2<\/sup>2x dx<\/li>\n<\/ol>\nAnswer:
\n1. We have sin x sin 2x sin 3x
\n= 1\/2 (2 sin x sin 3x) sin 2x
\n= 1\/2 (cos 2x – cos 4x) sin 2x
\n= 1\/4 (2 sin 2x cos 2x – 2 cos 4x sin 2x)
\n= 1\/4 [sin 4x – (sin 6x – sin 2x)]
\n= 1\/4(sin 4x + sin 2x – sin 6x)
\n\u222bsin x sin 2x sin 3x dx
\n= \\(\\frac{1}{4}\\) \u222b(sin 4x + sin 2x – sin 6x) dx
\n= –\\(\\frac{1}{16}\\) cos 4x – \\(\\frac{1}{8}\\) cos 2x + \\(\\frac{1}{24}\\) cos 6x + c.<\/p>\n
2. sec2<\/sup>x cos2<\/sup>2x = \\(\\frac{\\left(2 \\cos ^{2} x-1\\right)^{2}}{\\cos ^{2} x}\\)
\n= \\(\\left(\\frac{2 \\cos ^{2} x}{\\cos x}-\\frac{1}{\\cos x}\\right)^{2}\\) = (2cosx – secx)2<\/sup>
\n= 4cos2<\/sup>x + sec2<\/sup>x – 4
\n= 2(1 + cos2x) + sec2<\/sup>x – 4
\n= 2cos2x + sec2<\/sup>x – 2
\n\u222bsec2<\/sup> x cos2<\/sup> 2x dx = \u222b(2 cos 2x + sec2<\/sup> x – 2)dx
\n= sin 2x + tan x – 2x + c.<\/p>\nQuestion 2.
\nFind \\(\\int \\frac{2+\\sin 2 x}{1+\\cos 2 x} e^{x} d x\\)?
\nAnswer:
\n
\n= \u222bex<\/sup> [sec2<\/sup> x + tan x]dx
\n= \u222bex<\/sup>[tanx + sec2<\/sup>x]dx = ex<\/sup> tanx + c.<\/p>\nQuestion 3.
\nEvaluate \\(\\int \\frac{\\sec ^{2} x d x}{\\sqrt{\\tan ^{2} x+4}}\\)?
\nAnswer:
\nPut tanx = u, sec2<\/sup>xdx = dy
\n<\/p>\nQuestion 4.
\nFind the following integrals.
\n
\nAnswer:
\n(i) I = \\(\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\sin x}{1+\\cos ^{2} x} d x\\)
\nPut cosx = t \u21d2 -sin xdx = dt
\nWhen x = 0 \u21d2 t = cos0 = 1,
\n<\/p>\n
(ii) I = \\(\\int_{0}^{1} x e^{x^{2}} d x\\)
\nPut x2<\/sup> = t \u21d2 2xdx = dt
\nWhen x = 0 \u21d2 t = 0,
\nx = 1 \u21d2 t = 1
\nI = \\(\\frac{1}{2} \\int_{0}^{1} e^{t} d t\\) =
\n
\n= [e1<\/sup> – e0<\/sup>] = e – 1.
\n
\nPut sin x = t \u21d2 cos xdx = dt
\nWhen x = 0 \u21d2 t = sin0 = 0,
\n<\/p>\n(iv) I = \\(\\int_{0}^{2} x \\sqrt{x+2} d x\\)
\nPut x + 2 = t2<\/sup> \u21d2 dx = 2tdt
\nWhen x = 0 \u21d2 t = \\(\\sqrt{2}\\), x = 2 \u21d2 t = 2
\n<\/p>\n(v) I = \\(\\int_{0}^{\\frac{\\pi}{2}} \\sqrt{\\sin x} \\cos x d x\\)
\nPut sin x = t \u21d2 cos xdx = dt
\nWhen x = 0 \u21d2 t = sin0 = 0,
\n
\n
\nPut tan x = t \u21d2 sec2<\/sup> xdx = dt
\nWhen x = 0 \u21d2 t = tan 0 = 0,
\n<\/p>\nQuestion 5.
\n(i) If f (x) is an odd function, then \\(\\int_{-a}^{a} f(x)\\) = ?
