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deesuwalka

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Posts posted by deesuwalka

  1. [latex] \int\limits^2_0 \dfrac{x^2}{x^2-2x+2} dx [/latex]

     

    [latex] =\int\limits^2_0 \dfrac{x^2-2x+2+2x-2}{x^2-2x+2} dx [/latex]

     

    [latex] =\int\limits^2_0 1+ \dfrac{2x-2}{x^2-2x+2} dx [/latex]

     

    [latex] =\int\limits^2_0 dx+\int\limits^2_0 \dfrac{2x-2}{x^2-2x+2} dx [/latex]

     

    Let [latex] t=x^2-2x+2 [/latex]

    [latex] dt=(2x-2)dx [/latex]

     

    [latex] =\bigg[x\bigg]^2_0+\int\limits^2_0\dfrac{dt}{t} [/latex]

     

    [latex] =2+\bigg[\ln\,t\bigg]^2_0\;\;\implies\bigg[\ln(x^2-2x+2)\bigg]^2_0 [/latex]

     

    [latex] =2+\bigg[\ln\,2-\ln\,2\bigg] [/latex]

    [latex] =2+0=2 [/latex]

  2. [latex] \frac{x+y}{x-y}\times\frac{x-y}{x+y} [/latex]

     

    [latex] =\frac{\not x+\not y}{\not x-\not y}\times\frac{\not x-\not y}{\not x+\not y} [/latex]

     

    [latex]= 1 \times 1[/latex]

     

    [latex]= 1 [/latex]

  3. How many equations would you say this expression represents, 1 2 or 3 or more?

     

    [math] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }[/math]

    We can read the equations three types,

     

     

    [latex] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} } [/latex]

     

     

    [latex] \frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{z- z_{1} }{ z_{2}- z_{1} } [/latex]

     

     

    [latex] \frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} } [/latex]

     

    So, I think it would be 3 equations.

  4. Use order of operation, i.e.,PEMDAS (Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction)

     

     

    [latex] 36\div 6(2+2+2)= ? [/latex]

     

     

    First, Parenthesis-

     

    [latex] =36\div6(6) [/latex]

     

    Now, between multiplication and division, we apply operation which comes first, here division comes first so we apply division-

     

    [latex]= 6(6) [/latex]

     

    Now, applying multiplication-

     

    [latex]= 6(6)\;\;\implies 6\times6=36 [/latex]

     

    I hope it' ll help.

  5. We substitute [latex] \sqrt{9-x^2} [/latex] by [latex] x= 3sin\theta [/latex] because a trigonometric substitution</a> helps us to get a perfect square under the radical sign. This simplifies the integrand function.http://www.actucation.com/calculus-2/indefinite-integration/basic-methods-of-integration/integration-by-substitution-of-trigonometric-functions

     

    Now, we can simplified it easily,

     

    [latex] \int\sqrt{9-x^2}\;\;\implies\int\sqrt{3^2-x^2} [/latex]

     

     

    [latex] \int\sqrt{3^2-x^2}\;\implies\sqrt{3^2-(3sin\theta)^2} [/latex]

     

     

    [latex] =\sqrt{9-9sin^2\theta} [/latex]

     

     

     

    [latex] \sqrt{9-9sin^2\theta}\;\;\implies\sqrt{9(1-sin^2\theta)}\;\;\implies\sqrt{9cos^2\theta}\;=\;\int 3cos\theta [/latex]

     

    I hope it' ll help.

  6. You can simplify it easily,

     

    [latex] \frac{48}{2y} [/latex]

     

    Write this as,

     

    [latex] \frac{48}{2}\times\frac{1}{y} [/latex]

     

    Now, simply divide 48 by 2 and then multiply,

     

    [latex] 24\times\frac{1}{y}=\frac{24}{y} [/latex]

     

    I hope it' ll help.

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