evaluate the following : `int\_(0)^(1)x^(2)sqrt(x^(3)+1)dx`
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let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. a - ii, b - iv, c - i, d - iii if , passing through (1, 2), satisfies the differential equation , then 2, 0 ( any two integers with one of them as 0). given = let, tan-1x3 = z⇒ ⇒ so,where c is the integrating constant. given, x3 – 1, x – 1 as pair of polynomialsneed to find the quotient and remainder⇒ to find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b⇒ since, the polynomial is of third degree we can write the q(x) as x2 + ax + b∴ p(x) = (x – a)( x2 + ax + b) + c⇒ x3 + 1 = (x + 1)(x2 + ax + b) + c⇒ x3 + 1 = (x3 + x2 + ax2 + ax + bx + b) + c⇒ x3 + 1 = x3 + (a + 1)x2 + (b + a)x + (c + b)∴ a + 1 = 0, b + a = 0, c + b = 1⇒ a = – 1, b = 1⇒ c + b = 1⇒ c = 0quotient = x2 + ax + b = x2 – x + 1remainder = 0 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. multiply 2 in both lhs and rhs,integrating on both the sides, write a pair of intergers whose sum gives 2x²+3x+5=0 has solutions by the quadratic formula x=(-3±√(9-4×2×5))/4 =(-3±√(-31))/4
there are no real solutions, since the discriminant is -31, whose square root is imaginary.