\n(a) 0
\n(b) 1
\n(c) 2\\(\\int_{0}^{a} f(x)\\) dx
\n(d) 2a
\nEvaluate
\n(ii) \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2} \\sin ^{99} x \\cdot \\cos ^{100} x d x\\)
\n(iii) \\(\\int_{-1}^{1} e^{|x|} d x\\)
\nAnswer:
\n(i) (a) 0.<\/p>\n
(ii) Here, f(x) = sin99<\/sup>x.cos100<\/sup>x .then,
\nf(-x) = sin99<\/sup>(- x).cos100<\/sup>(- x) = – sin99<\/sup> x. cos100<\/sup> x = -f(x)
\n\u2234 odd function \u21d2 \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2} \\sin ^{99} x \\cdot \\cos ^{100} x d x=0\\).<\/p>\n(iii) Here, f(x) = e|x|<\/sup>, f(-x) = e|-x|<\/sup> = e|x|<\/sup> = f(x)
\n\u2234 even function.
\n
\nwe have |x| = x, 0 \u2264 x \u2264 1
\n<\/p>\nQuestion 6.<\/p>\n
\n- Show that cos2<\/sup> x is an even function. (1)<\/li>\n
- Evaluate \\(\\int_{-\\pi \/ 4}^{\\pi \/ 4} \\cos ^{2} x d x\\) (2)<\/li>\n<\/ol>\n
Answer:
\n1. Let f(x) = cos2<\/sup>x \u21d2 f(-x) = cos2<\/sup> (-x) = cos2<\/sup> x = f(x) even.<\/p>\n2.
\n<\/p>\n
Question 7.
\nFind the following integrals.
\n
\nAnswer:
\n<\/p>\n
<\/p>\n
Question 8.
\nFind the following integrals.
\n
\nAnswer:
\n
\n
\nAdd (1) and (2)
\n
\n<\/p>\n
<\/p>\n
<\/p>\n
Question 9.
\nFind the following integrals.<\/p>\n
\n- \\(\\int \\frac{1}{3+\\cos x} d x\\)<\/li>\n
- \\(\\int \\frac{2 x}{x^{2}+3 x+2} d x\\)<\/li>\n<\/ol>\n
Answer:
\n1. \\(\\int \\frac{1}{3+\\cos x} d x\\)
\nPut t = tanx\/2 \u21d2 dt = 1\/2 sec2<\/sup> x\/2 dx
\n<\/p>\n2. \\(\\int \\frac{2 x}{x^{2}+3 x+2} d x\\) = \\(\\int \\frac{2 x}{(x+2)(x+1)} d x\\)
\n
\n2x = A(x + 1) + B (x + 2)
\nwhen x = -1, -2 = B ; B = -2
\nwhen x = -2, -4 = -A ; A = 4
\n<\/p>\n
= 4log(x + 2) – 2log (x + 1) + C.<\/p>\n
Plus Two Maths Integrals Four Mark Questions and Answers<\/h3>\n
Question 1.
\nFind the following integrals.
\n
\nAnswer:
\n
\nx2<\/sup> + x +1 = A(x2<\/sup> + 1) + (Bx + C)(x + 2)
\nPut x = -2 \u21d2 4 – 2 + 1 = 5A \u21d2 A = \\(\\frac{3}{5}\\)
\nEquating the coefficients of x2<\/sup>
\n\u21d2 1 = A + B \u21d2 B = 1 – \\(\\frac{3}{5}\\) = \\(\\frac{2}{5}\\)
\nEquating the constants
\n\u21d2 1 = A + 2C \u21d2 2C = 1 – \\(\\frac{3}{5}\\) = \\(\\frac{2}{5}\\) \u21d2 C = \\(\\frac{1}{5}\\)
\n<\/p>\n
\n\u21d2 1 = A(x – 1) + B(x + 3)
\nPut x = 1 \u21d2 1 = 2A \u21d2 A = \\(\\frac{1}{2}\\)
\nPut x = -3 \u21d2 1 = -4B \u21d2 B = – \\(\\frac{1}{4}\\)
\n<\/p>\n
\nEquating the constants; \u21d2 1 = A
\nEquating the coefficients if t;
\n\u21d2 0 = A + B \u21d2 B = -1
\n<\/p>\n
Question 2.
