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geerky42 Group Title

\[\Large\displaystyle \int t^2 \left( t - \dfrac{2}{t} \right) dt\]

  • one year ago
  • one year ago

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  1. UnkleRhaukus Group Title
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    \[=\int (t^3-2t)\mathrm dt\]\[=\int t^3\mathrm dt-2\int t\mathrm dt\]

    • one year ago
  2. geerky42 Group Title
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    Hmm, I should have done that. Thanks.

    • one year ago
  3. some_someone Group Title
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    I[t^(2)(t-(2)/(t)),t] Remove all extra parentheses from the expression. I[,,t^(2)(t-(2)/(t)),t] To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is t. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. I[,,t^(2)(t*(t)/(t)-(2)/(t)),t] Complete the multiplication to produce a denominator of t in each expression. I[,,t^(2)((t^(2))/(t)-(2)/(t)),t] Combine the numerators of all expressions that have common denominators. I[,,t^(2)((t^(2)-2)/(t)),t] Divide each term in the numerator by the denominator. I[,,t^(2)((t^(2))/(t)-(2)/(t)),t] Reduce the expression (t^(2))/(t) by removing a factor of t from the numerator and denominator. I[,,t^(2)(t-(2)/(t)),t] To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is t. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. I[,,t^(2)(t*(t)/(t)-(2)/(t)),t] Complete the multiplication to produce a denominator of t in each expression. I[,,t^(2)((t^(2))/(t)-(2)/(t)),t] Combine the numerators of all expressions that have common denominators. I[,,t^(2)((t^(2)-2)/(t)),t] Divide each term in the numerator by the denominator. I[,,t^(2)((t^(2))/(t)-(2)/(t)),t] Reduce the expression (t^(2))/(t) by removing a factor of t from the numerator and denominator. I[,,t^(2)(t-(2)/(t)),t] Multiply t^(2) by each term inside the parentheses. I[,,t^(3)-2t,t] To find the integral of t^(3), find the anti-derivative. The formula for the anti-derivative of a basic monomial is I[x^(n)=(x^(n+1))/((n+1)),x]. (t^(4))/(4)+I[,,-2t,t] The indefinite integral also has some unknown constant. This can be proven by completing the opposite operation (derivative) in which C would go to 0. The value of C can be found in cases when an initial condition of the function is given. (t^(4))/(4)+I[,,-2t,t]+C To find the integral of -2t, find the anti-derivative. The formula for the anti-derivative of a basic monomial is I[x^(n)=(x^(n+1))/((n+1)),x]. (t^(4))/(4)-t^(2) The indefinite integral also has some unknown constant. This can be proven by completing the opposite operation (derivative) in which C would go to 0. The value of C can be found in cases when an initial condition of the function is given. (t^(4))/(4)-t^(2)+C

    • one year ago
  4. some_someone Group Title
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    so, (t^(4))/(4)-t^(2)+C

    • one year ago
  5. geerky42 Group Title
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    It's a little too much, but thank you for your effort.

    • one year ago
  6. UnkleRhaukus Group Title
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    @some_someone Say what?

    • one year ago
  7. some_someone Group Title
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    I evaluated the integral, isnt that what you wanted @geerky42

    • one year ago
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