(lim)┬(n→∞)⁡〖(sin π/2n.sin 2π/2n.sin 3π/2n…….sin ((n-1)π)/2n)^(1/n) 〗 is equal to: a)1/2 b) 1/3

(lim)┬(n→∞)⁡〖(sin π/2n.sin 2π/2n.sin 3π/2n…….sin ((n-1)π)/2n)^(1/n) 〗 is equal to: a)1/2 b) 1/3

   • Definite Integration Class 12   (lim)┬(n→∞)⁡〖(sin π/2n.sin 2π/2n.sin 3π/2n…….sin ((n-1)π)/2n)^(1/n) 〗 is equal to: a) 1/2 b) 1/3 c) 1/4 d) None of these Definite Integral Definition If an integral has upper and lower limits, it is called a Definite Integral. There are many definite integral formulas and properties. Definite Integral is the difference between the values of the integral at the specified upper and lower limit of the independent variable. It is represented as; ∫ab f(x) dx Properties of Definite Integrals Proofs Property 1: p∫q f(a) da = p∫q f(t) dt This is the simplest property as only a is to be substituted by t, and the desired result is obtained. Property 2: p∫q f(a) d(a) = – q∫p f(a) d(a), Also p∫p f(a) d(a) = 0 Suppose I = p∫q f(a) d(a) If f’ is the anti-derivative of f, then use the second fundamental theorem of calculus, to get I = f’(q)-f’(p) = – [f’(p) – f’(q)] = – q∫p(a)da. Also, if p = q, then I= f’(q)-f’(p) = f’(p) -f’(p) = 0. Hence, a∫af(a)da = 0. Property 3: p∫q f(a) d(a) = p∫r f(a) d(a) + r∫q f(a) d(a) If f’ is the anti-derivative of f, then use the second fundamental theorem of calculus, to get; p∫q f(a)da = f’(q)-f’(p)… (1) p∫rf(a)da = f’(r) – f’(p)… (2) r∫qf(a)da = f’(q) – f’(r) … (3) Let’s add equations (2) and (3), to get p∫r f(a)daf(a)da + r∫q f(a)daf(a)da = f’(r) – f’(p) + f’(q) = f’(q) – f’(p) = p∫q f(a)da Property 4: p∫q f(a) d(a) = p∫q f( p + q – a) d(a) Let, t = (p+q-a), or a = (p+q – t), so that dt = – da … (4) Also, note that when a = p, t = q and when a = q, t = p. So, p∫q wil be replaced by q∫p when we replace a by t. Therefore, p∫q f(a)da = –q∫p f(p+q-t)dt … from equation (4) From property 2, we know that p∫q f(a)da = – q∫p f(a)da. Use this property, to get p∫q f(a)da =p∫q f(p+q-t)da Now use property 1 to get p∫q f(a)da = p∫q f(p + q – a )da Property 5: ∫p0f(a)da = ∫p0f(p-a)da Let, t = (p-a) or a = (p – t), so that dt = – da …(5) Also, observe that when a = 0, t =p and when a = p, t = 0. So, ∫p0 will be replaced by ∫po when we replace a by t. Therefore, ∫p0f(a)da = – ∫0pf(p – t)da … from equation (5) From Property 2, we know that ∫qpf(a)da = -∫pqf(a)da. Using this property , we get ∫p0f(a)da = ∫p0f(p-t)dt Next, using Property 1, we get ∫a0f(a)da = ∫p0f(p – a)da Property 6: ∫2p0f(a)da = ∫p0f(a)da + ∫p0f(2p – a))da From property 3, we know that ∫qpf(a)da = ∫rpf(a)da + ∫qrf(a)da Therefore, ∫2p0f(a)da = ∫p0f(a)da + ∫2ppf(a)da = I1 + I2 … (6) Where, I1 = ∫p0f(a)da and I2 =∫2ppf(a)da Let, t = (2p – a) or a = (2p – t), so that dt = -da …(7) Also, note that when a = p, t = p, and when a =2p, t= 0. Hence, ∫0a when we replace a by t. Therefore, I2 = ∫2ppf(a)da = – ∫0pf(2p-0)da… from equation (7) From Property 2, we know that ∫qpf(a)da =- ∫pqf(a)da. Using this property, we get I2 = ∫p0f(2p-t)dt Next, using Property 1, we get I2 = ∫a0f(a)da + ∫a0f(2p-a)da Replacing the value of I2 in equation (6), we get ∫2p0f(a)da = ∫p0f(a)da + ∫pof(2p – a)da Property 7: ∫2a0f(a)da = 2 ∫a0f(a)da … if f(2p – a) = f(a) and ∫2a0f(a)da = 0 … if f(2p- a) = -f(a) we know that ∫2p0f(a)da =∫p0f(a)da + ∫p0f(2p – a)da … (8) Now, if f(2p – a) = f(a), then equation (8) becomes ∫2p0f(a)da = ∫p0f(a)da + ∫p0f(a)da =2∫p0f(a)da And, if f(2p – a) = – f(a), then equation (8) becomes ∫2p0f(a)da = ∫p0f(a)da -∫p0f(a)da = 0 Property 8: ∫p−pf(a)da = 2∫p0f(a)da … if f(-a) =f(a) or it is an even function and ∫a−af(a)da = 0, … if f(-a) = -f(a) or it is an odd function. Using Property 3, we have ∫p−pf(a)da = ∫0−af(a)da + ∫p0f(a)da = I1 + I,2 …(9) Where, I1 =∫0−af(a)da, I2 =∫p0f(a)da Consider I1 Let, t = -a or a = -t, so that dt = -dx … (10) Also, observe that when a = -p, t = p, when a = 0, t =0. Hence, ∫0−a will be replaced by ∫0a when we replace a by t. Therefore, I1 = ∫0−af(a)da = – ∫0af(-a)da … from equation (10) From Property 2, we know that∫qpf(a)da = – ∫pqf(a)da, use this property to get, I1 =∫0−pf(a)da = ∫p0f(-a)da Next, using Property 1, we get I1 = ∫0−pf(a)da = ∫p0f(-a)da Replacing the value of I2 in equation (9), we get ∫p−pf(a)da = I1 + I2 = ∫p0f(-a)da + ∫p0f(a)da =2 ∫p0f(a)da … (11) Now, if ‘f’ is an even function, then f(– a) = f(a). Therefore, equation (11) becomes ∫p−pf(a)da = ∫p0f(a)da +∫p0f(a)da =2∫p0f(a)da And, if ‘f’ is an odd function, then f(–a) = – f(a). Therefore, equation (11) becomes ∫p−pf(a)da = – ∫a0f(a)da + ∫p0f(a)da = 0 Now, let us evaluate Definite Integral through a problem sum. Example Question 1: Evaluate ∫2−1f(a3 – a)da Solution: Observe that, (a3 – a) ≥ 0 on [– 1, 0], (a3 – a) ≤ 0 on [0, 1] and (a3 – a) ≥ 0 on [1, 2] Hence, using Property 3, ∫0−1f(a3 – a)da +∫10f(a – a3 )da +∫21f(a3 – a)da Solving the integrals, we get ∫2−1f(a3–a)da=x44–(x22)]−10+[(x22–(x44))01+[x44−(x22)]12 = – [14 – 12] + [ – 14] + [ 4 – 2] -[14 -12 = 114