\nFind the following integrals.<\/p>\n
\n- \u222b e2x<\/sup> sin3xdx<\/li>\n
- \u222b x sin-1<\/sup>xdx<\/li>\n<\/ol>\n
Answer:
\n1. I = \u222be2x<\/sup> sin3xdx = \u222b sin 3x \u00d7 e2x<\/sup>dx
\n<\/p>\n<\/p>\n
2. \u222b x sin-1<\/sup>xdx = \u222b sin-1<\/sup>x \u00d7 xdx
\n<\/p>\nQuestion 3.
\n(i) Which of the following is the value of \\(\\int \\frac{d x}{\\sqrt{a^{2}-x^{2}}}\\)? (1)
\n
\n(ii) Evaluate \\(\\int \\frac{2 x}{x^{2}+3 x+2} d x\\) (3)
\nAnswer:
\n(i) [sin-1<\/sup>\\(\\frac{x}{a}\\) + c]<\/p>\n(ii)
\n
\n\u21d2 2x = A(x + 1) + B(x + 2) \u21d2
\nPut x = -2 and x = -1, we get A = 4, B = -2
\n<\/p>\n
Question 4.<\/p>\n
\n- Choose the correct answer from the bracket.
\n\u222bex<\/sup> dx = — (e2x<\/sup> + c, e-x<\/sup> + c, e2x<\/sup> + c) (1)<\/li>\n- Evaluate: \u222b ex<\/sup> sin x dx<\/li>\n<\/ol>\n
Answer:
\n1. ex<\/sup> + c<\/p>\n2. I = \u222bex<\/sup> sinxdx = sinx.ex<\/sup> – \u222bcos x.ex<\/sup>dx
\n= sin x.ex<\/sup> – (cos x.ex<\/sup> – \u222b(- sin x).ex<\/sup> dx)
\n= sinx.ex<\/sup> – cosxex<\/sup> – \u222bsinx.ex<\/sup>dx
\n= sin x.ex<\/sup> – cos xex<\/sup> – I
\n2I = sin x.ex<\/sup> – cos xex<\/sup>
\nI = \\(\\frac{1}{2}\\)ex<\/sup>(sinx – cosx) + c.<\/p>\nQuestion 5.
\n(i) f(x)\u222bg(x) dx – \u222b(f'(x)\u222bg(x) dx)dx (1)
\n(a) \u222bf'(x)g{x)dx
\n(b) \u222bf(x)g'(x)dx
\n(c) \u222b\\(\\frac{f(x)}{g(x)}\\)dx
\n(d) \u222bf(x)g(x)dx
\n(ii) Integrate sin-1<\/sup>\\(\\sqrt{\\frac{x}{a+x}}\\)dx w.r.to x. (3)
\nAnswer:
\n(i) (d) \u222bf(x)g(x)dx<\/p>\n(ii) \u222bsin-1<\/sup>\\(\\sqrt{\\frac{x}{a+x}}\\)dx,
\nPut x = a tan2<\/sup>\u03b8, \u03b8 = tan-1<\/sup>\\(\\sqrt{\\frac{x}{a}}\\)
\n\u21d2 dx = 2a tan\u03b8 sec2<\/sup>\u03b8 d\u03b8
\nI = \u222bsin-1<\/sup>\\(\\left(\\frac{\\tan \\theta}{\\sec \\theta}\\right)\\) 2a tan\u03b8 sec2<\/sup>\u03b8 d\u03b8
\n= \u222bsin-1<\/sup>(sin\u03b8)2a tan\u03b8 sec2<\/sup>\u03b8 d\u03b8
\n= 2a\u222b\u03b8 tan\u03b8 sec2<\/sup>\u03b8 d\u03b8
\nPut tan\u03b8 = t, \u03b8 = tan-1<\/sup> t \u21d2 sec2<\/sup>\u03b8 d\u03b8 = dt
\n= 2a \u222b tan-1<\/sup> t (t) d\u03b8
\n
\n= a[tan2<\/sup>\u03b8.\u03b8 – tan\u03b8 + \u03b8] + c
\n= a[\u03b8(1 + tan2<\/sup>\u03b8) – tan\u03b8] + c
\n<\/p>\nQuestion 6.
\nMatch the following. (4)
\n
\nAnswer:
\n<\/p>\n
Question 7.
\nEvaluate \\(\\int \\frac{x}{\\sqrt{x+a}+\\sqrt{x+b}} d x\\)?
\nAnswer:
\n<\/p>\n
Question 8.
\nMatch the following.
\n
\nAnswer:
\n1.
