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  1. u=1-2t du=-2dt Int 1/(u^9) du However, du is -2dt; we have to multiply by -2, but to balance...

    1 Answers · Science & Mathematics · 02/12/2008

  2. u = (x-1) (you can solve this for x too!) du = dx x = u + 1 integral(1 to 2)[x√(x-1) dx] Substitute the... can now integrate... evaluate(0 to 1) [(2/5)u^(5/2)+(2/3)u^(3/2)] Evaluating... (2/5 + 2/3) - (0 + 0) = 16/15

    5 Answers · Science & Mathematics · 30/07/2009

  3. ..., ∫ [(t + 2t²) /√t] dt = let: √t = u t = u² dt = 2u du thus, substituting... + 3x + 7)^(- 3) (2x + 3) dx = let: x² + 3x + 7 = u differentiate both sides: d(x² + 3x + 7) = du...

    1 Answers · Science & Mathematics · 13/02/2011

  4. Yes. U is the subspace generated by the set {u,v,w}. If the set {u,v,w} is linearly independent, then that is a basis of U as well. U is sometimes also called the span of {u,v,w} or simply U=span(u...

    3 Answers · Science & Mathematics · 01/01/2009

  5. 5(10-u) = 6(-5u+10) 50 - 5u = -30u + 60 -5u + 30u = 60 - 50 25u = 10 u = 10/25 = 2/5 :)

    6 Answers · Science & Mathematics · 18/02/2011

  6. 1 / u = 1 / v − 1 / f Isolate all terms with v by adding (1 / f) on both sides: 1 / u + 1 / f = 1 / v Multiply by v on both sides: v(1 / u + 1 / f...

    2 Answers · Science & Mathematics · 13/02/2012

  7. Hello, ∫ {3 /[u √(2 - u²)]} du = first divide and multiply the integrand by u: ∫ {3...differentiate both sides: d(u²) = d(2 - t²) 2u du = - 2t dt u du = - t dt then, substituting: ∫ {3 /[u² √(2 - u²...

    1 Answers · Science & Mathematics · 25/10/2010

  8. ∫ 2u²/√(u²+4) du u = 2tanθ du = 2sec²θ dθ ∫ 2(2tanθ)²(...dθ 4 sec(θ)tan(θ) - 4 ln|secθ + tanθ| + C θ = arctan(u/2) 4 sec(arctan(u/2))tan(arctan(u/2)) - 4 ln|sec(arctan(u/2)) + tan...

    2 Answers · Science & Mathematics · 04/10/2009

  9. If u between 3π/2 < u < 2π , u will be in quandrant IV If u/2 , u/2 will be 3π/4 < u/2 < π , so that u/2 in...

    1 Answers · Science & Mathematics · 06/05/2011

  10. a^u = a^u e^u/e^u a = e^ln(a) a^u = [e^ln(a)]^u a^u = e^[ln(a)*u] ∫a^u du = ∫e^[uln(a)] du Let v = uln(a) --> dv = ln(a) du ∫a^u du = [1/ln(a)]∫e^v dv ∫a^u du = [1/ln(a)] e^v + C ∫a^u du = [1/ln...

    2 Answers · Science & Mathematics · 09/12/2008

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