \\(\\int_{0}^{b} \\frac{h}{b} x d x\\)
\n= \\(\\frac{h}{b}\\left(\\frac{x^{2}}{2}\\right)_{0}^{b}=\\frac{h}{b} \\cdot \\frac{b^{2}}{2}=\\frac{1}{2} b h .\\)<\/p>\n

\"Inter<\/p>\n

Question 58.
\nFind the area bounded between the curves y2<\/sup> – 1 = 2x and x = 0.
\nSolution:
\n\"Inter
\n= \\(\\left(-\\frac{y^{3}}{3}+y\\right)_{0}^{1}\\) = 1 – \\(\\frac{1}{3}\\) = \\(\\frac{2}{3}\\)<\/p>\n

Question 59.
\nFind the area enclosed by the curves y = 3x and y = 6x – x2<\/sup>.
\nSolution:
\n\"Inter
\ny = 6x – x2<\/sup>
\nThe straight line y = 3x meets the parabola
\ny = 6x – x2<\/sup>
\n3x = 6x = x2<\/sup>
\nx2<\/sup> – 3x = 0
\nx(x – 3) = 0
\nx = 0 or 3
\nRequired area = \\(\\int_{0}^{3}\\left(6 x-x^{2}-3 x\\right) d x\\)
\n= \\(\\int_{0}^{3}\\left(3 x-x^{2}\\right) d x=\\left(\\frac{3 x^{2}}{2}-\\frac{x^{3}}{3}\\right)_{0}^{3}\\)
\n= \\(\\frac{27}{2}\\) – \\(\\frac{27}{3}\\) = \\(\\frac{27}{6}\\) = \\(\\frac{9}{2}\\)<\/p>\n

Question 60.
\nFind the area enclosed between y = x2<\/sup> – 5x and y = 4 – 2x.
\nSolution:
\nEquations of the curves are
\ny = x2<\/sup> – 5x ………. (1)
\ny = 4 – 2x …………….. (2)
\nx2<\/sup> – 5x = 4 – 2x
\nx2<\/sup> – 3x – 4 = 0
\n(x + 1)(x – 4) = 0
\nx = -1, 4
\n\"Inter
\n= 44 – \\(\\frac{64}{3}\\) – \\(\\frac{3}{2}\\) – \\(\\frac{1}{3}\\)
\n= \\(\\frac{264-128-9-2}{6}\\) = \\(\\frac{125}{6}\\)<\/p>\n

\"Inter<\/p>\n

Question 61.
\nFind the area bounded between the curves y = x2<\/sup>, y = \\(\\sqrt{x}\\).
\nSolution:
\n\"Inter
\n= \\(\\left(\\frac{2}{3} x \\sqrt{x}-\\frac{x^{3}}{3}\\right)_{0}^{1}\\)
\n= \\(\\frac{2}{3}\\) – \\(\\frac{1}{3}\\) = \\(\\frac{1}{3}\\)<\/p>\n

Question 62.
\nFind the area bounded between the curves y2<\/sup> = 4ax, x2<\/sup> = 4by (a > 0, b > 0).
\nSolution:
\nEquations of the given curves are
\ny2<\/sup> = 4ax …………………… (1)
\nx2<\/sup> = 4by ……………………. (2)
\nFrom equation (2), y = \\(\\frac{x^{2}}{4 b}\\)
\nSubstituting in (1) \\(\\left(\\frac{x^{2}}{4 b}\\right)^{2}\\) = 4ax
\nx4<\/sup> = (16 b2<\/sup>) |4ax|
\n\"Inter
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Definite Integrals Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2B Definite Integrals Important Questions Question 1. [T.S. Mar. 16; May 06] Solution: I = = ln 10 – ln 5 = ln (10\/5) = …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35251"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35251"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35251\/revisions"}],"predecessor-version":[{"id":35252,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35251\/revisions\/35252"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35251"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35251"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35251"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}