\n<\/p>\n
2. \u222bsec x(sec x + tan x)dx = \u222b(sec2<\/sup> x + sec x. tan x)dx
\n= tanx + secx + c.<\/p>\n3. \u222be3x<\/sup>dx = \\(\\frac{e^{3 x}}{3}\\) + c.<\/p>\n4. \u222b(sin x + cos x)dx = sin x – cosx + c.<\/p>\n
Question 9.
\nConsider the integral I = \\(\\int \\frac{x \\sin ^{-1} x}{\\sqrt{1-x^{2}}} d x\\)?<\/p>\n
\n- What substitution can be given for simplifying the above integral? (1)<\/li>\n
- Express I in terms of the above substitution. (1)<\/li>\n
- Evaluate I. (2)<\/li>\n<\/ol>\n
Answer:
\n1. Substitute sin-1<\/sup> x = t.<\/p>\n2. We have, sin-1<\/sup> x = t \u21d2 x = sint
\nDifferentiating w.r.t. x; we get,
\n\\(\\frac{1}{\\sqrt{1-x^{2}}}\\)dx = dt
\n\u2234 I = \u222bt sin t dt.<\/p>\n3. I = \u222bt sin t dt = t.(-cost) -\u222b(-cost)dt = -t cost + sint + c
\n= -sin-1<\/sup> x. cos (sin-1<\/sup> x) + sin(sin-1<\/sup> x) + c
\nx – sin-1<\/sup> x.cos(sin-1<\/sup> x) + c.<\/p>\nQuestion 10.
\nEvaluate \\(\\int_{0}^{\\pi \/ 4} \\log (\\tan x) d x\\).
\nAnswer:
\n<\/p>\n
Question 11.
\nFind the following integrals.<\/p>\n
\n- \\(\\int \\frac{\\sec ^{2} x}{\\cos e c^{2} x} d x\\) (2)<\/li>\n
- \\(\\int \\frac{1}{x^{2}-6 x+13} d x\\) (2)<\/li>\n<\/ol>\n
Answer:
\n1. \\(\\int \\frac{\\sec ^{2} x}{\\cos e c^{2} x} d x\\) = \\(\\int \\frac{\\sin ^{2} x}{\\cos ^{2} x} d x\\) = \u222btan2<\/sup> xdx
\n= \u222b(sec2<\/sup>x – 1)dx = tanx – x + c.<\/p>\n2. \\(\\int \\frac{1}{x^{2}-6 x+13} d x\\)
\n<\/p>\n
Question 12.
\nMatch the following. Justify your answer.
\n
\nAnswer:
\n<\/p>\n
Question 13.
\n(i) \u222bsin2x dx = ? (1)
\n(a) 2 cos x + c
\n(b) -2 sin x + c
\n(c) \\(\\frac{\\cos 2 x}{2}\\) + c
\n(d) \\(-\\frac{\\cos 2 x}{2}\\) + c
\n(ii) Evaluate \u222bex<\/sup> sin 2x dx (3)
\nAnswer:
\n(i) (d) \\(-\\frac{\\cos 2 x}{2}\\) + c.<\/p>\n(ii) Consider I = \u222bex<\/sup> sin 2x dx
\n= \u222bsin 2x. ex<\/sup>dx = sinx.ex<\/sup> – 2\u222bcos 2x. ex<\/sup>dx
\n= sin 2x.ex<\/sup> – 2 (cos 2x.ex<\/sup> + 2\u222bsin 2x. ex<\/sup>dx)
\n= sin 2x. ex<\/sup> – 2 cos 2x ex<\/sup> – 4 \u222bsin 2x. ex<\/sup>dx
\n= sin 2x. ex<\/sup> – 2 cos 2x ex<\/sup> – 4I
\n5 I = sin 2x. ex<\/sup> – 2 cos 2x ex<\/sup>
\nI = \\(\\frac{e^{x}}{5}\\) (sin 2x – 2 cos 2x).<\/p>\nQuestion 14.<\/p>\n
\n- Resolve \\(\\frac{x^{2}+1}{x^{2}-5 x+6}\\) into partial fractions. (2)<\/li>\n
- Hence evaluate \u222b\\(\\frac{x^{2}+1}{x^{2}-5 x+6}\\). (2)<\/li>\n<\/ol>\n
Answer:
\n1.
\n<\/p>\n
2.
\n
\n5x – 5 = A(x – 2) + B(x – 3)
\nx = 2, 5 = -B, B = -5
\nx = 3, 10 = A, A = 10
\n(1) \u21d2 I = \u222b 1dx + \u222b\\(\\frac{10}{x-3}\\) dx – \u222b\\(\\frac{5}{x-2}\\) dx
\n= x + 10log(x – 3) – 5log(x – 2) + c.<\/p>\n
Question 15.
\nEvaluate \\(\\int_{0}^{4}\\) xdx as a limit of sum.
\nAnswer:
\nBy definition,
\n\\(\\int_{a}^{b}\\) f(x) dx =
\n(b – a)\\(\\lim _{n \\rightarrow \\infty} \\frac{1}{n}\\){f(a) + f(a + h) +…….+f(a + {n – 1)h)}
\nHere, a = 0, b = 4, f(x) = x, h = \\(\\frac{4-0}{n}=\\frac{4}{n}\\) \u21d2 nh = 4
\n<\/p>\n
Question 16.<\/p>\n
\n- Define the real valued function f(x) = |x2<\/sup> + 2x – 3| (2)<\/li>\n
- Evaluate \\(\\int_{0}^{2}\\)|x2<\/sup> + 2x – 3|dx. (2)<\/li>\n<\/ol>\n
Answer:
\n1. f(x) = |x2<\/sup> + 2x – 3| = |(x – 1) (x + 3)|
\nWe have;
\n<\/p>\n2. I = \\(\\int_{0}^{2}\\)|x2<\/sup> + 2x – 3|dx
\n<\/p>\nQuestion 17.
\nConsider the function f(x) = |x|+|x + 1|<\/p>\n
\n- Define the function f (x) in the interval [-2, 1]. (2)<\/li>\n
- Find the integral \\(\\int_{-2}^{1}\\) f(x) dx (2)<\/li>\n<\/ol>\n
Answer:
\n1. Given, f(x) = |x|+|x + 1|.
\nWe have,
\n
\nCombining these two functions, we get the function f(x).
\n<\/p>\n
2.
\n<\/p>\n
Question 18.
\nEvaluate \\(\\int_{\\sqrt{6}}^{\\sqrt{3}} \\frac{d x}{1+\\sqrt{\\tan x}} d x\\). (4)
\nAnswer:
\n<\/p>\n
Plus Two Maths Integrals Six Mark Questions and Answers<\/h3>\n
Question 1.
\n(i) Fill in the blanks. (3)
\n(a) \u222b tan xdx = —
\n(b) \u222b cos xdx = —
\n(c) \u222b\\(\\frac{1}{x}\\)dx = —
\n(ii) Evaluate \u222bsin3<\/sup> xcos2<\/sup> xdx (3)
\nAnswer:
\n(i) (a) log|secx| + c
\n(b) sinx + c
\n(c) log|x| + c.<\/p>\n(ii) \u222bsin3<\/sup> xcos2<\/sup> xdx = \u222bsin2<\/sup> xcos2<\/sup> x sin xdx
\n= \u222b(1 – cos2<\/sup> x)cos2<\/sup> x sin xdx
\nPut cos x = t \u21d2 – sin xdx = dt
\n\u2234 \u222b(1 – cos2<\/sup> x)cos2<\/sup> xsin xdx = -\u222b(1 – t2<\/sup> )t2<\/sup>dt
\n= \u222b(t4<\/sup> – t2<\/sup>)dt = \\(\\frac{t^{5}}{5}-\\frac{t^{3}}{3}\\) + c
\n= \\(\\frac{\\cos ^{5} x}{5}-\\frac{\\cos ^{3} x}{3}\\) + c.<\/p>\nQuestion 2.
\nFind the following integrals.
\n
\nAnswer:
\n(i) I = \u222b(3x – 2)\\(\\sqrt{x^{2}+x+1} d x\\)
\nLet 3x – 2 = A(2x + 1) + B
\n\u21d2 3 = 2 A \u21d2 A = \\(\\frac{3}{2}\\)
\n\u21d2 -2 = A + B \u21d2 -2 = \\(\\frac{3}{2}\\) + B
\n\u21d2 B = -2 – \\(\\frac{3}{2}\\) = – \\(\\frac{7}{2}\\)
\n<\/p>\n
\nUsing (2) and (3) in (1) we have;
\n<\/p>\n
(ii) I = \\(\\int \\frac{2 x-3}{x^{2}+3 x-18} d x\\)
\nLet 2x – 3 = A(2x + 3) + B
\n\u21d2 2 = 2A \u21d2 A = 1
\n\u21d2 -3 = 3A + B \u21d2 -3 = 3 + B \u21d2 B = -6
\n<\/p>\n
<\/p>\n
(iii) I = \\(\\int \\frac{5 x+2}{1+2 x+3 x^{2}} d x\\)
\nLet 5x + 2 = A{6x + 2) + B
\n\u21d2 5 = 6 A \u21d2 A = \\(\\frac{5}{6}\\)
\n\u21d2 2 = 2A + B \u21d2 2 = \\(\\frac{5}{3}\\) + B \u21d2 2 – \\(\\frac{5}{3}\\) = \\(\\frac{1}{3}\\)
\n<\/p>\n
<\/p>\n
(iv) I = \\(\\int \\frac{5 x+3}{\\sqrt{x^{2}+4 x+10}} d x\\)
\nLet 5x + 3 = A(2x + 4) + B
\n\u21d2 5 = 2A \u21d2 A = \\(\\frac{5}{2}\\)
\n\u21d2 3 = 4A + B \u21d2 3 = 10 + B \u21d2 B = -7
\n<\/p>\n
\nUsing (2) and (3) in (1) we have;
\n<\/p>\n
Question 3.
\nConsider the expression \\(\\frac{1}{x^{3}-1}\\)<\/p>\n
\n- Split it into partial fraction. (2)<\/li>\n
- Evaluate \u222b \\(\\frac{1}{x^{3}-1}\\) dx (4)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
\n1 = A (x2<\/sup> + x + 1) + (Bx + c)(x + 1),
\nPut x = -1 \u21d2 1 = A(1 + 1 + 1) \u21d2 A= \\(\\frac{1}{3}\\)
\nEquating like terms.
\n0 = A + B \u21d2 B = – \\(\\frac{1}{3}\\), 1 = A + C \u21d2 C = \\(\\frac{2}{3}\\)
\n<\/p>\n2.
\n
\nPut, x – 2 = D (2x – 1) + E ,
\n1 = 2 D \u21d2 D = \\(\\frac{1}{2}\\),
\n-2 = -D + E \u21d2 E = –\\(\\frac{3}{2}\\)
\n<\/p>\n
<\/p>\n
Question 4.
\n(i) Match the following (4)
\n
\n(ii) Consider the function f(x) = \\(\\frac{x^{4}}{x+1}\\) Evaluate \u222bf(x)dx (2)
\nAnswer:
\n(i)
\n<\/p>\n
(ii) Here the numerator is of degree 4 and denominator of degree 1. So to make it a proper fraction we have to divide Nr by Dr.
\n<\/p>\n
Question 5.<\/p>\n
\n- Evaluate the as \\(\\int_{0}^{2}\\)x2<\/sup>dx the limit of a sum. (3)<\/li>\n
- Hence evaluate \\(\\int_{-2}^{2}\\)x2<\/sup>dx (1)<\/li>\n
- If \\(\\int_{0}^{2}\\) f(x)dx = 5 and \\(\\int_{-2}^{2}\\) f(x)dx = 0, then \\(\\int_{-2}^{0}\\) f(x)dx = …….. (2)<\/li>\n<\/ol>\n
Answer:
\n1. Here the function is f(x) = x2<\/sup>, a = 0, b = 2 and h = \\(\\frac{b-a}{n}=\\frac{2}{n}\\)
\n\\(\\int_{0}^{2}\\)x2<\/sup>dx =
\n<\/p>\n2. \\(\\int_{-2}^{2}\\) x2<\/sup>dx = 2 \\(\\int_{0}^{2}\\)x2<\/sup>dx = \\(\\frac{16}{3}\\)<\/p>\n3.
\n<\/p>\n
Question 6.
\nFind \u222b\\(\\sqrt{\\tan x}\\)xdx.
\nAnswer:
\nGiven;
\nI = \u222b\\(\\sqrt{\\tan x}\\)xdx,
\nPut tanx = t2<\/sup> \u21d2 sec2<\/sup>xdx = 2tdt \u21d2 dx = \\(\\frac{2 t d t}{1+t^{4}}\\)
\n<\/p>\n<\/p>\n
<\/p>\n
Question 7.
\n(i) Match the following. (2)
\n
\n(ii) Integrate \\(\\frac{\\sec ^{2} x}{5 \\tan ^{2} x-12 \\tan x+14}\\) w.r.to x. (4)
\nAnswer:
\n(i)
\n<\/p>\n
<\/p>\n
<\/p>\n
Question 8.<\/p>\n
\n- Evaluate \\(\\int_{0}^{1} \\sqrt{x} d x\\) (1)<\/li>\n
- If \\(\\int_{0}^{a} \\sqrt{x} d x=2 a \\int_{0}^{\\pi \/ 2} \\sin ^{3} x d x\\), find the value of a. (3)<\/li>\n
- Hence find \\(\\int_{a}^{a+1}\\)x dx. (2)<\/li>\n<\/ol>\n
Answer:
\n1.
\n<\/p>\n
2. Given;
\n<\/p>\n
3. When a = 0
\n
\nWhen a = 4
\n<\/p>\n
Question 9.
\n(i) Let f (x) be a function, then \\(\\int_{0}^{a}\\) f(x) dx = ? (1)
\n(a) 2 \\(\\int_{0}^{a}\\) f(x – a) dx
\n(b) \\(\\int_{0}^{a}\\) f(a – x) dx
\n(c) f(a)
\n(d) 2\\(\\int_{0}^{a}\\) f(a – x) dx
\nEvaluate
\n
\nAnswer:
\n(i) (b) \\(\\int_{0}^{a}\\) f(a – x) dx<\/p>\n
(ii)
\n
\n(1) + (2)
\n
\n\u21d2 I = 1.<\/p>\n
<\/p>\n
Question 10.
\nFind the following integrals.<\/p>\n
\n- \u222b\\(\\frac{2 e^{x}}{e^{3 x}-6 e^{2 x}+11 e^{x}-6} d x\\)<\/li>\n
- \u222b\\(\\frac{(3 \\sin x-2) \\cos x}{5-\\cos ^{2} x-4 \\sin x} d x\\)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
\n\u21d2 1 = A(t – 2)(t – 3) + B(t – 1)(t – 3) + C(t – 1)(t – 2)
\nPut t = 1 \u21d2 1 = A(-1)(-2) \u21d2 A = \\(\\frac{1}{2}\\)
\nPut t = 2 \u21d2 1 = B(1)(-1) \u21d2 B = -1
\nPut t = 3 \u21d2 1 = B(2)(1) \u21d2 B = \\(\\frac{1}{2}\\)
\n<\/p>\n
<\/p>\n
2. I = \u222b\\(\\frac{(3 \\sin x-2) \\cos x}{5-\\cos ^{2} x-4 \\sin x} d x\\)dx
\nPut sin x = t \u21d2 cosxdx = dt
\n
\n\u21d2 3t – 2 = A(t – 2) + B
\nEquating the coefficients if t; \u21d2 3 = A
\nEquating the constants
\n\u21d2 -2 = -2A + B \u21d2 -2 = -6 + B \u21d2 B = 4
\n<\/p>\n
Question 11.<\/p>\n
\n- Find \u222b\\(\\frac{1}{x^{2}+a^{2}}\\)dx (1)<\/li>\n
- Show that 3x + 1 = \\(\\frac{3}{4}\\)(4x – 2) + \\(\\frac{5}{2}\\) (2)<\/li>\n
- Evaluate \\(\\int \\frac{3 x+1}{2 x^{2}-2 x+3} d x\\) (3)<\/li>\n<\/ol>\n
Answer:
\n1. \u222b\\(\\frac{1}{x^{2}+a^{2}}\\)dx = 1\/a tan-1<\/sup> x\/a + c.<\/p>\n2. 3x + 1 = A \\(\\frac{d}{d x}\\)(2x2<\/sup> – 2x + 3) + B
\n= A(4x – 2) + B
\n3 = 4A; A = 3\/4
\n1 = -2A + B
\n1 = -3\/2 + B, B = 1 + 3\/2 = 5\/2
\n\u2234 3x + 1 = 3\/4(4x – 2) + 5\/2<\/p>\n3.
\n<\/p>\n
We hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"
Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals. Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Maths\u00a0Chapter Wise Questions Chapter Chapter 7 Chapter […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\n
Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals - A Plus Topper<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n